Electrostatic Potential and

Capacitance
2
“Just like how a sponge
stores water, a capacitor
stores electric charge,
ready to be released
when needed, reflecting
the concept of potential
energy in electrical
systems.”

SYLLABUS &
WEIGHTAGE

Years
List of Concept Names
2020 2022 2023 2024 2025

(Electrostatic potential, potential due to a point charge,

potential due to an electric dipole, potential due to a — 3 Q* 1 Q (1 M) 1 Q (1 M) —
system of charges)

(Equipotential surfaces, potential energy of a system

of charges, potential energy in an external field, 1 Q (3 M) 2 Q* 1 Q (5 M) 1 Q (5 M) —
electrostatics of conductors)

(Dielectrics and polarization, capacitors and capacitance,

the parallel plate capacitor, effect of dielectric on 1 Q (1 M)
1 Q (5 M) 3 Q* 1 Q (3 M) 1 Q (4 M)
capacitance, combination of capacitors, energy stored in 1 Q (5 M)
a capacitor)

* All questions are of MCQ type and carries equal marks.

 Electric Field & Potential The work done to move a charge from Potential Energy due to
− dV −∂V −∂V −∂V infinity to a point in an electric field. A single charge A system of charges
=E = Ex = , Ey = , Ez
dr ∂x ∂y ∂z r 1 Q1 Q2 U = qV(r), where V(r) is
  ∆U = qq qq q q 
∆V = − ∫ E.dr Q1 Q2 4πε 0 r external potential at r. U= k  1 2 + 1 3 + 2 3 
 r12 r13 r23 

Potential Energy of a Dipole

  Equipotential Surface ● A dielectric is an electrical insulator that can be
● U = P ⋅ E = PE cos θ
● Constant value polarized by an applied electric field.
● W = PE (Cosq1 – Cosq2) v2
of potential at ● Dielectric in Capacitors
v2 v1 v2
● Uminimum (Stable Equilibrium) q = 0°, Cosq = 1, all the points of + – + – + –
v1
v1 + – + – + –
U = –PE the surface. +
+ – – + –
v
● ● Field lines and v2 v v1 2 E + – + – + –
1
Umaximum (Unstable Equilibrium) q = 180, Cosq k
+ – + – + k –
= –1, U = + PE equipotential v2 v1 v1 + – + – + 1 k2 k3 –
v2 – +
surface are + – + –
CONCEPT MAP

v2
+ – + – + –
What is Electric Potential Difference? perpendicular to + – + – + –
The difference in electric potential between two each other at every location. t t
1 t2 t3

Electric Potential Energy

● No work is done in moving charge on an
points (A & B) in an electric field. VAB = VA – VB k A 0 A 0 A
equipotential surface. c'  0 , c'  c' 
d t t1 t2 t3
d t   
● It is the work done by an external agent in moving k k1 k2 k3
a unit positive test charge from infinity to the
desired point. S.I. unit J/C. Electric Potential ELECTRIC
POTENTIAL &

59
W Parallel Plate Capacitor
● V= ; Where, W= work done, q = test charge CAPACITANCE
Q Electric field in
σ σ I A
What is the value − =0 1 s
● Region I: E =
Electric Potential due to of energy stored in 2ε 0 2ε 0 E III d
Dielectric in capacitor

A point A system of A dipole capacitor? σ σ 2 -s

● Region II: E = − =0 II
charge charges P 1 Q2 Capacitance 2ε 0 2ε 0
=U = CV 2 A
r1P q 2 2C
1 q
q Q  CV σ σ σ Q
P r Q r2P 2 Also, energy density d ● Region III: E = + = =
P q3 –q +q  A 2ε 0 2ε 0 ε 0 ε 0 A
r3P 1 C 0
= ε0 E 2 V d
2 (Where σ is surface charge density)
Q 1 Q kp Cos θ
VP = V= Vp =

Electrostatics of Conductors
r2
4πε0 r 4πε0
∑ ri
i Grouping of capacitors in
kp
=Vaxial = and Vequitorial 0 Series Combination Parallel Combination
r2
● Capacitance, ● Capacitance,
To Access One

1 1 1  CC 
Scan This QR Code
Shot Revision Video

● Inside a conductor, electrostatic field is zero. Ceq = nC [if C1 = C2 = ...Cn = C]

   Ceq   1 2  ,V  V1  V2
● At the surface of a charged conductor, electrostatic field must be Ceq C1 C2  C1  C2 
normal to the surface at every point.
σ ● Voltage divider rule: ● Charge divider rule, Q ∝ C:
● Electric field at the surface of the charged conductor, E = n .
ε0 1 C2  C1   C1   C2 

Electrostatic Potential and Capacitance

● Electrostatic shielding is the process of protecting a region from V , V1  V , V2   V Q1    Q, Q2   Q
C C1  C2  C1  C2  C
 1  C 2   C1  C2 

P
external electric field by surrounding it with a conductive material.

W
1 ELECTROSTATIC POTENTIAL

NCERT Definitions (Commonly asked in 1 mark)

U Conservative Force: A force field where the work done in moving an object between two points is independent of the path taken.
U Electric Field: A region around a charged object where another charged object experiences a force.
U Electrostatic Potential (V): Work done by an external force in bringing a unit positive charge from infinity to a point is called
electrostatic potential at that point.
U Work Done: The energy transferred to or from an object by means of a force acting on the object as it moves, work = Fd Cosθ
U Potential Energy: The energy stored in a system of charges due to their positions in an electric field.
U Test Charge (q): A charge that is small enough not to disturb the electric field in which it is placed. q5
U Point Charge Potential: The potential at a point in space due to the presence of a single q1
point charge. q4
r5P
U Electric Dipole: A pair of equal and opposite charges separated by a small distance. r4P
r1P
U Dipole Moment: A vector quantity that measures the separation and direction of an electric
dipole. is calculated as the product of the charge and the distance between the charges. r3P
P
U Potential Due to a System of Charges: The potential energy associated with a unit positive r2P q3
q2
charge at a point P due to configuration of multiple charges in space.

Important Facts

Work done to move a charge in an electrostatic field is independent of the path

01
~ conservative nature of electrostatic force.

02 By convention, potential energy and electrostatic potential are zero at infinity.

The electric potential at a point due to a point charge is inversely proportional to the distance from the
03 charge.

04 Electric dipole has a dipole moment vector pointing from the negative to the positive charge.

05 The potential on the perpendicular bisector of an electric dipole is zero.

06 The potential inside a conductor is same as that on its surface in electrostatic equilibrium.

P
W CBSE Class – XII PHYSICS 60
 The electric potential due to a point charge is positive for positive charges and negative for negative
07 charges.

08 The electric potential at a point in space is a scalar quantity, not a vector.

Important Formulae

1. Work done by external forces in moving a charge q from (in m), ∈o = electrical permittivity (8.85 × 10–12 C2/N – m2)
R to P is
3. Electric potential due to a short dipole
∫ R Fext ⋅ dr =
P
− ∫ Fext ⋅ dr ; Where, F = external
P
=
WRP
R ext 1 p
force (in N) =
(i) At axial point, V ⋅ 2 Where p = dipole
4πε0 r
axis
q moment (in C – m)
R r P
2. Electric potential due to a point charge (ii) At equatorial point, V = 0.
Q 4. Potential due to a system of charges
V=
4π ∈0 r ; Where V = Electric potential (in V), Q =
1  q1 q2 q 
electric charge (in C), r = position vector from the origin V = V1 + V2 +  + Vn =  + ++ n 
4πε 0  r1P r2 P rnP 

Classification

Electrostatic Potential
Due to a Point Charge Due to an Electric Dipole Due to a System of Charges
Definition The electrostatic potential created The potential created by two equal and The collective electrostatic potential due
by a single point charge. opposite charges separated by a distance. to multiple charges.
Formula 1 Q ; where V is electric 1 P ⋅ rˆ 1 qi
V= V= where, P is dipole V =∑ where, qi and ri
4πε 0 r 4πε0 r 2
4 πε 0 ri

Potential, Q is charge, r is moment individual charges and distance from the

distance from point charge and ∈o charges to the point of interest
is electrical permittivity in free respectively.
space (8.85 ×10–12C2/N m2)

Difference Between (Commonly asked in 2-3 marks)

Electrostatic Field (E) vs. Electrostatic Potential (V)
Electrostatic Field (E) Electrostatic Potential (V)
Vector quantity, indicating both magnitude and direction. Scalar quantity, indicating only magnitude.
Vector field that describes the force experienced by a positive Work done per unit charge in moving a charge from infinity to a
test charge. point in space.
Unit: N/C. Unit: J/C or Volts.
F W
E= ; For a charge q experiencing force F V= ; Work done to move charge q
q q

Direction is from positive to negative charges. Do not have direction.

61 Electrostatic Potential and Capacitance P

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 Potential Due to a Point Charge vs. Potential Due to an Electric Dipole
Potential Due to a Point Charge Potential Due to an Electric Dipole
p cos(θ)
V= ; For a dipole moment p at a point making an angle θ
V=
Q
; For a charge Q at distance r 4πε 0 r 2
4πε0 r with the dipole axis.
Symmetrical in all directions for a given distance. Varies with the orientation of the dipole and the observation point.

Real Life Application Based Questions

1. Which principle electrostatic generators use to generate high-voltage electricity, and what are their applications?
Ans. Electrostatic generators like the Van de Graaff generator create potential difference between two points It uses the principles
of electrostatic potential to generate high-voltage electricity. These generators are used in particle accelerators, nuclear physics
experiments, and X – ray machines.
2. How do the batteries in many electronic devices work in context of electric potential ?
Ans. Batteries store electrical energy and release it as needed. The voltage of a battery is a measure of its electric potential, determining
its capacity to do work.
3. How do charging devices for electric vehicles and mobile phones charge batteries?
Ans. Charging devices are designed to manage the electric potential effectively to ensure safe and
efficient charging of batteries. They adjust the voltage and current according to the battery’s
needs and its current state of charge. This regulation is crucial as it prevents overcharging and
undercharging, which can both damage the battery and affect its lifespan.
4. Why do you feel a shock when touching a metal doorknob after walking across a carpet?
Ans. When you walk across a carpet, especially in dry conditions, your body accumulates excess electrons due to friction, leading to a
high electrostatic potential. Touching a metal doorknob, which conducts electricity and is at a lower potential, allows the excess
charge to discharge rapidly, creating the sensation of a shock.
5. Why are electrostatic precipitators used in industrial settings like power plants?
Ans. Electrostatic precipitators are used in power plants to remove fine particles from exhaust gases. By charging particles in the smoke
with a high electrostatic potential, they are attracted to plates of the opposite charge within the precipitator, thereby removing them
from the gas stream and reducing pollution.

Myth Buster

U Myth: The electric potential due to an electric dipole is zero at all points along the axis of the dipole.
Fact: The electric potential due to an electric dipole is not zero along the axis of the dipole; it varies with distance and angle from
the dipole axis.
U Myth: Electrostatic potential is always positive.
Fact: Electrostatic potential can be positive, negative, or zero. The sign depends on the nature of the charge creating the potential.
For a positive charge, the potential is positive, and for a negative charge, it is negative. The potential is zero at points infinitely
distant from all charges.
U Myth: The potential at a point due to a charge Q is independent of the sign of Q.
kQ
Fact: The potential due to a point charge Q at a distance r is given by V = . The sign of Q affects whether the potential is
r
positive or negative. For a positive charge, the potential is positive, and for a negative charge, it is negative.
U Myth: Electric potential and electric potential energy are the same.
Fact: Electric potential is the potential energy per unit charge. Potential energy depends on the system of charges, whereas
electric potential at a point is a property of the electric field at that point and does not depend on the test charge placed in the field.

P
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 competency BASED SOlved Examples
1. Assertion (A): Work done in moving a charge around a
Multiple Choice Questions
(1 M) closed path, in an electric field is always zero.
Reason (R): Electrostatic force is a conservative force. (Re)
1. The electric potential V at any point (x, y, z) is given by
V = 3x2 where x is in meters and V in volts. The electric 2. Assertion (A): The potential at a point is characteristic of
field at the point (1 m, 0, 2m) is:  (Un) the electric field at a point only whereas electric potential
energy at a point is characteristic of the charge–field system.
(a) 6 V/m along – x axis (b) 6 V/m along + x axis
Reason (R): The potential is independent of a test charge
(c) 1.5 V/m along – x axis (d) 1.5 V/m along + x axis placed in the field and the electric potential energy is due
2. A + 3.0 mC charge Q is initially at rest at a distance of to an interaction between the electric field at the point and
r1 = 10 cm from a + 5.0 mC charge q fixed at the origin. the charged particle placed in the field at that point. (Un)
The charge Q is moved away from q to a new position
Subjective Questions
at r2 = 15 cm. In this process work done by the field is
(Ev)
(a) –45 J (b) 36 J Very Short Answer Type Questions (2 M)
(c) – 36 J (d) 45 J 1. The work done in moving a charge particle between two
3. An electric dipole having a dipole moment of 4 × 10–9 C m points in an uniform electric field, does not depend on
is placed in a uniform electric field such that the dipole the path followed by the particle. Why? (Un)
is in stable equilibrium. If the magnitude of the electric Ans. We know that the electrostatic force is a conservative force.
field is 3 × 103 N/C, what is the work done in rotating i.e the work done by the force in moving a test charge from
the dipole to a position of unstable equilibrium? (Un) one point to another is independent of the path connecting
(a) zero (b) 1.2 × 10–5J the two points.  (2 M)
(c) 2.4 × 10–5J (d) – 1.2 × 10–5J 2. A test charge q is made to move in the electric field of
a point charge Q along two different closed paths. First
4. A positively charged particle is released from rest in an
path has sections along and perpendicular to lines of
uniform electric field. The electric potential energy of
electric field. Second path is a rectangular loop of the
the charge (Un) (NCERT Exemplar)
same area as the first loop. How does the work done
(a) remains constant because the electric field is uniform. compare in the two cases? (Un) (NCERT Exemplar)
(b) increases because the charge moves along the electric field.
(c) decreases because the charge moves along the electric field.
(d) decreases because the charge moves opposite to the electric
field. Q
q
5. Consider a uniform electric field in the ẑ direction. The
potential is constant (Un) (NCERT Exemplar)
(a) in all space.
(b) for any x for a given z. a
b
(c) for any y for a given z.
(d) on the x – y plane for a given z. Ans. In the presence of an electric field, the work done on a test
charge q as it moves along a closed path depends on the
path taken and the configuration of electric field. As the
Assertion and Reason
(1 M) electric field is conservative, work done will be zero in both
the cases. (2 M)
Direction: The following questions consist of two statements.
Assertion (A) and Reason (R). Answer these questions by 3. Do free electrons travel to region of higher potential or
selecting the appropriate option given below: lower potential? (Un) (NCERT Exemplar)
(a) Both A and R are true, and R is the correct explanation of A. Ans. Free electrons travel to regions from higher potential to
(b) Both A and R are true, but R is not the correct explanation of    A. lower potential. This movement is driven by the force
exerted on the electrons by the electric field. Electrons
(c) A is true, but R is false experience force in direction opposite to electric field
(d) A is false, but R is true. vector. (2 M)

63 Electrostatic Potential and Capacitance P

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 4. A regular hexagon of side 10 cm has a charge 5 mC at 3 2
each of its vertices. Calculate the potential at the centre − =
0 x = 9 cm. (1 M)
x 15 − x
of the hexagon. (Ev) (NCERT Exemplar)
Ans. The given figure shows six equal amount of charges, q, at If x lies on the extended line OA, the required condition is
the vertices of a regular hexagon. 3 2
− =
0
+q +q x x − 15
A B
l x = 45 cm. (1 M)
Thus, electric potential is zero at 9 cm and 45 cm away
F C from the positive charge on the side of the negative charge.
+q +q Note that the formula for potential used in the calculation
required choosing potential to be zero at infinity.  (1 M)
E D
+q +q 3. Two-point charges are placed along the x-axis as shown.
Where,
–X –q + 2q +x
Charge, q = 5μC = 5 × 10–6 C
Let side of the hexagon, l = 10 cm Along the line joining the two charges, how many points
Distance of each vertex from centre O, d =10 cm are possible at which the total potential due to the charges
6× q is zero? Predict the probable locations.  (Un)
Electric potential at point O, V = (1 M)
4π ∈ d Ans. Location 1: As the two charges are opposite, in between
Where, e0 = Electrical permittivity of free space the two charges, there is a location, where the individual
electric potentials balance. As the charges are not of equal
6 × 5 × 10−6
∴V = magnitude, this point of zero potential occurs a little closer
4π× 8.85 × 10−12 × 0.1 to the smaller charge, that is, –q.  (1½ M)
= 2.7 × 106 V  (1 M) Location 2: To the left of the smaller charge, –q, there is
another location at which the individual potentials due to
Short Answer Type Questions (2 M) –q and +2q will exactly balance each other. (1½ M)
1. (a) Calculate the potential at a point P due to a charge –X –q + 2q +x
of 4 × 10–7 C located 9 cm away. (1 M)
(b) Hence obtain the work done in bringing a charge
Total potential is
of 2 × 10–9 C from infinity to the point P. Does the
answer depend on the path along which the charge zero at these points
is brought? (2 M) (Ev) (NCERT Exemplar) 4. A conducting wire connects two charged conducting spheres
1 Q 4 × 10 −7 C such that they attain equilibrium with respect to each other.
Ans. (a) V == 9 × 109 Nm 2 C −2 × The distance of separation between the two spheres is very
4πε 0 r 0.09 m
large as compared to either of their radii. (Un)
= 4 × 104 V (1 M)
q1 q2
(b) W = qV = 2 × 10 C × 4 × 10 V
–9 4
r1 r2
= 8 × 10–5 J (1 M)
No, work done will be path independent. Any arbitrary
infinitesimal path can be resolved into two perpendicular
displacements: One along r and another perpendicular to r. Find the ratio of the magnitudes of the electric fields at
The work done corresponding to the later will be zero. (1 M) the surfaces of the two spheres.
2. Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm Ans. At equilibrium, the potential on the surface of a larger
apart. At what point on the line joining the two charges is sphere = potential on the surface of a smaller sphere.
the electric potential zero? Take the potential at infinity kq1 kq2
to be zero. (Ev) (NCERT Exemplar) =
V =
r1 r2
Ans. Let O be the origin where positive charge is located.
q1 r1
O P A So, = (1 M)
q2 r2
3×10–8 C –2 × 10–8 C
Since the two charges are very far from each other, the
Let P be the required point on the x-axis where the potential
electric fields on the surfaces of the two spheres will be:
is zero. If x lies between O and A, we have
kq1 kq2
1  3 × 10 −8 2 × 10 −8  =E1 = and E2 (1 M)
 − = r12 r22
(15 − x ) × 10−2 
−2
0
4πε 0  x × 10 E1 r2
The ratio of the electric fields is, = (1 M)
where x is in cm. That is, E2 r1
P
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 5. Test charge q moves along a path P → Q → R → S in a Ans. Electric field in this region is given by:
uniform electric field region directed along + x-axis. The − dV
E=
coordinates of the points are as follows: dx
P(a, b, 0), Q(2a, 0, 0), R(a, –b, 0) and S(0, 0, 0). (An) Here, dV/dx is the slope of the given graph.
y dV −10
P(a, b) →E =
− =
−5volt / m
dx 2
dV
Thus, E =
− =
5volt / m (1 M)
dx
Q
S x Charge on the particle is 2 μC, thus the force on the charge
(0, 0) T (2a, 0)
is
F = qE = 2 × 10–6 × 5 = 10–5 N (1 M)
→E As the electric field is in the direction in which the electric
R(a, –b)
potential decreases, we can say that the direction of the
(a) Identify the plane of motion of the test charge.(1 M) electric field is along +x direction as V is decreasing as we
move towards +x direction. (1 M)
(b) Determine the work done by the electric field as
the test charge moves from P to S. Use diagram, if Now, as the particle has a positive charge, it will experience
a force in the direction of the electric field i.e., +x direction.
necessary. (2 M)
7. Calculate potential on the axis of a ring due to charge Q
Ans. (a) As the z-coordinate of each of the points is zero, the
uniformly distributed along the ring of radius R.
plane of motion from P to Q to R to S is in x-y plane.(1 M) (Ev)(NCERT Exemplar)
(b) Since E is conservative, the work done is path
independent, so replace the path P → Q → R → S with a Ans.
dq
simpler path as P → T → S.
Work done along P → T = 0, as the path is perpendicular to x2 + a2
a
the direction of E. (1 M)
Work done along T → S = -qEa x P

So total work done = 0 - qEa = -qEa (1 M)

(1 M)
OR
Let us take Point P to be at a distance x from the centre of
Since E is conservative, the work done is path independent, the ring.
so replace the path P → Q → R → S with a simpler path
dq dq
P → S. =
Electric potential, V k=e∫ ke ∫ (1 M)
r x + a2
2

W = qE. PS. cos (90 + θ) (1 M)

Q
Here, (90 + θ) is the angle between the electric field E and ke
displacement vector PS. θ is the ∠SPT.
V=
x2 + a2
∫
0
dq

W=
−qE a 2 + b 2 sinθ =V
ke Q
(1 M)
x2 + a2
a
=
−qE a 2 + b 2
a + b2
2
(1 M) Long Answer Type Questions
= –qEa 1. Given figure shows a charge array known as an electric
quadrupole. For a point on the axis of the quadrupole,
6. The variation of electric potential in a region is shown
obtain the dependence of potential on r for r/a>> >1 and
in the graph below. Find the magnitude and direction of contrast your results with that due to an electric dipole
the force on a particle having a charge of +2μC just after and an electric monopole (i.e., a single charge).  (Ev)
it is released at a point x = 1 m in this region.  (An) A B
a a C
P
V (in volt) +q –q –q +q

10 Ans. Given, AC = 2a, BP = r

AP = r + a and PC = r – a
A a B a C
P
x +q –q –q +q
0 2
(in m) r  (1 M)

65 Electrostatic Potential and Capacitance P

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 The potential at P is V. Therefore, the total electric potential due to the disk is then
\V = Potential at P due to A+ Potential at P due to B + obtained by summing or integrating the potentials due to all
Potential at P due to C the elemental rings at point P,
1  q 2q q  k ⋅ 2πr σdr
= − + (1 M) dV = (1 M)
4πε0  AP BP CP 
 r 2 + x2
V R k 2µr σdr
 1 1 
=
1
⋅q
2
− +  ∫ 0 dV = ∫ 0 r 2 + x 2
4πε0  ( r + a ) r ( r − a ) 
q  r (r − a ) − 2(r + a )(r − a ) + r (r + a )  σ R r
=
4πε0   V= ∫ dr
2ε 0 (r )
1/ 2
r (r + a )(r − a ) 
0 2
+ x2
q  r − ra − 2r + 2a + r + ra 
2 2 2 2
=   σ  2 R

4πε0  r (r 2 − a 2 ) =V  r + x 2 
 2ε0  0
q ⋅ 2a 2 q ⋅ 2a 2 σ  2
= = (1 M)
4πε0 r (r 2 − a 2 )  a2  =V R + x 2 − x  (1 M)
4πε0 .r ⋅ r 2 1 − 2  2ε 0  
 r  Q
[we know that πR2σ = Q (charge on disc), So, σ = ]
According to the question, R2
2πR 2 σ  2
r q ⋅ 2a 2 =V R + x 2 − x 
If >> 1, a << r. Therefore , V = 4πε0 R 2  
a 4πε0 ⋅ r 3
2Q  2
V∝
1 =V R + x 2 − x  (1 M)
r3 4πε 0 R 2  

As, we know that electric potential at a point on axial line

due to an electric dipole is Hints & Explanations
1
V∝
r2 Multiple Choice Questions
1
In case of electric monopole, V ∝ (1 M) 1. (a) Electric potential V = 3x2
r
Then, we conclude that for larger r, the electric potential dV
E= −
due to a quadrupole is inversely proportional to the cube of dx
the distance r, while due to an electric dipole, it is inversely
d
proportional to the square of r and inversely proportional to E= −
dx
( )
3x 2
the distance r for a monopole (1 M)
2. Calculate potential on the axis of a disc of radius R due E = –6x
to a charge Q uniformly distributed on its surface. At the point (1, 0, 2)
(Ap) (NCERT Exemplar)
Electric field E = 6 × 1= –6 V/m
Ans. Given,
Thus, electric field at (1m, 0, 2m) is 6Vm–1 along negative
Radius of disk = R x-axis.
Magnitude of charge distributed on the disk = Q 2. (d) Initially when both charges are 10 cm apart
Suppose an elementary part of a ring of radius r of thickness
1 ( 3µC )( 5µC )
dr on a disc of radius R. Potential energy of system,= U=
i
4πε0 10 × 10−2 m
Charge on the ring is dq, then potential dV due to ring at
P, will be, Finally, both charges are 15cm apart
kdq 1 (3µ)(5µC )
dV = Potential energy of system, Uf =
r′ ∴ r′ = r + x (½ M)
2 2
4πε0 15 × 10−2 m
dq is the charge on the ring = σ × area of the ring Work done by external force,
dq = σ(2πrdr) = 2πrσ⋅dr  (½ M) = Wext = Uf – Ui
Now apply potential at P due to element, 1 ( 3µC )( 5µC ) 1 ( 3µC )( 5µC )
= −
kdq −2
4πε0 15 × 10 m 4πε0 10 × 10−2 m
dV =
r′
k ⋅ 2πr σdr 1 ( 3µc )( 5µc )  1 1 
dV = (1 M)
=  15 − 10 
r +x
2 2 4πε0 10−2  

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 9 × 109 × 3 × 10−6 × 5 × 10−6  10 − 15  5. (d) In a uniform electric field, the electric potential is
=  150  = −45 J constant along x – y plane, as it will be equipotential surface
10−2  
for the field along z direction.
Work done by field, Wfield = –Wext = 45 J This means that the potential is same at every point with the
3. (c) 2.4 × 10 –5 J same z-coordinate, regardless of the y – coordinate or the
4. (c) When a positively charged particle is released from position in the x – y plane.
rest in a uniform electric field, it experiences a force in the
direction of the electric field. Since the force acts in the
Assertion and Reason
same direction as the displacement of the particle, work is 1. (a) As the electrostatic force is conservative, no work is
done by the electric field on the particle, causing it to gain performed when moving a charge along a closed path in an
kinetic energy. electric field. A conservative force, such as the electrostatic
As the particle moves along the electric field lines, it loses force, conserves mechanical energy. This force is solely
electric potential energy (which is converted into kinetic influenced by the positions of the charges, disregarding
energy) because it moves to regions of lower electric their velocity. Consequently, the electrostatic force’s work
potential. Electric potential decreases in the direction of the in moving a charge along a closed path is determined solely
electric field, so the electric potential energy of the charge by the initial and final positions of the charge, independent
decreases as it moves along the field lines. of the path taken.
Therefore, the electric potential energy of the charge 2. (a) Both A and R are true and R is the correct explanation
decreases because the charge moves along the electric field. of A

67 Electrostatic Potential and Capacitance P

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2 EQUIPOTENTIAL SURFACES AND POTENTIAL ENERGY

NCERT Definitions (Commonly asked in 1 mark)

U Equipotential Surfaces: These are imaginary surfaces in a region of space where electric potential is constant. They are always
perpendicular to the electric field lines. They indicate regions of equal electric poiential.
v2
v2 v1 v2
v1
v1

v2 v v2 E
v1
1

v2 v1 v1
v2
v2

U Electric Potential Energy of a System of Charges: The work done in assembling a collection of charges by bringing them
from infinity.It determines the stability and interactions within the system of charges.
r23
q3 q2

r13 r12

q1
U Potential Energy in an External Field: Energy associated with an object or system of objects due to their position in an
external electric field.
U Electrostatic sheilding: It is the process of protecting a region from external electric fields by surrounding it with a conductive
material.

Important Facts

01 Equipotential surfaces near a single point charge are concentric spheres.

02 The work done in moving a charge over an equipotential surface is zero.

03 The direction of the electric field is always perpendicular to the equipotential surfaces.

For a dipole, potential energy depends on the magnitude, direction of dipole moment and the electric field
04 strength.

Potential energy is minimum when dipole moment is aligned with electric field and maximum when
05 anti- aligned.

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Important Formulae
1. Relation between Electric Field and potential
δV
E = − ; Where E = Electric field (in V/m), ẟV = change in Potential (in V), ẟl = perpendicular distance of the surface of two
δl
given points (in m)

2. Potential energy of charge q at a distance r from origin

V = qV(r); Where V(r) is external potential at a particular point
3. Potential energy of a system of two charges in an external field y
Potential energy of the system = the total work done in assembling the configuration q1 A

= q1V ( r1 ) + q2V ( r2 ) +
q1q2 r1
2
4πε 0 r12 
r1
4. Potential energy of dipole B
 q2
 
U =− p ⋅ E =− pEcosθ ; Where p= dipole moment (in C – m) 
r2
5. Work required to rotate a dipole in a uniform electric field from an initial orientation, o x
W = U2 – U1 = pE(cosθ1 – cosθ2)
Where U = Potential energy (in J),

Classification

System of charges Equipotential surface Digram

v2
v2 v1 v1 v2
v1 v1
Point charge Spherical v2 v v E
1 v1 2
v v v2
v2 1 1
v2

E
Uniform electric field Parallel planes

Dipole Spherical – +

Identical Positive
Spherical + +
charges

Infinite line charge Coaxial cylindrical r

+q

69 Electrostatic Potential and Capacitance P

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 Difference Between (Commonly asked in 2-3 marks)

Electric Field Lines Equipotential Surfaces

Imaginary lines representing the direction and strength of the Imaginary lines/surface indicating regions of equal electric
electric field at each point. potential.
Point in the direction of the electric field at each point in space. Perpendicular to the electric field lines.
Represented by lines that indicate the direction a positive test Represented by surfaces that connect points with the same electric
charge would move if placed in the field. potential.
Density is dense where the electric field is stronger, and Density depends on the rate of change of potential; closer together
sparse where it is weaker. where the change is greater.

Electrostatic Potential Vs Electrostatic Potential Energy

Electrostatic Potential Electrostatic Potential Energy
Scalar quantity representing the work done per unit charge in The work done in moving a charge from infinity to a point in an
bringing a charge from infinity to a point in the field. electric field without acceleration.
Depends on both the position in the electric field and the magnitude
Dependent on the position in the electric field.
of the charge.
Measured in volts (V). Measured in joules (J).
Does not depend on the test charge. Directly proportional to the magnitude of the test charge.
Represents the potential of a field to do work. Represents the actual work done in moving a charge.

Mnemonics

U Potential Energy: U = qV
“U queen Victoria”

(U) Potential energy (q) Charge (V) Potential

Real Life Application Based Questions

1. Why is it essential for all the electrical appliances and outlets in a home to be connected to a common ground, creating an
equipotential surface?
Ans. Connecting all electrical appliances and outlets to a common ground ensures that they share the same electrical potential, thus
forming an equipotential surface. This is crucial for safety, preventing voltage differences between various objects that could lead
to electrical shocks. When everything is at the same potential, the risk of current flowing through a person touching two different
surfaces with different potentials is minimized, protecting against electrical hazards.
2. Why do birds sitting on high-voltage power lines not get electrocuted, considering the high potential energy associated with
these power lines?
Ans. Birds do not get electrocuted on power lines because they are only touching one wire at a time, thus not creating a voltage
difference across their bodies that would allow current to flow through them. The potential energy of the system remains constant
relative to the bird, as it is at the same electrical potential as the wire it is perched on. There’s no path for the current to flow from
one point of higher potential to another point of lower potential through the bird, hence no electrocution occurs.

Myth Buster

U Myth: Equipotential surfaces represent physical surfaces that exist in the real world.
Fact: Equipotential surfaces are imaginary surfaces in space where all points have the same electric potential. They help visualize
the electric field and potential distribution but do not necessarily correspond to physical surfaces.
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U Myth: Equipotential surfaces can intersect each other.
Fact: Equipotential surfaces never intersect each other. If they did, at the point of intersection, there would be two different
values of potential, which is impossible. The potential at every point on an equipotential surface is constant, and each surface
represents a unique potential value.
U Myth: The potential energy of a system of charges depends on the path taken to assemble the system.
Fact: The potential energy of a system of charges is path-independent. This is because electrostatic forces are conservative. The
work done in assembling a charge configuration from infinity only depends on the initial and final configurations, not on the path
taken to bring them together.

competency BASED SOlved Examples

5. Which of the following is not the property of an
Multiple Choice Questions equipotential surface? 
(1 M)
(Re)
(a) They do not cross each other.
1. An electric dipole consisting of charges +q and -q
separated by a distance L is in stable equilibrium in (b) The work done in carrying a charge from one point to
 another on an equipotential surface is zero.
a uniform electric field E . The electrostatic potential
(c) For a uniform electric field, they are concentric spheres.
energy of the dipole is (Re)
(d) They are imaginary spheres.
(a) qLE (b) zero
6. Equipotential surfaces (Un) (NCERT Exemplar)
(c) –qLE (d) –2qEL
(i) are closer in regions of large electric fields compared to
2. Two charges 14μC and -4μC are placed at (–12 cm, 0, 0) regions of lower electric fields.
 B (ii) will be more crowded near sharp edges of a conductor.
and (12 cm, 0, 0) in an external electric field E =  2  ,
r  (iii) will be more crowded near regions of large charge
6 2
where B = 1.2 × 10 N/(cm ) and r is in metres. The densities.
electrostatic potential energy of the configuration is (Ev) (iv) will always be equally spaced.
(a) 97.9 J (b) 102.1 J (a) i and ii (b) ii and iv
(c) 2.1 J (d) -97.9 J (c) i, ii and iii (d) only iv
3. Equipotential at a large distance from a collection of 7. The work done to move a charge on an equipotential
charges whose total sum is not zero are:  (Re) surface is  (Un)
(a) spheres (b) planes (a) Less than 1 (b) Greater than 1
(c) ellipsoids (d) parabolas
(c) Zero. (d) Infinity.
4. Four charges –q, –q, +q and +q are placed at the corners
8. In a region of constant potential
of a square of side 2 L is shown in figure. The electric
(Un) (NCERT Exemplar)
potential at point A midway between the two charges +q
and +q is:  (Un) (a) the electric field is uniform
+q –q (b) the electric field is zero
(c) there can be no charge inside the region.
(d) the electric field shall necessarily change if a charge is
2L placed outside the region.
A
Assertion and Reason
(1 M)
+q –q
Direction: The following questions consist of two statements.
Assertion (A) and Reason (R). Answer these questions by
1 2q  1  1 2q  1 
(a)  1−  (b)  1+ 
selecting the appropriate option given below:
4πε 0 l  5 4πε 0 L  5 (a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
1 g  1 
(c)  1−  (d) Zero (c) A is true, but R is false
4πε 0 2 L  5 (d) A is false, but R is true.

71 Electrostatic Potential and Capacitance P

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 1. Assertion (A): The direction of electric field is always the other end of the field lines originated from the charges
perpendicular to equipotential surface. inside. This contradicts the original assumption. Hence, the
entire volume inside must be equipotential.  (1 M)
Reason (R): Work is done by the electric force in moving
a charge between any two points on an equipotential 4. A charge of +2μC is moved from point A to point B in
surface is zero.  (Un) an electric field. The potential difference between the
two points is 50V. Calculate the change in the potential
2. Assertion (A): A charged metallic shell may have charged
energy of the charge during the process. (Ev)
particle inside it.
Ans. The change in potential energy (ΔU) of the charge can be
Reason (R): There can’t exist electric field lines inside calculated using the formula:
the conductor.  (Re)
ΔU = q. ΔV (1 M)
Substituting the given values, we get:

Subjective Questions ΔU=(2×10–6C) × (50V) = 100×10–6J =100μJ  (1 M)

Very Short Answer Type Questions  (2 M) Short Answer Type Questions (3 M)

1. (i) Draw equipotential surfaces for an electric dipole. 1. (a) Draw the equipotential surfaces corresponding to a
(1 M) (CBSE, 2019) (Re) uniform electric field in the z-direction. (1 M)
(ii) What is the potential due to dipole at line bisecting (b) Derive an expression for the electric potential at any
the dipole length? (1 M) point along the axial line of an electric dipole. (2 M)
Ans. (i) Equipotential surface due to electric dipole: (Ev) (CBSE, 2019)
Equipotential surface Ans. (a) The equipotential surfaces corresponding to a uniform
electric field in the z-direction is as shown in fig.
Electric field

equipotential surfaces
+ in X - Y Plane

d Y
 (1 M)
X (1 M)
(ii) The potential due to the dipole is zero at the line
bisecting the dipole length. (1 M) (b) Expression for the electric potential at any point along
the axial line of electric dipole:
2. Two charges 2mC and –2mC are placed at points A and
B 6 cm apart. O
P
–q a a +q
(a) Identify an equipotential surface of the system.
(1 M) r
(b) What is the direction of the electric field at every r–a
point on this surface? (1M) r+a
(An) (NCERT Exemplar) Electrostatic potential at point P due to +q charge
Ans. (a) An equipotential surface is the plane on which total
1 +q
potential is zero. V+a =
4πε 0 ( r − a )
It is at the midpoint of the line AB and perpendicular to it.
(1 M) Electrostatic potential at point P due to -q charge
(b) The direction of the electric field at every point on this 1 −q
V+0 =
surface is normal to the plane in the direction of AB.(1 M) 4πε0 ( r + a )
3. Prove that a closed equipotential surface with no charge Potential at point P according to the principle of superposition
within itself must enclose an equipotential volume. (1 M)
(Un) (NCERT Exemplar) 1 +q 1 −q
V = V+ q + V− q = + V
Ans. Let us assume that in a closed equipotential surface with no 4πε 0 ( r − a ) 4πε 0 ( r + a )
charge the potential is changing from position to position.
Let the potential just inside the surface is different to that of q  1 1  q r + a −r + a
= − V= V
the surface causing a potential gradient (dV/dr)  (½ M) 4πε 0  r − a r + a  4πε 0  r 2 − a 2 

∴ E= −
dV q × 2a kP
=
(½ M) =
dr
It means there will be field lines pointing inwards or outwards
4πε 0 (r 2 − a 2 ) (r 2
− a2 ) (1 M)
from the surface. These lines cannot be again on the surface,  1 
as the surface is equipotential. It is possible only when Where, p = 2aq  P = Dipole moment and k = 
 4 π ∈o 

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 2. X and Y are two equipotential surfaces separated by a Ans. (a) Remains the same. The potential at a location depends
distance of 2 m in a uniform electric field of 10 V/m. on the source charge. It is independent of the test charge at
Surface X has a potential of 10 V (Ev) (CBSE SQP, 2023) the location.  (1½ M)
(a) Calculate the potential of surface Y. (1½ M) (b) Increases. The initial potential energy of q1 & q2 charge
(b) What is the work done in moving a +2 C charge from system was negative. When the test charge is changed from
surface Y to surface X along path 1? How will this work positive to negative, the potential energy becomes positive.
change when the charge is moved along Path 2? Give So it increases.  (1½ M)
a reason for your answer.  (1½ M) 4. A proton (e) approaches a short fixed electric dipole (p)
moving along the dipole axis as shown in the figure. At a
Path 1 large distance from the dipole, the kinetic energy of the
proton was K0 = 400 eV.

The graph below shows the variation of kinetic energy

Y X E (K) of the proton at points close to the dipole. Find the
value of r0 (r0>> length of the dipole). In the graph r is
the distance from the centre of the dipole. (Ap)
2m K(eV)
10 V Path 2
Vy
Ans. (a) Given E = 10 V/m 100
Vx = 10 V
Δr=2m
  r0
| V | E  r (½ M) O r
0.2 (in m)
= 10 × 2 = 20 V
Ans. Given:
Since, the potential decreases in the direction of the electric
field, the potential at surface Y will be more than the Initial kinetic energy of proton = K0 = 400 eV
potential at surface X. As the proton approaches the dipole its kinetic energy
V = 20 + 10 = 30V  (1 M) reduces and the potential energy increases, however, the
total energy is conserved.
(b) Given: q = 2 C
Electric potential due to the dipole at axial point = p/4π∈0r2
Work done in moving charge from Y to X along Path
1 = (Vx – Vy)q Let P and K be the potential and kinetic energy of the proton
at any instant, respectively. From the graph, at r = 0.2 m, K
W = (10 – 30) × 2 = 100 eV  (1 M)
W = –20 × 2 = – 40 J  (½ M) Applying conservation of energy:
Work done in moving charge along Path 2 will be the same K + P = K0 (1 M)
as work done along Path 1.  (½ M)
100 + ep/(4π∈0 r ) = 400
2
This is because the work done between two surfaces is
This implies = 300 × 0.22 = 12
independent of the path since the force acting on the charge
is conservative in nature.  (½ M) At r = r0 the kinetic energy is zero and thus we have,
3. Given are two charges, q1, a negative source charge, and ep/4π∈0r20 = 400
q2, a test charge. The test charge q2 is initially positive 12 12 3
= =
400, r02 = r m ≈ 0.17 m (1 M)
and then changed into a negative charge of the same r 20 400 0 10
magnitude.  (Un) 5. Two point charges of magnitude +q and -q are placed at
(–d/2, 0, 0) and (d/2, 0, 0), respectively. Find the equation
q1 of the equipotential surface where the potential is zero.
(Ap) (NCERT Exemplar)
Ans. Let the plane be at a distance x from the origin.
P z
q2
y
h
(a) Will the potential at the position of charge q2 due to
x
the source charge q1 (i) remain the same, (ii) increase
or (iii) decrease? (1½ M) q –q
(b) Will the potential energy of the q1 & q2 charge system –d/2 o x d/2
(i) remain the same, (ii) increase or (iii) decrease?
Give an explanation in each case. (1½ M) The potential at the point P is

73 Electrostatic Potential and Capacitance P

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 1 q 1 q (iv) Work needed to bring –q to D when +q at A, –q at B,
− (1 M) and +q at C. This is given by (charge at D ) × (potential
4πε 0 ( x + d / 2) + h 
2 2 1/ 2
4πε 0 ( x − d / 2) 2 + h 2 1/ 2
    at D due to charges at A, B and C )
If this is to be zero. 
−q2 1 
=
 2 −   (1 M)
1
=
1 4πε 0 d 2
1/ 2 1/ 2
( x + d / 2) 2 + h 2  ( x − d / 2) 2 + h 2  Add the work done in steps (i), (ii). (iii) and (iv). The total
work required is
Or, (x – d/2)2 + h2 = (x + d/2)2 + h2
−q2   1   1  
⇒ x2 – dx + d2/4 = x2 + dx + d2/4 = ( 0) + (1) + 1 −  + 2− 
4πε 0 d   2  2  
Or, 2dx = 0
⇒x=0
The equation is that of a plane x = 0  (2 M)
=
−q2
4πε 0 d
(
4− 2 )
 (1 M)

Long Answer Type Questions  (5 M)

The work done depends only on the arrangement of the
charges, and not how they are assembled. By definition,
1. Four charges are arranged at the corners of a square this is the total electrostatic energy of the charges.
ABCD of side. Find the work required to put together (b) The extra work necessary to bring a charge q0 to the
this arrangement.  (4 M) point E when the four charges are at A, B, C and D is q0×
(b) A charge q0 is brought to the centre E of the square, (electrostatic potential at E due to the charges at A, B, C
the four charges being held fixed at its corners. How and D). The electrostatic potential at E is clearly zero since
much extra work is needed to do this? (1 M) potential due to A and C is cancelled by that due to B and
 (Ev) (NCERT Exemplar) D. Hence, no work is required to bring any charge to point
E.  (1 M)
+q –q
A B 2. (a) Determine the electrostatic potential energy of a
system consisting of two charges 7μC and –2μC (and
with no external field) placed at (–9 cm, 0, 0) and (9
cm, 0, 0) respectively.  (1 M)
D C (b) How much work is required to separate the two
–q +q
charges infinitely away from each other?  (2 M)
Ans. (a) Since the work done depends on the final arrangement (c) Suppose that the same system of charges is now
of the charges, and not on how they are put together, we placed in an external electric field E = A(1/r2 );
calculate work needed for one way of putting the charges A = 9 × 105 NC–1 m2. What would the electrostatic
at A, B, C and D. Suppose, first the charge +q is brought energy of the configuration be? (2 M)
to A, and then the charges –q. +q, and –q are brought to  (NCERT Exemplar)
B, C and D, respectively. The total work needed can be
1 q1q2 7 × ( −2) × 10 −12
calculated in steps: Ans. (a) U = =
9 × 109 × =
−0.7 J 
4πε 0 r 0.18
(i) Work needed to bring charge +q to A when no charge is  (1 M)
present elsewhere: this is zero.  (1 M)
(b) W = U2 – U1 = 0 – U = 0 – (– 0.7) = 0.7 J. (2 M)
(ii) Work needed to bring –q to B when +q is at A. This is
(c) The mutual interaction energy of the two charges
given by (charge at B) × (electrostatic potential at B due
remains unchanged. In addition, there is the energy of
to charge +q at A)
interaction of the two charges with the external electric
 q(−q) −q 2  field. We find,
=
+q 
+   (1 M)
 4πε 0 d 4πε 0 d  q1V ( r1 ) + q2V ( r2 )= A
7 µC
+A
−2µC
0.09 m 0.09 m  (1 M)
(iii) Work needed to bring charge +q to C when +q is at A and
and the net electrostatic energy is
–q is at B. This is given by (charge at C ) × (potential at
7 µC −2µC
q1V ( r1 ) + q2V ( r2 ) +
C due to charges at A and B ) q1q2
= A +A − 0.7 J
4πε 0 r12 0.09 m 0.09 m
−q2  1 
=
1 − 
4πε 0 d 2 = 70 – 20 – 0.7 = 49.3 J  (1 M)

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 3. (a) When a group of charges is situated at a considerable
Hints & Explanations distance, it can be treated as a point charge. In this scenario,
the equipotential surface for a point charge maintains
Multiple Choice Questions a consistent distance from the point in all directions.
1. (c) – qLE Consequently, the equipotential points surrounding a point
Given charge form a spherical surface.
+q 
E 1 2q  1 
4. (a) 1−
L 4πε 0 l  
5

–q 5. (c)
6. (c) In regions with a stronger electric field, the equipotential
Dipole moment is
surfaces are closer together because the potential changes
p = qL more rapidly over a shorter distance. Near sharp edges
Applying the formula of potential energy of dipole in or points of a conductor, the electric field is stronger,
uniform electric field. leading to a more rapid change in potential. As a result,
U = –pEcosq equipotential surfaces will be closer together near sharp
In stable equilibrium, q =0 edges. In regions with a higher charge density, there is a
so stronger electric field, resulting in equipotential surfaces
being closer together to accommodate the more significant
Umin = –pE = –qLE
potential changes.
2. (a) Given, q1 = 14μC, q2 = –4μC
7. (c) Work done in equipotential surface is zero as W = q (VA
 B
External electric field, E =  2  – VB) and VA = VB
r 
or V =
B 8. (b) We know, the electric field intensity E and electric
r dv
B = 1.2 × 106 N/(cm2) potential V are related as E = −
dr
Potential energy of two charges in external field is given by For V = constant, dv/dr implies that electric field intensity,
q1q2 E = 0.
U = q1V1 + q2V2 +
4πε0 r12

1.2 × 106 1.2 × 106 Assertion and Reason

=14 × 10 −6 × − 4 × 10 −6 × − 9 × 109
0.12 0.12 1. (a)
14 × 10 −6 × 4 × 10 −6
× 2. (b) A charge conductor may initially have a charged
0.24 particle inside it but since E field inside it is always zero so
= 97.9 J it gets redistributed on the outer surface.

75 Electrostatic Potential and Capacitance P

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3 DIELECTRIC AND CAPACITORS

NCERT Definitions (Commonly asked in 1 mark)

U Polar Molecules: Polar molecules are molecules where the distribution of electrons is uneven, resulting in a permanent dipole
moment. This occurs when the atoms in the molecule have different electronegativities. As a result, polar molecules have a
positive and a negative pole.
Examples: Water (H2O), Ammonia (NH3), and Hydrogen Chloride (HCl).
H
O
Polar
Cl P H H P

HCl H2O
U Nonpolar Molecules: Nonpolar molecules are molecules where the distribution of electrons is symmetrical, resulting in no
permanent dipole moment. This occurs when the atoms in the molecule have similar electronegativities, causing the electron
cloud to be distributed evenly around the molecule.
Examples: Oxygen (O2), Nitrogen (N2), and Carbon dioxide (CO2)

Non-polar H H O
O
O
H2
CO2
U Dielectrics: Insulating materials that, when placed in an electric field, become polarized and reduce the field’s strength.
U Linear Isotropic Dielectrics: Linear isotropic dielectrics are materials that exhibit a linear relationship between the electric field
and polarization, and their properties are the same in all directions.
U Dielectric Constant (εr): A measure of the ability of a dielectric material to increase the capacitance of a capacitor compared to
a vacuum.
U Dielectric Strength: The maximum electric field that a dielectric material can withstand without breaking down (i.e., becoming
conductive).
U Polarization: The process in which the positive and negative charges in a dielectric are slightly displaced in opposite directions
under an electric field.
U Electric Susceptibility (χe): A dimensionless parameter that quantifies the extent of polarization of a dielectric material in
response to an applied electric field.
U Electric Dipole Moment: A measure of the separation of positive and negative charges in a system, which contributes to the
polarization in dielectrics.
U Capacitors: Devices that store electrical energy in an electric field, typically consisting of two conductors separated by an
insulator or dielectric.
U Capacitance: A measure of a capacitor’s ability to store charge, defined as the ratio of the electric charge on each conductor to
the potential difference between them.
U Parallel Plate Capacitor: A type of capacitor with two parallel conducting plates separated by a dielectric material.

Conductive d
plates

Dielectric
U Energy Stored in a Capacitor: The energy stored in a capacitor is a function of the capacitance and the voltage across it.
U Electric Field in Capacitors: A region around a charged capacitor where electric forces can be felt.
U Breakdown Voltage: The maximum voltage a dielectric can withstand without becoming a conductor.

P
W CBSE Class – XII PHYSICS 76
Important Facts

Capacitance is directly proportional to the permittivity of the dielectric and the area of the plates, and
01 ε0 A
inversely proportional to the distance between the plates. C =
d

02 Dielectrics increase capacitance by reducing the electric field strength between the plates.

03 In a parallel plate capacitor, the electric field is uniform between the plates and zero outside.

The combination of capacitors in series results in a lower total capacitance than any individual capacitor
04 in the combination.

The combination of capacitors in parallel results in a higher total capacitance than any individual capacitor
05 in the combination.

The breakdown voltage of a dielectric is crucial in determining the maximum voltage a capacitor can
06 handle.

07 Polarization in dielectrics occurs due to the alignment of dipole moments in the material.

The presence of a dielectric in a capacitor affects both its capacitance and its charging and discharging
08 behavior.

Important Formulae
1. Capacitance of the capacitor
Q ; Where Q = charge of the capacitor (in C), V= potential difference (in V)
C=
V
2. Polarization for linear isotropic dielectrics
P = ε0χe E ; Where P = Polarisation (in C.m2), and χe= electric susceptibility of the dielectric medium, ε0 = Vaccum permitivitty

(8.85 ×10–12 F/m)
3. The Parallel Plate Capacitor
I Area A
I. Electric field in outer region I (region above the plate 1),
σ σ 1 s
E= − = 0 ; Where s = surface charge density (in C/m2)
2ε 0 2ε 0 E d
II. Electric field in outer region II (region below the plate 2),
σ σ 2 -s
E= − =0 II
2ε 0 2ε 0

77 Electrostatic Potential and Capacitance P

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 III. Electric field in the inner region between the plates 1 and 2,
σ σ σ Q
E= + = =
2ε 0 2ε 0 ε 0 ε 0 A

IV. Capacitance
Q ε A
C = == 0 ; Where, d = separation between plates (in m)

V d
4. Effect Of Dielectric on Capacitance
I. Dielectric constant
ε Cmedium
= =
Cair , Where, ε0 = permittivity of free space, ε = permittivity of medium
K
ε0
II. Capacitance of parallel plate capacitor
Aε K
(i) C = 0 in medium of dielectric constant K
d
Aε0
(ii) C = if space between plate partially filled with dielectric of thickness t.
 1
d − t 1 − 
 K
5. Combination of capacitors
1 1 1 1
(i) In series, = + + , q1 = q2 = q3 ,V = V1 + V2 + V3
C C1 C2 C3
(ii) In parallel, C = C1 + C2 + C3, q = q1 + q2 + q3, V1 = V2 = V3 = V
6. Energy stored by capacitor
1 Q2 1
=
I.
U =
CV 2
= QV
2 2C 2
II. Energy stored in the capacitor for parallel plate capacitor
1 Q 2 ( Aσ) 2 d
= = ×
2 C 2 ε0 A

III. Relation between electric field and surface charge density between plates of capacitor.
σ
E=
ε0
IV. Energy stored in capacitor, U = (1/2)ε0E2 × Ad
V. Energy density of electric field
U  1
=   ε 0 E 2 [Energy per unit volume]
Ad  2 

Classification

Capacitor Vs Capacitance
Aspect Capacitor Capacitance
A measure of the ability of a capacitor to store electrical
A device that stores electrical energy in the form of an
charge, quantitatively defined as the ratio of the electric
Definition electric field. It consists of two conductors separated by
charge on each conductor to the potential difference
an insulator.
between them.
Physical A physical device constructed with materials that can A property or characteristic of a capacitor, not a physical
Existence conduct electricity separated by an insulating layer. object.
Its effectiveness depends on the material between the Depends on the physical structure of the capacitor
Dependency plates (dielectric), the area of the plates, and the distance (shape, size, separation of plates) and the dielectric
between them. material.

P
W CBSE Class – XII PHYSICS 78
 Unit Not applicable. Farads (F), where 1 Farad = 1 Coulomb/Volt.
Used in circuits to block direct current while allowing
Used to quantify how much charge a capacitor can store,
Usage alternating current to pass, in filters, in energy storage,
indicating its efficiency in energy storage applications.
etc.
Capacitors come in various types and sizes depending Capacitance is a fixed value for a given capacitor but
Variability on the application, such as ceramic, electrolytic, and changes with the type of dielectric material and its
supercapacitors. dimensions.

Combination of Capacitance
Series Parallel
1 1 1 1
Capacitance decreases in series, = + + ... Capacitance adds up in parallel, Ctotal = C1 + C2 +...Cn
Ctotal C1 C2 cn

The total voltage across capacitors is the sum of individual

The voltage across each capacitor is the same
voltages
Used to increase the overall voltage rating Used to increase the overall capacitance
Reduces the overall capacitance which can be useful in Increases the overall capacitance, useful in energy storage
timing and filtering circuits applications
C
C C C
C

– + –
K
V
–
+ –
K V

Difference Between (Commonly asked in 2-3 marks)

Capacitor Vs Dielectric
Feature Capacitor Dielectric
An electrical component that stores electric
Definition A material that resists the flow of electric current.
charge.
Increases the capacitance of a capacitor and reduces
Primary Role Stores and releases electrical energy in circuits.
clectric field strength.
Consists of two conductive plates separated by
Composition Insulating materials like glass, paper, or air.
an insulating layer.
Electrical Does not allow current to flow, can be polarised in
Can store charge for a given potential difference.
Property electric fields.

Capacitance with dielectric or without dielectric

Without Dielectric With Dielectric
Capacitance is lower as it depends on the vacuum
Capacitance increases due to the presence of the dielectric
permittivity, plate area, and separation
The electric field is stronger as there is no medium to reduce The dielectric material reduces the electric field strength between
its strength the plates
Energy stored is lower for a given voltage Energy stored increases for the same voltage
Breakdown voltage is lower, limiting the maximum voltage
Breakdown voltage increases, allowing higher voltages to be applied
that can be applied
Suitable for applications requiring lower capacitance Ideal for high-capacity storage and insulation purposes

79 Electrostatic Potential and Capacitance P

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 Real Life Application Based Questions
1. How do capacitors improve the energy efficiency of air conditioning units?
Ans. In air conditioning systems, capacitors are used to store and release electrical energy to help start and run the motors efficiently.
These capacitors help smooth out electrical current fluctuations, providing a stable power supply and reducing the energy required
to maintain motor speed. This leads to enhanced energy efficiency and reduced wear on the components.
2. In photography, how does a camera flash utilize the properties of capacitors?
Ans. Camera flashes use capacitors to store a large amount of charge at high voltage, which is then rapidly released to produce a brief,
intense burst of light. The capacitor charges up slowly and discharges in a fraction of a second to generate the necessary light to
illuminate a scene briefly and brightly.
3. Why are capacitors with a high dielectric strength important in medical devices like defibrillators?
Ans. Defibrillators require capacitors with high dielectric strength because they need to deliver a controlled and quick high-voltage
shock to a patient’s heart. The dielectric strength determines the capacitor’s ability to withstand high voltages without breaking
down, ensuring the safety and effectiveness of the shock administered during critical life-saving procedures.
4. How does the combination of capacitors influence the sound quality in audio equipment?
Ans. In audio equipment, different combinations of capacitors are used to filter and stabilize voltage supply to various components.
By smoothing out fluctuations in the power supply, capacitors help in reducing noise and improving sound quality, ensuring that
audio signals are clear and undistorted.

Myth Buster
U Myth: Dielectrics can conduct electricity just like conductors when placed in an electric field.
Fact: Dielectrics do not conduct electricity. Instead, they insulate against electric flow. When an electric field is applied, dielectrics
become polarized, meaning their molecular charges shift slightly, creating dipoles that oppose the external electric field. This
polarization strengthens the material’s insulating properties.
U Myth: The capacitance of a capacitor is dependent on the amount of charge stored.
Fact: The capacitance of a capacitor is determined by its physical characteristics such as the area of the plates, the distance
between them, and the type of dielectric material used. It does not depend on the amount of charge stored. Capacitance is a
measure of a capacitor’s ability to store charge for a given potential difference (voltage).
U Myth: Energy stored in a capacitor is only dependent on the charge it holds.
Fact: The energy stored in a capacitor is dependent on both the charge it holds and the Capacitance. It is calculated by the formula
1
E = CV 2 , where E is the energy, C is the capacitance, and V is the voltage. This formula shows that the energy is proportional
2
to the square of the voltage as well as the capacitance.

Mnemonics

U Energy Stored in a Capacitor U Capacitors in Parallel

“U Can See the CV Square” helps to remind the formula, C Capacitors in Parallel
 1 2
 U = 2 CV  I Increase
 
U Capacitors in Series P Same potential difference

C Capacitors in Series

I Inverse

S Same charge

P
W CBSE Class – XII PHYSICS 80
 competency BASED SOlved Examples
5. Three capacitors C 1, C 2 and C 3 are connected in a
Multiple Choice Questions
(1 M) combination as shown below.
C2 = 8mF
1. Which of the diagrams correctly represents the electric
field between two charged plates if a neutral conductor
is placed in between the plates? (An)(CBSE, 2022)
C1 = 4mF
(a) (b) C3 = 4mF

10 V
(c) (d)
Identify the correct statement(s). (Un)
(i) The charge on capacitor C1 is greater than that on
2. A variable capacitor is connected to a 200 V battery. If its capacitor C2.
capacitance is changed from 2 F to X F, the decrease in
energy of the capacitor is 2 × 10–2 J. The value of X is – (ii) The charge on capacitor C1 is the same as that on
capacitor C3.
(Ev) (CBSE, 2022)
(iii) The charge on capacitor C1 is 30 μC.
(a) 1mF (b) 2mF
(a) Only (i) is correct.
(c) 3mF (d) 4mf
(b) Only (iii) is correct.
3. A car battery is charged by a 12 V supply, and energy
stored in it is 7.20 × 105J. The charge passed through the (c) Both (i) and (iii) are correct.
battery is – (Ev) (CBSE, 2022) (d) Both (i) and (ii) are correct.
(a) 6.0 × 104C (b) 5.8 × 103J
Assertion and Reason
6
(c) 8.64 × 10 J 6
(d) 1.6 × 10 C (1 M)
4. In one kind of computer keyboard, each key is mounted Direction: The following questions consist of two statements.
on one end of a plunger. The other end of the plunger Assertion (A) and Reason (R). Answer these questions by
is attached to a movable metal plate. Refer to the figure selecting the appropriate option given below:
given.
(a) Both A and R are true, and R is the correct explanation of A.
B (b) Both A and R are true, but R is not the correct explanation
Key
of A.
(c) A is true, but R is false
Movable plate (d) A is false, but R is true.
Dielectric 1. Assertion (A): thin uncharged metallic plate placed in
Fixed plate between the two charged plates of a capacitor results in
The dielectric material between the two plates is made of an arrangement equivalent to two capacitors in a series
a soft material and is compressible. The combination of combination. The equivalent capacitance of this combination
the two plates and the dielectric between them constitutes stays the same irrespective of the position of the metallic
a capacitor. plate in between the plates of the capacitor.
Each key on the keyboard when pressed is recognized Reason (R): The change in the position of the central
due to which one of the following factors?  (An) metallic plate, results in the decrease in plate separation of
(a) The pressing of the key increases the capacitance of the one capacitor that is compensated by the increase in plate
capacitor below the key due to a decrease in separation separation for the other. (Un)
between the plates.
2. Assertion (A): The charge-to-voltage ratio increases on
(b) The decrease in the thickness of the soft dielectric layer
insertion of a dielectric material between the capacitor plates,
decreases the capacitance of the capacitor below the key.
when either the voltage or charge is kept constant.
(c) The momentary decrease in the space between the plates
of the capacitor is detected as a mechanical sound signal Reason (R): The capacitance of a capacitor increases when
of a specific frequency. it is filled with a dielectric material with a dielectric constant
(d) all of the above greater than 1. (Un)

81 Electrostatic Potential and Capacitance P

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 σ 4πb 2 q1 b 2 b
Subjective Questions
V 100v q
=E = 104 N/C 1 = 1 2 ⋅ = ⋅ 2 =
d 0.01 m σ 2 4πa q2 q2 a a
Very Short Answer Type Questions (2 M) (1 M)
∴ The ratio of electric fields at the surfaces of two spheres
1. Find the total charge stored in the network of capacitors
connected between A and B as shown in figure: E1 σ1 b
= =
(Ev) (CBSE, 2020) E2 σ 2 a (1 M)
4mF 2mF 2. A 200μF parallel plate capacitor having plate separation
of 5 mm is charged by a 100 V dc source. It remains
A B connected to the source. Using an insulated handle, the
distance between the plates is doubled and a dielectric
slab of thickness 5 mm and dielectric constant 10 is
6mF 3mF introduced between the plates. Explain with reason, how
the (i) capacitance, (ii) electric field between the plates,
(iii) energy density of the capacitor will change?
(Ev) (CBSE, 2019)
Ans. The effective separation between the plates with air in
3V
t 5
Ans. The given network can be redrawn as shown here. between is, 2d − t + = 10 − 5 + = 5.5 mm
k 10
C1 = 4mF C3 = 2mF
(i) new capacitance
ε0 A 10
C '= = C ≈ 182µF
5.5mm 11
ε A
C2 = –6mF C4 = 3mF [ As C =
200 µF ] & C =0
5 min (1 M)

(ii) change in the electric field or effective new electric field

V 100 N
E= = = 104
,m
d 0.01 C
3V (1 M) (1 M)
(iii) Energy density
The equivalent capacitance of the network is given as:
1
1 1 1 1 1 U=
ε0 E 2 ,  (1 M)
= + = + 2
C ( C1 + C2 ) ( C3 + C4 ) ( 4 + 6) µF ( 2 + 3) µF
1
=× 8.85 × 10−12 × (104 ) 2

 1 1 3 10 2
=  + = µF ⇒ C = µF
 10 5  10 3 4.4 × 10–4 J/m3

3. A slab of material of dielectric constant K has the same
∴ Total charge on the network
area as the plates of a parallel-plate capacitor but has a
 10  thickness (3/4)d, where d is the separation of the plates.
Q = CV =  µF  × 3V =10µC (1 M)
 3  How is the capacitance changed when the slab is inserted
between the plates? (Ev)
Short Answer Type Questions (3 M) Ans. Let E0 = V0/d be the electric field between the plates when
1. Two charged conducting spheres of radii a and b are there is no dielectric and the potential difference is V0. If the
connected to each other by a wire. Find the ratio of the dielectric is now inserted, the electric field in the dielectric
electric fields at their surfaces. (Ev) (CBSE, 2023) will be E = E0/K. The potential difference will then be
Ans. Consider charges q1 and q2 with capacitances C1 and C2 1  E 3 
=V E0  d  + 0  d 
for two spheres. The flow of charge occurs from the sphere 4  K 4 
at higher potential to the one at lower potential until their
1 3  K +3 (1 M)
= E0 d  + = V0
 4 4 K 
potentials become equal.
4K
After sharing, the charges on two spheres would be
The potential difference decreases by the factor (K + 3)/4K
q1 C1V
= …(i) while the free charge Q0 on the plates remains unchanged.
q2 C2V (1 M)
C1 a The capacitance thus increases
Also =
C2 b (1 M) Q0 4 K Q0 4K
=
C = = C0 (1 M)
From (i) Ratio of the surface charge on the two spheres V K + 3 V0 K +3

P
W CBSE Class – XII PHYSICS 82
 4. A network of four 10μF capacitors is connected to a 500 Kq
V supply, as shown. E1 =
(a / 2) 2
Determine (a) the equivalent capacitance of the network. Kq
(1½ M) E2 =
(a / 2) 2
(½ M)
(b) the charge on each capacitor. (Note, the charge on a
capacitor is the charge on the plate with higher potential, Enet = E1 – E2
equal and opposite to the charge on the plate with lower ∵ E1 = E2
potential.) (1½ M) (Ev) Enet = 0  (1 M)
+Q –Q
+ – (ii) No, the electric potential is a scalar quanitity and its
+ – C value along the line joining of two equal postive charges
B + –
+ – cannot be zero.  (1½ M)
C2
Long Answer Type Questions  (5 M)
+
+
+
+
–Q – – – – +Q
C1 C3
+Q – – – – –Q
+
+
+
+

1. (a) D escribe briefly the process of transferring the

charge between the two plates of a parallel plate
C
+ 4 – capacitor when connected to a battery. Derive an
+ –
A + –
D expression for the energy stored in a capacitor. (3 M)
+ –
+Q –Q (b) A parallel plate capacitor is charged by a battery to a
500 v potential difference V. It is disconnected from battery
and then connected to another uncharged capacitor
Ans. (a) In the given network, C1, C2 and C3 are connected in
of the same capacitance. Calculate the ratio of the
series. The effective capacitance C′ of these three capacitors
energy stored in the combination to the initial energy
is given by
on the single capacitor. (2 M)
1 1 1 1
= + + (½
½ M) (Ev) (CBSE, 2019)
C ′ C1 C2 C3
Ans. (a) When the two plates of a parallel-plate capacitor are
For C1= C2= C3 = 10μF, C' = (10/3)μF. The network connected to a battery, it pulls electrons from the positive
has C and C4 connected in parallel. Thus, the equivalent plate of the capacitor and gives them to the negative plate.
capacitance C of the network is (1 M)
 10  Thus, the battery transfers positive charge from negative
C = C ′ + C4 =  + 10 µF = 13.3µF (1 M)
 3  to positive plate the work done in this transfer process is
(b) Charge on capacitors, C1, C2 and C3 is the same, say stored in form of electrostatic energy in the capacitor.
Q. Let the charge on C4 be Q′. Now, since the potential Energy stored in capacitor: Let charge on the capacitor at
difference across AB is Q/C1, across BC is Q/C2, across CD any instant be q. Then, the potential of capacitor is q/c,
is Q/C3, we have where C is the capacitance. Suppose an additional charge
dq is given to capacitor, then the work done in giving this
Q Q Q
+ + =
500 V (½
½ M) additional charge is
C1 C2 C3
q
Also, Q/C4 = 500 V. dW= × dq (1 M)
C
This gives for the given value of the capacitances, Total work done in giving charge Q to capacitor is
10 q −Q
Q= 500 V × µF= 1.7 × 10 −3 C and =q Q q  1 q2 
q −Q 1 q2
3 =W ∫= dW ∫ = dq  = W
= q 0=
 2 C  q =0
q 0C 2 C
Q′ = 500 V × 10μF = 5.0 × 10–3C (1 M)
5. Consider two identical point charges located at points which is the form of energy stored. Using Q = CV, we have
(0,0) and (a,0). 1 2
U= W= Cv (1 M)
(i) Is there a point on the line joining them at which the 2
electric field is zero?  ½ M)
(1½ which is the expression for energy stored in a capacitor.
(ii) Is there a point on the line joining them at which the (b) When a parallel-plate capacitor is charged by a battery
electric potential is zero? (1½
½ M) (An) (CBSE, 2023) to a potential difference V is disconnected from battery and

Justify your answers for each case. connected to an uncharged capacitor of same capacitance,
charge flows from the charged capacitor to the uncharged
Ans. (i) Yes, electric field is zero at mid point.
capacitor till their potentials become equal. The equal
Electric field being a vector quantity, its resultant is zero. potential of the two capacitors is called common potential.
The net electric field at a/2: The energy stored in capacitor 1 is

83 Electrostatic Potential and Capacitance P

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 1 1 2 C123 = 200 pF
=U = C1V12 cv
2 2
When two capacitors are combined:
Charge on combination q is q1 + q2 = q
300 V (1 M)
Capacitance c is c1 + c2 = C + C = 2C.
Potential difference is
C4 = 100 pF
c1v1 + c2 v2 cV + c × 0 cV V
V′
= = = = . (1 M)
c1 + c2 c+c 2c 2 The equivalent capacitance =
200
pF
3
Energy stored in combination is
200
1 1 V 
2
1  V 2  cV 2 charge on C4   1012  300  2  108 C , (½ M)
U = c(v) 2 = × 2C ×   = (2c)  = 3
2 2 2 2  4  4
200  1012  300
cV 2 potential difference across C4   200 V
3  100  1012
u' 4 1
Than= = (1 M)
u cV 2
2 potential difference across C1 = 300 – 200 = 100 V
2 charge on C1 = 100 × 10 – 12 × 100 = 1 × 10–8 C (½ M)
2 (i) Derive an expression for the capacitance of a parallel potential difference across C2 and C3 series combination =
plate capacitor with air present between the two 100 V
plates.  (2 M) potential difference across C2 and C3 each = 50V
(ii) Obtain the equivalent capacitance of the network charge on C2 and C3 each = 200 × 10–12 × 50 = 1 × 10–8 C
shown in figure. For a 300 V supply, determine the (1 M)
charge on each capacitor. (3 M) 3. (a) A camera usually operates at 1.5 V and this potential
(An) (CBSE SQP, 2023) difference is not sufficient to emit light energy
100 pF using flash. For this purpose, the flash circuit of the
camera has a capacitor that is charged to 300 V-330
V using various electrical components. If the voltage
C1 generated across the plates of the capacitor is 300 V
and the capacitance of the parallel plate capacitor
used is 100 μF, then find the energy released when
200 pF the trigger button on the camera is pressed. (1 M)
200 pF
+ (b) How much charge does the 100 μF capacitor charged
C2 C3 to 300 V hold? (1 M)
100 pF 300 V
– (c) If the distance between the parallel plate capacitor
of capacitance 100 μF is increased two times, then
C4 calculate the capacitance of the capacitor. (1 M)
Ans. (i) Derivation of the expression for the capacitance (d) The graph below shows the variation of charge ‘q’
Let the two plates be kept parallel to each other separated with potential difference ‘V’ for a parallel plate
by a distance d and cross-sectional area of each plate is A. capacitor ‘C’ for scenarios P and Q.
Electric field by a single thin plate, E = σ/2∈o (½ M) Scenario P - the space between the capacitor ‘C’ is
filled with air.
Total electric field between the plates, E = σ/∈o = Q/A ∈o
(½ M) Scenario Q - the space between the capacitor ‘C’ is
filled with a substance of dielectric constant K.
Potential difference between the plates, V = Ed = [Q/A ∈o ] d.
(½ M) Which of the two lines A or B corresponds to scenario
Q? Give a reason for your answer.  (2 M)
Capacitance, C = Q/V = A∈0/d (½ M)
(Un &An) (CBSE SQP, 2023)
(ii)
C1 = 100 pF q A

C23 = 100 pF B

300 V

C4 = 100 pF
V

P
W CBSE Class – XII PHYSICS 84
Ans. (a) V = 300 V E due to a plane sheet of charge = σ/eo is independent of
C = 100 μF the distance from the sheet. Charge density σ on the plate
remains the same because the charge on capacitor plates
Energy = 1/2 CV2
remains the same. So, E also remains the same.
= 1/2 × 100 × 10–6 (300)2
OR
= 4.5 J  (1 M)
As E = V/d = Q/Cd = Q/eo  A
(b) q = CV
Since Q and A remain unchanged, E remains the same.
q = 100 × 10–6 × 300 = 0.03 C (1 M)
(1 M)
(c) Capacitance of a parallel plate capacitor C = (∈0 A)/d
(e) Energy stored in the capacitor increases.
C = 100 μF
Energy stored is proportional to both charge and potential
d′ = 2d difference. Charge is constant but the potential difference
C′ = (∈0 A)/d′ has increased.  (1 M)
C′ = (∈0 A)/2d = 100/2 = 50 μF  (1 M)
Hence, if the distance between the plates of the capacitor Hints & Explanations
is increased two times the capacitance of the capacitor
decreases by 1/2 ie becomes 50 μF. Multiple Choice Questions
(d) The slope of the q vs V graph gives the capacitance of
a parallel plate capacitor.
When the space between the plates of a capacitor is filled 1. (d)
with a substance of dielectric constant K, its capacitance
increases K times. (1 M)
Greater the slope of the q vs V graph, the higher the 2. (a) Energy of a capacitor
capacitance. 1
U = − CV 2
As line A has a greater slope it represents greater capacitance 2
and corresponds to scenario Q. (1 M) 1 1
−2 × 10 −2 J = ⋅ X ⋅ (200) 2 − ⋅ ( 2µF ) ⋅ (200) 2
4. A parallel plate capacitor of capacitance C is charged to 2 2
a potential V by a battery. Q is the charge stored in the
4 × 104
capacitor. −2 × 10 −2 − X − 2µ F
2
The battery is then disconnected, and the distance
between the plates of the capacitor is increased by a small 10–6 = X2μF
amount. What changes will occur in each of the following X = 2μF – 10–6
quantities? Will they increase, decrease or remain the
X = 2μF – 1μF = 1μF
same? Give an explanation in each case.
3. (a) ∴ΔU = ΔQ × V
(a) Capacitance (1 M)
∆U
(b) Charge (1 M) ⇒ ∆Q =
V
(c) Potential difference (1 M)
7.20 × 105
(d) Electric field (1 M) =
(e) Energy stored in the capacitor (1 M) 12
Ans. (a) Capacitance decreases. = 6.0 × 104C
Capacitance is inversely proportional to the distance of 4. (a) The pressing of the key increases the capacitance of
separation.  (1 M) the capacitor below the key due to a decrease in separation
(b) Charge remains the same. between the plates.
The battery is disconnected. So charge cannot move into or 5. (c) Both (i) and (iii) are correct.
out of the plates of the capacitor. (1 M)
Assertion and Reason
(c) Potential difference increases.
1. (a) Both assertion and reason are true and reason is the
As Q = CV correct explanation for assertion.
Charge Q is constant, C decreases, so V increases.  (1 M) 2. (a) Both A and R are true and R is the correct explanation
(d) Electric field remains the same. of A

85 Electrostatic Potential and Capacitance P

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 miscellaneous Exercise
Which of the following statements incorrectly describes the
Multiple Choice Questions work done in moving the unit test charge in the presence of
(1 M)
an electric field in the above context?
1. Given below are the representations of uniform electric and (a) Work done along path I is zero.
gravitational fields.
(b) Total work done along path II and then along B – C – A
A A is zero.
(c) Work done along path II is more than the work done
along path III.
d d
(d) Total work done along path III and then along B – C – A
q m is EQUAL to the total work done along path II and then
B B along B – C – A.
3. An electron is introduced in a region of an electric field. The
charge starts accelerating in the direction opposite to that of
the field.
E g Which of the following statements is true? (Un)
(a) The field does positive work on the electron and its
potential energy increases.
(a) (b) (b) The field does positive work on the electron and its
potential energy decreases.
In fig (a), a positive charge q moves from A to B in the (c) The field does negative work on the electron and its
direction parallel to electric field E. The charge-field system potential energy increases.
undergoes a change in its electrical potential energy.
(d) The field does negative work on the electron and its
In fig (b), a mass particle m moves from A to B in the potential energy decreases.
direction parallel to gravitational field g. The mass-field 4. Some charge is being given to a conductor. Then its potential:
system undergoes a change in gravitational potential energy. (Re)
Identify the statement that correctly states the changes in the (a) Is maximum at surface
energies of the above two systems.  (Un) (b) Is maximum at centre
(a) The charge-field system loses electrical potential energy (c) Is same throughout the conductor
whereas the mass-field system gains gravitational (d) Is maximum somewhere between surface and centre
potential energy. 5. A capacitor of 8µF is charged to a potential of 1000V. The
(b) The charge-field system gains electrical potential energy energy stored in the capacitor is (Ev)
whereas the mass-field system loses gravitational (a) 10 J (b) 12 J
potential energy.
(c) 2 J (d) 4 J
(c) Both the charge-field system and the mass-field system
6. The ratio of electric field and electric potential at midpoint
lose their respective potential energies. of an electric dipole, for separation l is (Ev)
(d) Both the charge-field system and the mass-field system
gain their respective potential energies. (a) 1 (b) 3l

2. In Fig (a), a positive charge Q is located at a point. A unit test
charge q moves along path I in one complete circle around Q. (c) 2 (d) ∞

In Fig (b), II and III represent the paths along which a unit
7. The equivalent capacitance for the given combination is (Ev)
test charge is moved from point A to B in the presence of an
electrostatic field.  (Un)
I II B

Q r εo A
q III (a) 3εo A (b)
d d

A C (c) 4εo A (d) None of these

(a) (b) d

P
W CBSE Class – XII PHYSICS 86
 8. The capacitance of a parallel plate capacitor depends on
Subjective Questions
(Re)
(a) Separation between the plates
Very Short Answer Type Questions (2 M)
(b) Charge
(c) Potential difference across plates 1. Two positive point charges of 18μC and 15 μC are 1m apart.
What is the work done in bringing them 0.4m apart? (Ev)
(d) All of these
2. Define the concept of electric potential and explain how it
9. Amount of work done in moving an electric charge Q1 once
is related to the work done in moving a charge in an electric
round a circle of radius R with a charge Q2 at the center of
field.  (Un)
the circle is (Ev)
3. Describe the behavior of the electric potential due to an
Q1Q 2
(a) (b) ∞ electric dipole and explain how it varies with distance from
4πε o R the dipole.  (Un)
Q1Q 2
(c) (d) Zero Short Answer Type Questions (3 M)
4πεo R 2
10. Energy per unit volume for a capacitor having area A and 1. The potential of large liquid drop when eight liquid drops are
separation d kept at potential difference V is given by (Ev) combined is 20 V. What will be the potential of each drop?
(Ev)
2 1 V2
(a) 1 ε0 V (b) 2. A parallel plate capacitor C with plates of unit area and
2 d2 2ε 0 d 2 separation d is filled with a liquid of dielectric constant K
Q2 = 2 The level of liquid is d/3 initially. Suppose the liquid
(c) 1 CV 2 (d) level decreases at a constant speed v. what will be the time
2 2C
constant as a function of time t? (Un)

Assertion and Reason

(1 M) C
Direction: The following questions consist of two statements. d d R
Assertion (A) and Reason (R). Answer these questions by 3
selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A.
V
(b) Both A and R are true, but R is not the correct explanation of    A.
(c) A is true, but R is false
3 Three capacitors each of capacitance 9pF are connected in
(d) A is false, but R is true.
series: (Un)
1. Assertion (A): The electric potential is constant everywhere
inside a charged conductor and is equal to its value at the (a) What is the total capacitance of the combination? (1.5 M)
surface. (b) What is the potential difference across each capacitor if
the combination is connected to 120 V supply? (1.5 M)
Reason (R): Constant work has to be done to move a test
charge from the interior of a charged conductor to its surface. 4. An electrical technician requires a capacitance of 2µF in a
(Un) circuit across a potential difference of 1 kV. A large number
of 1µF capacitors are available to him, each of which can
2. Assertion (A): The total charge stored in a capacitor is zero. withstand a potential difference of not more than 400 V.
σ Suggest a possible arrangement that requires a minimum
Reason (R): The field just outside the capacitor is (s is
ε0 number of capacitors. (Un)
the charge density). (Re)
5. A 4µF capacitor is charged by a 200 V supply. It is
3. Assertion (A): The electric potential due to a positive charge
then disconnected from the supply and is connected to
is positive at all points in space.
another uncharged 2µF capacitor. How much electrostatic
Reason (R): The electric potential is a scalar quantity that energy of the first capacitor is lost in the form of heat and
represents the potential energy per unit charge at a point in
electromagnetic radiation?  (Ev)
an electric field. (Un)
6. Two parallel plate capacitors, denoted as X and Y, share
4. Assertion (A): The electric potential due to a negative charge
identical plate areas and plate separations. Capacitor X
is negative at all points in space.
has air as the dielectric material between its plates, while
Reason (R): Negative charges have a lower potential energy capacitor Y contains a dielectric medium with a relative
compared to positive charges in an electric field. (Un) permittivity (dielectric constant) of εr = 4. (Ev)

87 Electrostatic Potential and Capacitance P

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 X Y (a) The equivalent capacitance (1 M)
(b) The potential difference across each capacitor (1 M)
(c) The charge on each capacitor (1 M)
(d) If both the capacitors are connected in parallel, then
calculate charge on each capacitor. (2 M)
+ –
OR
15 V (a) How is the electric field due to a charged parallel plate
(a) Determine the capacitance of each capacitor when the capacitor affected when a dielectric slab is inserted
equivalent capacitance of the combined system is 4 μF. between the plates fully occupying the intervening
(1 M) region?  (2 M)
(b) Compute the potential difference across the plates of (b) A slab of material of dielectric constant K has the same area
capacitors X and Y. (1 M) as the plates of a parallel plate capacitor but has thickness
1
(c) Calculate the ratio of the electrostatic energy stored in d , where d is the separation between the plates. Find
capacitor X to that in capacitor Y. (1 M) 2
the expression for the capacitance when the slab is
7. You are given three capacitors of 2μF, 3μF and 4μF
respectively. (Ev) inserted between the plates. (3 M)
2. (a) Find the expression for the electric field intensity and
(a) Form a combination of all these capacitors of equivalent
the electric potential, due to a dipole at a point on the
13
capacitance µF . (1 M) equatorial line.  (3 M)
3
(b) What is the maximum and minimum value of the (b) (i) Define the capacitance of a capacitor.
equivalent capacitance that can be obtained by (ii) A slab of material of dielectric constant has the
connecting these capacitors?  (2 M) same area as the plates of a parallel plate capacitor
but thickness 3d/4. Find the ratio of the capacitance
8. Two charges 2µC and –2µC are placed at points A and B, 6
with dielectric inside it to its capacitance without
cm apart.
dielectric. (2M) (An)
(a) Identify an equipotential surface of the system.  (2 M)
3. Given figure shows a charge array known as an electric
(b) What is the direction of the electric field at every point quadrupole. For a point on the axis of the quadrupole, obtain
on this surface? (1 M)(Un) the dependence of potential on r for r/a>> >1 and contrast
9. Two identical flat metallic surfaces, labelled as A and B, are your results with that due to an electric dipole and an electric
positioned in parallel within the air with a separation of 1.0 cm, monopole (i.e., a single charge).  (Re & Ap)
as depicted in the diagram. Surface A is charged positively
A a B a C
with a potential of 10 V, while the outer surface of B is P
+q –q –q +q
grounded.
B OR
A If two similar large plates, each of area A having surface
charge densities +σ and –σ are separated by a distance d in
cm
X 1.0 Z air, find the expressions for
(a) field at points between the two plates and on outer side
Y of the plates. Specify the direction of the field in each
case. (3 M)
(a) Determine the magnitude and direction of the uniform (b) the potential difference between the plates. (1 M)
electric field existing between points Y and Z. Calculate
(c) the capacitance of the capacitor so formed. (1 M)
the work done in transferring a charge of 20 μC from
point X to point Y.  (2M)(Un) 4. (i) Two capacitors, each having a capacitance of 12 pF, are
(b) Is it possible to have a nonzero electric potential within connected sequentially to a 50 V battery. What is the total
a region where the electric field strength is zero? electrostatic energy stored in this arrangement?  (2 M)
(1 M)
(ii) If they were connected in parallel to the same battery
Long Answer Type Questions  (5 M) instead, what would be the new energy stored in the
combination?  (2 M)
1. Two capacitors of capacitance C1 = 6 m F and C2 = 3 m F
are connected in series across a cell of emf 18 V. Calculate: (iii) Additionally, determine the amount of charge drawn
(Ev) from the battery in both scenarios.  (1 M)
P
W CBSE Class – XII PHYSICS 88
 (ii) In Sarah’s experiment, if the initial capacitance of the
Case Based Questions
capacitor without the dielectric is 100 pF, and the capacitance
Sarah is a physics student conducting an experiment to study the with the dielectric is 300 pF, what is the dielectric constant
behavior of a dielectric slab placed between two parallel plates in a of the material used?
capacitor setup. She sets up the apparatus by connecting the plates
to a power supply and inserting a dielectric slab between them. She (a) 2 (b) 3
measures the capacitance of the system before and after inserting the
(c) 4 (d) ⅓
dielectric slab. A dielectric slab is a substance which does not allow
the flow of charges through it but permits them to exert electrostatic (iii) Which property of a dielectric allows it to permit electrostatic
forces on one another. When a dielectric slab is placed between the forces but not the flow of charges?

plates, the field Eo polarises the dielectric. This induces charge –Q
on the upper surface and +Q on the lower surface of the dielectric. (a) Conductivity (b) Resistance

These induced charges set up a field E p inside the dielectric in the (c) Permittivity (d) Insulation

opposite direction of E o as shown. (An) (iv) What is the charge on the upper surface of the dielectric
when a charge of 5 μC is induced on the lower surface, and
+ + + + + + +Q the dielectric constant is 2?  (1 M)
– – Qp
–
Ep
– – – –
E = E0 – Ep
–
Dielectric Slab (a) 5 μC (b) 10 μC
+ + + + + + + Qp (c) 2.5 μC (d) 20 μC
– – – – – – –Q
OR
(i) What is the primary purpose of inserting a dielectric slab
In a capacitor with a dielectric, if the electric field within the
between the plates of a capacitor in Sarah’s experiment?
dielectric is 500 V/m and the distance between the plates is
(a) To allow the flow of charges
(b) To reduce the electrostatic forces 0.02 m, what is the potential difference across the plates?
(c) To increase the capacitance (a) 5 V (b) 10 V
(d) To study charge flow between plates
(c) 2 V (d) 50 V

Answer keys
Multiple Choice Questions
1. (c) 2. (c) 3. (b) 4. (c) 5. (d) 6. (d) 7. (a) 8. (a) 9. (d) 10. (a)
Assertion and Reason
1. (c) 2. (c) 3. (a) 4. (b)
Case-Based Type Questions
Case Based-I
(i) (c) (ii) (b) (iii) (d)
(iv) (a) OR (b)

89 Electrostatic Potential and Capacitance P

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 Hints & Explanations
3. (a) The assertion is true, and the reason is the correct
Multiple Choice Questions explanation. The electric potential due to a positive charge
is positive at all points in space because it represents the
1. (c) Both the charge-field system and the mass-field system work done in bringing a positive test charge from infinity
lose their respective potential energies. to that point. Since work is done against the repulsive force
2. (c) Work done along path II is more than the work done of the positive charge, the potential is positive.
along path III. 4. (b) The assertion is true, but the reason is not the correct
3. (b) The field does positive work on the electron and its explanation. The electric potential due to a negative charge
potential energy decreases. is negative at all points in space because it represents the
4. (c) Inside the conductor. E = 0. So potential remains same. work done in bringing a positive test charge from infinity to
that point. Since work is done by the attractive force of the
5. (d) U = 1/2 CV2
negative charge, the potential is negative.
6. (d) Potential at mid point due to both negative and positive
charge = 0
Subjective Questions
⸫ V′ = 0
E
∴ ∞ , whatever will be the magnitude of electric field.
= Very Short Answer Type Questions
V'
7. (a) In this case 3 parallel plate capacitor are joined in 1. ∆U = U2 – U1 (1 M)
parallel, q1q 2 qq
=
∆U − 1 2
4πεo r2 4πεo r1

q1q 2  1 1
=
∴ ∆U  0.4 − 1 
4πε o  
⸫ CP = C1 + C2 + C3
∆U = 18 × 15 × 10–12 × 9 × 109 (1.5)
o A o A o A
CP    Work done = Change in Potential energy
d d d
W = ∆U = 3.6 × 10-6 J (1 M)
3 A 2. Electric potential at a point in an electric field is the amount
CP  o
d of work done in bringing a unit positive charge from infinity
o A to that point. It is a scalar quantity and is measured in volts
8. (a) C  (V). The work done in moving a charge q in an electric
d
field from point A to point B is given by the equation
⸫ C∝
1 W = q(VB - VA), where VA and VB are the electric potentials
d at points A and B respectively. (2 M)
9. (d) Electrostatic force is a conservative force so work done 3. The electric potential due to an electric dipole varies
in carrying an electric charge Q, once around a circle is with distance r from the dipole according to the equation
zero. V = (p cosθ)/(4πε₀r²), where p is the magnitude of the dipole
1 1  0 A  2 moment, θ is the angle between the position vector r and
F
CV 2  V the dipole moment vector, and ε₀ is the permittivity of free
10. (a) Energy density  2  2 d  space. The potential is maximum along the axis of the
Ad Ad Ad
dipole and decreases as 1/r² away from the dipole axis.
(2 M)
Assertion and Reason
Short Answer Type Questions
1. (c) A is true but R is false
1. Volume of 8 drops = volume of the bigger drop
2. (c) Charge stored on two plates are +Q and –Q.
= +Q – Q = 0 4 3 4
πr × 8 = πR3 ⇒ 2r = R (1 M)
3 3
qenclosed
=and, φ = ( qenclosed 0) According to the charge conservation, 8q = Q
ε0 q
Potential of one small drop, V ' =
f=0⇒E=0 4πεo r

P
W CBSE Class – XII PHYSICS 90
 Q As n cannot be a fraction, we must take n = 3. If C0 is
Similarly potential of big drop V = 4πε R C
o capacitance of each capacitor, the capacitance of a row 0
V' q R n
= × As m rows are connected in parallel, net capacitance
V Q r
V ' q 2r mC0
= × C=
20 8q r n
20 Given, C = 2 µF and C0 = 1 µF, = n = 3
=
V = 5V
4 (2 M) m × (1µF ) 2×3
\ 2 µF = or m = =6
2. At any instant t, level of the liquid = d/3 – vt 3 1
Minimum number of capacitors, N = mn = 3 × 6 = 18
The arrangement can be regarded as a series combination of
(1 M)
two capacitances with A = unit area.

∴ C1 = 0
ε 5. Given, C1 = 4µF = 4 × 10−6 F , V1 = 200 V
2d (1 M)
+ vt Initial energy of first capacitor,
3
2ε0 1 1
and C2 = U1 = C1V12 = × (4 × 10−6 ) × (200) 2 =8 × 10−2 J
d 2 2
− vt
3 (1 M)
2d When another uncharged capacitor C2 = 2µF, is connected
C1 K=1 + vt across first capacitor,
3
d Common potential,
C1 K=2 − vt
3
q1 + q2 C1V1 + 0 4 × 10−6 × 200 400
=V = = = volt
2d
+ vt
d
− vt C1 + C2 C1 + C2 (4 + 2) × 10−6 3
1 1 1
= + = 3 + 3
Ceq C1 C2 ε0 2ε0 (1 M)
Final energy,
6ε 0 (1 M)
or Ceq = 2
3vt + 5d 1 1  400 
U 2 = (C1 + C2 )V 2 = × (4 + 2) × 10−6 ×  
2 2  3 
6ε0 R
=
Time constant, τ RC
= (1 M) 16
= × 10−2 J =5 ⋅ 33 × 10−2 J
eq
3vt + 5d
3
3. Given C1 = C2 = C3 = 9pF Energy loss,
When capacitors are connected in series, the equivalent ∆U =U1 − U 2 =8 × 10−2 − 5.33 × 10−2 =2.67 × 10−2 J
capacitance CS is given by
(1 M)
1 1 1 1
= + + (1 M) 6. Let the area of the plates be A and distance between the
CS C1 C2 C3
plates be d
1 1 1 1 3 1
= + + = = ∴ Capacitance of capacitor X
CS 9 9 9 9 3
A∈
CS = 3pF (½ M) C1 = 0

d
(b) In series, charge on each capacitor remains the same, so
charge on each capacitor Also, capacitance of capacitor Y
q = CSV = (3 × 10 F) × (120 V) = 3.6 × 10 coulomb
–12 –10
∈r A ∈0 4 A ∈0
= C2 = = 4C1
(1 M) d d
Potential difference across each capacitor,
(a) Equivalent capacitance in series combination,
q 3.6 × 10−10
=
V = = 40V (½ M)
C1 9 × 10−12 C × C2 (½ M)
Ceq = 1
4. The only way to raise the potential difference is by linking C 1 + C2
capacitors in series. To achieve this configuration, let’s
C × 4C1
consider having m rows connected in parallel, with each ∴ 4µF = 1
row having n capacitors connected in series. In this setup, C1 + 4C1
the total number of capacitors N = m × n (1 M) 4C
4µF = 1 ⇒ C1 =5µF
V 1kV 1000 V 5
V = nV0, i.e, n== = = 2.5
V0 400 V 400 V ∴ C2 = 4 × 5µF = 20µF (½ M)
(1 M)

91 Electrostatic Potential and Capacitance P

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 (b) Consider the total charge passing through the circuit C1 × C2 × C3
as “Q.” Ceq =
C1C2 + C2 C3 + C3C1
∴Q=C V i.e.
2 × 3× 4
eq =
2 × 3 + 3× 4 + 4 × 2
∴ Q = 4 µF × 15 V = 60mC (½ M)
12
Ceq= C min= µF (1 M)
Potential drop across capacitor X 13

Q 60µC
=
V = = 12 V
5µF
1
C1
Mistakes 101 : What not to do!
Potential drop across capacitor Y ● Students may fail to add the capacitance values together
when capacitors are connected in parallel.
Q 60µC ● Students may neglect to take the reciprocal of the sum of
=
V2 = = 3V (½ M)
C2 20µF the reciprocals of the capacitance values when capacitors
are connected in series.
(c) Electrostatic energy stored in a capacitor ● Students may omit the units (μF) in the final answer.

Q2
E= (½ M) 8. (a) Let P(x, y) be a point on zero potential surface. Let A
2C
(location of charge q = 2µC) be origin of coordinate system.
Q2 Distance,=
r1 x2 + y 2 ,
E1 2C1
C 20µF
∴ = 2 = 2 = =4 (½ M) Distance, r2 = (d − x) 2 + y 2
E2 Q C1 5µF
where d = 6 cm = 6 × 10–2 m.
2C2

y
Nailing the Right Answer
Remember to carefully consider the effect of the dielectric
material and constant, use the correct formulas, and
(1 M)
double-check calculations to avoid these common errors.
Potential at P due to charges, q1 = +2µC
7. (a) When C1 and C3 are in series and C2 is in parallel as and q2 = –2µC is given by
shown, then 1 q1 1 q2
V= + =0
4πε 0 r1 4πε 0 r2
C1 = 2mF C3 = 4mF 1 2 × 10−6 1 (−2 × 10−6 )
⇒ + =
0
4πε 0 x + y
2 2 4πε0 (d − x) 2 + y 2
1 1
A B ⇒ =
x2 + y 2 (d − x) 2 + y 2
⇒ x 2 + y 2 = (d − x) 2 + y 2
C2 = 3mF
d
⇒x= = 3cm 
2× 4 4 13 2
Ceq = +3= +3 = µF (1 M) So, plane passing through mid point of line joining A and B
2+4 3 3
has zero potential everywhere. (1 M)
(b) Maximum value of the equivalent capacitance is
(b) The direction of electric field is normal to surface PCQ
obtained when all capacitors are connected in parallel
everywhere as shown in figure 
combination,

i.e. Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9µF P

or Ceq = Cmax = 9µF (1 M) C E
and minimum value is obtained, when all capacitors are (1 M)
Q
connected in series combination,
P
W CBSE Class – XII PHYSICS 92
 After introduction of dielectric; the permittivity of medium
becomes Ke0.
Mistakes 101 : What not to do! so, final electric field between the plates of parallel plate
● Incorrect identification of the equipotential surface of the capacitor= E =
q E
system. AK ε K
1
● Misinterpretation of the direction of the electric field on the i.e., electric field reduces to times (1 M)
identified surface. K
(b) Consider a parallel plate capacitor, area of each plate
being A, the separation between the plates being d. Let a
dV 10V dielectric slab of dielectric constant K and thickness t < d
9. (a) Since =
E = = 1000V (1 M) be placed between the plates. The thickness of air between
dr 1 × 10 −2
since surface A is an equipotential surface i.e., DV = 0 the plates is (d – t). If charges on plates are +Q and –Q, then
surface charge density
∴ work done from X to Y= zero. (1 M) Q
dv σ=  (½ M)
(b) E = − A
dr The electric field between the plates in air,
dV σ Q
= 0=
; dV 0=
or V constant =
E =
dr ε0 ε0 A
So, we can have non-zero electric potential, where electric The electric field between the plates in the slab,
+Q –Q
field is zero.  (1 M) A B
Long Answer Type Questions K
1 1 1
=
1. (a) C +
C1 C2
C1C2 6×3
⇒C = = =2µF (1 M)
C1 + C2 6 + 3
(b) Q = CeqV
Substituting the values, we get t
Q = 2 mF × 18 V = 36 mC d (½ M)
V1 = Q/C1 = 36mC/6mF = 6 V σ Q
=
E2 = (½ M)
K ε0 K ε0 A
V2 = Q/C2 = 36mC/3mF = 12 V (1 M)
∴ The potential difference between the plates
(c) When capacitors are connected in series, the magnitude
of charge Q on each capacitor is the same. The charge on VAB = work done in carrying unit positive charge from one
plate to another
each capacitor will equal the charge supplied by the battery.
Thus, each capacitor will have a charge of 36 mC. (1 M) = ∑ E (as field between the plates is not constant)
Q Q
(d) Ceq = C1 + C2 = 6 + 3 = 9mF = E1 (d − t ) + E2=
t (d − t ) + t
ε0 A K ε0 A
As the capacitors are connected in parallel, therefore,
potential difference across each capacitor is same. Q  t 
∴=
VAB d −t + 
ε0 A  K (½ M)
∴Q1 = C1
∴ Capacitance of capacitor,
C 6 1
∴ Q1 = 1 = = µC =
C =
Q Q
V1 18 3 VAB Q  t 
d −t + 
C 3 1 ε0 A  K
∴ Q2 = 2 = = µC (2 M)
V2 18 6 ε0 A ε0 A
= =
or, C (½ M)
OR d −t +
t  1
d − t 1 − 
K  K
(a) Initial electric field between the plates of parallel plate ε0 A ε0 A
=
∴C =  d
σ q /A Here, t =   (½ M)
capacitor E= = =
q
(1 M) d 1 d 1   2
0
ε0 ε0 Aε0 d − 1 −  1 + 
2 K 2 K

93 Electrostatic Potential and Capacitance P

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 Where, p = 2qa, is the electric dipole moment. If
the point P is located far away from the dipole,
Nailing the Right Answer r >> a, then
 1 p
● Use the formula for calculating equivalent capacitance in a Eequa =− ⋅ 3 pˆ (1 M)
series circuit, recognize that the potential difference across 4πε0 r
each capacitor is the same, and use the given emf value to The electric potential due to the positive charge (+q) at
find the charge on each capacitor. point P is:
P
2. (a) Consider an electric dipole placed in a vacuum
consisting of charges -q and +q which are separated by a
distance 2a. Let P be a point on equatorial line which is at r 2 + a2 r 2 + a2
distance r from it.
r
OP = r
Now finding electric field at point P due to charge +q and a a
–q are –q +q
A O B
 1 q
=
E+ q ⋅ 2 , directed along BP 1 q
4πε0 r + a 2 V1 =
4πε0r + a2 2

The electric potential due to the negative charge (–q) at

 1 q
=
E− q ⋅ 2 , directed along PA point P is:
4πε0 r + a 2
1 −q
 V2 =
E+q 4πε0 r + a2 2

 Furthermore, these potentials possess equal magnitudes but

Eequaq differ in sign. Consequently, the total electric potential at
P
q point P is the result of adding these potentials together:

1 1  q q 
E–q V =V1 + V2 =  − 
(r 2
+ a2 ) 2 
4πε0  r + a 
r
2 2
r + a2
2

Therefore V = 0 (1 M)
q a O q
(b) (i) The amount of charge required on each plate to
–q +q
A a B increase the potential difference between them by unity is
P (1 M) the capacitance of the capacitor. Capacitance of a air filled
Since magnitude of both the charges will be equal. therefore: Aε 0
parallel plate capacitor = (1 M)
1 q d
E=
−q E=
+q ⋅ 2 (ii) Capacitance without dielectric,
4πε0 r + a 2
Aε
The components of E(+q) and E(–q) which are parallel to C= 0
the dipole axis will add up and the components which are d
perpendicular will cancel out. Total electric field Capacitance when filled with dielectric having thickness

Eequa = −( E− q cosθ + E+ q cosθ) pˆ 3d
4
−2 E− q cosθpˆ  E− q =
= E+ q  Aε 0 Aε 0 4ε 0 kA
C= = =
 t  3d 3d  d (k + 3)
1 q a d −t +  d − + 
=−2 ⋅ ⋅ pˆ  k   4 4k 
4πε0 r + a
2 2
r + a2
2
C' Aε 0 4k d 4k
  =
Ratio × = (1 M)
 cosθ =
a
 C d (k + 3) Aε 0 (k + 3)
 r 2 + a 2  3. Given, AC = 2a, BP = r
The negative sign indicates that total electric field at AP = r + a and PC = r – a
equatorial point is opposite in the direction of the dipole. A a B a C
 P
1 p +q –q –q +q
Eequa =
− ⋅ 2 pˆ
4πε0 (r + a 2 )3/2 r (1 M)

P
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 The potential at P is V. σ
(a) Magnitude of electric field intensities E=
1 E=
2
\V = Potential at P due to A+ Potential at P due to B + 2ε0
Potential at P due to C +s –s
+ –
1  q 2q q    
= − + (1 M)  
4πε0  AP BP CP 
 E1 E2
+ E1 – E2 E1


1  1 2 1  + E2 –
= ⋅q − + 
4πε0  ( r + a ) r ( r − a )  I II III
+ –
q  r (r − a ) − 2(r + a )(r − a ) + r (r + a ) 
=
4πε0  r (r + a )(r − a )  Plate – 1 Plate – 2
 
(i) In region I (outside)
q  r − ra − 2r + 2a + r + ra 
2 2 2 2
σ σ
=   EI = E2 − E1 = − = 0 (1 M)
4πε0  r (r 2 − a 2 )  2ε 0 2ε 0
(ii) In region II (inside)
q ⋅ 2a 2 q ⋅ 2a 2 σ σ σ
= = EII = E1 + E2 = + = (1 M)
4πε0 r (r 2 − a 2 )  a2  2ε o 2 ε o ε o
4πε0 .r ⋅ r 2 1 − 2 
(1 M)
 r  (iii) In region III (outside)
σ σ
According to the question, EIII = E1 − E2 = − =0 (1 M)
2ε 0 2ε 0
r q ⋅ 2a 2
If >> 1, a << r. Therefore , V = In the region II i.e., in the space between the plates, resultant
a 4πε0 ⋅ r 3 
electric field EII is directed normal to plates, from positive
1
V∝ to negative charge plate.
r3 (b) The potential difference between the plates is
σ Q
As, we know that electric potential at a point on axial line V = EII ⋅ d = d or V = d (1 M)
ε0 Aε0
due to an electric dipole is
(c) Capacitance of the capacitor so formed is
V∝
1 Q Q ε A
= =
C or C= 0 (1 M)
r2 V Qd / Aε0 d
1
In case of electric monopole, V ∝ (1 M) 4. Cs = C2 = 12pF, V = 50 V
r
Then, we conclude that for larger r, the electric potential (i) In series combination:
due to a quadrupole is inversely proportional to the cube of 12mF 12mF
the distance r, while due to an electric dipole, it is inversely
proportional to the square of r and inversely proportional to
the distance r for a monopole (1 M)

OR
Capacitor is based on the principle of electrostatic 50 V
induction. The capacitance of an insulated conductor 1  1 1
=  + 
increases significantly by bringing an uncharged earthed CS  12 12 
conductor near to it. This combination forms parallel plate
∴ Cs = 6 × 10–12 F (1 M)
capacitor.
1
Area Us = CV 2
+s –s 2
+ A –
1
Us = × 6 × 10 −12 × 50 × 50
+ – 2
+ – qs = Cs V
+ – ∵ Us = 75 × 10–10 J
d
qs = 6 × 10–12 × 50
= 300 × 10–12 C (1 M)

95 Electrostatic Potential and Capacitance P

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 (ii) Parallel combination then the potential difference is reduced across the capacitor
12pF plates, In this way, dielectric increases the capacitance of
the capacitor. (1 M)
ε0 A
(ii) (b) c =
d
ε A
12pF 100 = 0
d
kε A
50 V c′ = 0
d
Cp = (12 + 12) pF
 ε A
300 = k  0 
∴Cp = 24 × 10–12 F (1 M)  d 
1 300 = k × 100
Vs = × 24 × 10 −12 × 50 × 50
2 k = 3 (1 M)
= 3 × 10–8 J (1 M) (iii) (d) Insulation (1 M)
(iv) (a) The charge induced in the lower surface in the presence
(iii) qp = CpV
of dielectric will be same as that is upper surface. (1 M)
qp = 24 × 10–12 × 50
OR
= 1.2 × 10–9 C  (1 M)
(b) d = 0.02 m, E = 500 V/m
V
Case Based Questions E= ,
d

(i) 
(c) Inserting a dielectric increases the capacitance. The V=E×d
strength of the electric field is reduced due to the presence V = 500 × 0.02 =10 V (1 M)
of dielectric. If the total charge on the plates is kept constant,

P
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 Solutions
1
“Solutions play the vital role in our daily lives and
various industries. From the beverages we drink to
the medications we take, solutions are an integral
part of many processes and products. They are
essential in fields such as medicine, where saline
solutions are used for intravenous drips, and in
the automotive industry, where antifreeze solutions
help regulate engine temperatures. Understanding
solutions also aids in environmental science, helping
us address issues like water pollution and its impact
on ecosystems.”

SYLLABUS &
WEIGHTAGE

Years
List of Concept Names 2022
2020 2023 2024 2025
(Term-I)
Types of Solutions and Expressing Concentration
— — — — —
of Solutions

Solubility 1 Q (2 M) 1 Q* — 1 Q (2 M)* —

Vapour Pressure of Liquid Solutions & Ideal and

1 Q (1 M)
Non-Ideal Solutions — 3 Q* 1 Q (2 M) —
1 Q (2 M)
(Raoult’s Law)
Colligative Properties of Solutions and Abnormal
Molar Mass
(Relative Lowering of Vapour Pressure, 2Q
1 Q (2 M) 1 Q (1 M) 1 Q (3 M)
Elevation Of Boiling Point, Depression Of Freezing 4 Q* (1 M Each)
1 Q (3 M) 1 Q (3 M) 1 Q (4 M)
Point, Osmotic Pressure, Determination Of Molecular 1 Q (3 M)
Masses Using Colligative Properties, Abnormal
Molecular Mass, Van’t Hoff Factor)

* All questions were of MCQ type and carried equal marks.

* The mark allotment mentioned above includes internal choice questions as well.
 P
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Types Expressing Concentration of
Solutions
Gaseous
Gas-Gas → mixture of O2 and N2
Liquid-Gas → Chloroform with N2
● Mass by volume percentage (w/V): ● Normality: ● Molality:
Solid-Gas → Camphor in N2
Mass of solute No. of gram equivalent of solute × 1000 No. of moles of solute × 1000
× 100
Liquid Volume of solution Volume of solution (in mL) Mass of solvent (in g)

CBSE Class – XII CHEMISTRY

Gas-Liquid → O2 dissolved in water ● Volume percentage (V/V) Gram Equivalents of solute: ● Molarity:
Liquid-Liquid → Ethanol dissolved in water Volume of component Mass of solute
× 100 = No. of moles of solute × 1000
Solid-Liquid → Glucose dissolved in water Total volume of solution Equivalent weight
Volume of solution (in mL)
● Mass percentage (w/w) Equivalent mass of an element:
Solid Mass of component in solution Atomic mass
Gas-Solid → H2 in Pd × 100 =
Total mass of solution valency
CONCEPT MAP

Liquid-Solid → Amalgam of Hg with Na ● Parts per million: For trace quantities ● Mole Fraction:
Solid-Solid → Cu dissolved in gold
No. of parts of components × 106 No. of moles of components
Total no. of parts of all components of solution Total no. of moles of all components

Solubility
SOLUTIONS
Henry’s Law

20
Solid in liquid Partial pressure of gas in vapour Ideal and Non-ideal
phase (p) is proportional to the Solutions
● Endothermic, Δsol H > 0, Solubility
Gas in liquid mole fraction of gas (χ) in the
Effect of increases with increasing T. Increases with solution.
temperature ● Exothermic, Δsol H < 0, Solubility Effect of decrease in p = KHχ
temperature Ideal Solutions
decreases with increasing T. temperature KH → Henry’s constant
● Higher the value of KH, ● Obey Raoult’s law
Effect of Increases with lower the solubility. ● ΔHmix = 0, ΔVmix = 0
● Not significant effect Effect of
Pressure increase in ● Interactions A–A and B–B =
Pressure pressure A–B
● Ex: n-hexane and n-heptane
Azeotropes Ethyl bromide + Ethyl
The mixtures of liquids which boil at constant temperature like a pure liquid and possess same composition of components in chloride,
liquid as well as binary vapour phase are called constant boiling mixtures or azeotropic mixtures. Chlorobenzene + Bromo-
● Minimum boiling azeotropes: They are formed by those liquid pairs which show positive deviations from Raoult’s law. benzene, etc.
Ex: Ethanol-Water mixture.
● Maximum boiling azeotropes: They are formed by those liquid pairs which show negative deviations from Raoult’s law.
Ex: Nitric Acid-Water mixture.
Non-Ideal Solutions
To Access One

● Do not Obey Raoult’s law

Scan This QR Code
Shot Revision Video

Positive Deviation Negative Deviation

● ΔHmix > 0 ; ΔVmix > 0 ● ΔHmix < 0 ; ΔVmix < 0 ● ΔHmix and ΔVmix ≠ 0
● Interactions A–B < A–A or B–B ● Interactions A–B > A–A or B–B
● Ex: Acetone + Ethyl alcohol, Water + Ethyl alcohol, Carbon tetrachloride ● Ex: Acetone + Aniline, Acetone + Chloroform, HNO3 +
+ Chloroform, Chloroform + Ethanol, Acetone + Carbon disulphide, Water, Benzene + Chloroform, Pyridine + Glacial acetic
Acetone + Benzene, etc. 20 acid, etc.
 SOLUTIONS

Colligative Properties
Abnormal Molecular Mass

Definition Molar mass different from expected value

● Properties which depend on the number of solute particles and not on their Normal molar mass
van‛t Hoff factor (i) =
nature. Abnormal molar mass
Observed Colligative Property
=
Calculated Colligative Property
Types
i = 0 Solute undergoes neither association nor dissoci-
n2 P° − P ation in solution.
● Relative lowering of vapour pressure → χ2 = = 1 ° 1
n1 + n 2 P1 i < 1 Solute undergoes association in solution.
● Osmotic pressure → p = CRT i > 1 Solute undergoes dissociation in solution.
K b × 1000 × W2
● Elevation of boiling point → DTb = Kb × m =
M 2 × W1 Vapour Pressure of Liquid
DTb = Tb – T°b Solutions
K f × W2 × 1000
● Depression in freezing point → DTf = Kf × m =
M 2 × W1 Liquid-Liquid Solutions (Raoult’s Law)
DTf = T°f – Tf ● Raoult’s law: P1 = P°1 χ

21
Boiling point of ● Dalton’s law of partial pressure: PTotal = P1 + P2 +
Solvent
Solution ent
.......... Pn
1.013 bar solv
or 1 Atm id ● In vapour phase, Pi = yi PTotal
liqu
nt tion
nt solu
lve
So tion lve
lu so
So n III p°
ze 2
fro + p2
= p1
p total II
DTf
p°1 p2

Vapour pressure
Vapour pressure
DTb
T b° Tb Tf T f°
Temperature/K Temperature/K p1
Vapour pressure

x1 = 1 Mole fraction x1 = 0
Osmotic Pressure x2 = 0 x2 x2 = 1
● Osmosis: Flow of solvent molecules from pure solvent to solution through a Solids in Liquid Solutions
semi permeable membrane. Vapour pressure decreases of the solvent in the presence
● Osmotic pressure: Excess pressure that must applied to a solution to prevent of solute.
osmosis.
● Isotonic Solutions: Solutions with same osmotic pressure.
● Hypertonic Solution: In such solutions, water flow out of cell, and cell shrinks.
● Hypotonic Solutions: In such solutions, water flow into the cells and the cell
swells. 1 mol of Solvent
Pure Solvent
● Reverse Osmosis: The process of movement of solvent through a semi- 1 mol of Solute
permeable membrane from the solution to pure solvent by applying pressure
Solute

Solutions
more than osmotic pressure on the solution side is called reverse osmosis. Solvent

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 TYPES OF SOLUTIONS AND EXPRESSING CONCEN-
1 TRATION OF SOLUTIONS

NCERT Definitions (Commonly asked in 1 mark)

U Solution: A homogeneous mixture of two or more than two substances.

U Solvent: The component of a solution present in the largest amount.
U Solute: The component of a solution present in a lesser amount than the solvent.
U Gaseous Solutions: Solution where the solvent is the gas. For example, Mixture of oxygen and nitrogen gases.
U Liquid Solutions: Solutions where the solvent is a liquid. They can contain solid, liquid, or gas as solutes. For example, Oxygen
dissolved in water.
U Solid Solutions: Solutions where the solvent is a solid. These are typically mixtures of metals, known as alloys, like brass
(Copper and Zinc).
U Molarity (M): A concentration measure defined as the number of moles of solute per litre of solution.
U Molality (m): A concentration measure defined as the number of moles of solute per kilogram of solvent.
U Mole Fraction: A concentration measure defined as the ratio of the number of moles of a component to the total number of
moles of all the components present in the solution.

Important Facts

01 Mass percentage is commonly used in industrial chemical applications.

02 Volume percentage is often used for solutions containing liquids.

03 Mass by volume percentage is frequently used in medicine and pharmacy.

04 A concentration term used when solute is present in trace quantities. ~Parts per million

05 Parts per million (ppm) is used to measure concentration of pollutants in water and air.

06 Mass %, ppm, mole fraction and molality are independent of temperature.

07 A concentration term which is a function of temperature. ~Molarity

P
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 Important Formulae

1. Mass percentage (w/w): Mass of the component in the solution

Mass % of a component = × 100
Total mass of the solution

2. Volume percentage (V/V): Volume of the component

Volume % of a component = × 100
Total volume of solution

3. Parts per million (ppm): Number of parts of the component

× 106
Total number of parts of all components of the solution

4. Mole fraction of a component (χ): Number of moles of the component

Total number of moles of all the components

5. Molarity (M): Moles of solute

Volume of solution in litre

6. Molality (m): Moles of solute

Mass of solvent in kg

Classification

Types of Solutions (On the basis of physical states of solute and solvent)

Type of Solution Solute Solvent Common Examples

Gas Gas Mixture of oxygen and nitrogen gases

Gaseous Solutions Liquid Gas Chloroform mixed with nitrogen gas
Solid Gas Camphor in nitrogen gas
Gas Liquid Oxygen dissolved in water
Liquid Solutions Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water
Gas Solid Solution of hydrogen in palladium
Solid Solutions Liquid Solid Amalgam of mercury with sodium
Solid Solid Copper dissolved in gold

Real Life Application Based Questions

1. Explain how the concentration of fluoride in drinking water is an application of the parts per million (ppm) concept.
Ans. Fluoride concentration in drinking water is often kept around 1 ppm to prevent tooth decay, while 1.5 ppm causes the tooth to
become mottled and high concentrations of fluoride ions can be poisonous. This demonstrates how small amounts of a solute can
have significant health benefits.
2. In the food industry, why is the concentration of salt in brine solutions expressed in mass percentage (w/w)?
Ans. Expressing salt concentration in mass percentage ensures uniformity in preserving food. It helps in achieving the right osmotic
balance to inhibit microbial growth, thereby extending the shelf life of the product.

23 Solutions P
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 3. Why is the concentration of pollutants in the atmosphere often measured in parts per million (ppm)?
Ans. Measuring pollutants in ppm allows for precise monitoring of air quality. It helps in assessing the impact of pollutants on health
and the environment, even when they are present in very low concentrations.
4. What is the purpose of addition of antifreeze to car engines?
Ans. A 35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At this concentration, the antifreeze
lowers the freezing point of water to 255.4K (–17.6°C).

Myth Buster

U Myth: All solutions are liquid.

Fact: Solutions can be in any state of matter – solid, liquid, or gas.
U Myth: A solute is always a solid.
Fact: A solute can be a gas, liquid, or solid. For instance, oxygen gas dissolves in water to form a liquid solution, while ethanol
(a liquid) can dissolve in water as well.
U Myth: Molarity and molality are the same.
Fact: Molarity is the number of moles of solute per litre of solution, while molality is the number of moles of solute per kilogram
of solvent. Molarity depends on the volume of the solution, which can change with temperature, while molality depends on the
mass of the solvent, which is temperature-independent.
U Myth: The terms ‘dilute’ and ‘concentrated’ are quantitative measures of concentration.
Fact: ‘Dilute’ and ‘concentrated’ are qualitative descriptions that indicate relatively low or high amounts of solute, respectively.
They do not provide specific numerical values for concentration and can vary based on context and interpretation.

P
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 competency BASED SOlved Examples
(1 M) 1. Assertion (A): Camphor dissolved in nitrogen gas is an
Multiple Choice Questions example of a solid solution.
1. Which of the following statements is true regarding the Reason (R): Solid solutions are homogeneous mixtures
characteristics of solutions? (Un) where the solute is uniformly distributed within the solid
solvent. (Re)
(a) Solutions show varying composition and properties 2. Assertion (A): The molality of the solution change with
throughout the mixture. change in temperature.
(b) The solute determines the physical state of the solution.
Reason (R): The molality is expressed in units of moles per
(c) Solutions are heterogeneous mixtures of two or more 1000 gm of solvent. (Un)
components.
3. Assertion (A): Percent composition by mass is a way to
(d) Solutions have uniform composition and properties express the concentration of a solution.
throughout the mixture.
Reason (R): Percent composition by mass is calculated
2. If a student accidentally mixes oil with water while as the mass of the solute divided by the total mass of the
preparing a homogenous solution for an experiment, solvent, multiplied by 100. (Un)
what problem will they encounter and why? (Un)
(a) The solution will be homogenous because oil dissolves Subjective Questions
in water.
(b) The solution will be heterogeneous because oil does not Very Short Answer Type Questions  (2 M)
dissolve in water.
(c) The solution will turn into a solid solution due to the 1. Give example of the following:
density of oil. (a) Gaseous solution where, solute is liquid and solvent is gas
(d) No problem will be encountered as all liquids mix (b) Solid Solution where, solute is solid and solvent is solid
uniformly. (Un)
3. A glycerine solution, at 293 K, has a molality of 3.89 molal Ans. (a) Chloroform mixed with nitrogen gas (1 M)
and molarity of 5.33 M. (b) Copper dissolved in gold (1 M)
Which of these would be CORRECT for molarity and
2. Calculate the mass percentage of benzene (C6H6) and
molality of the same glycerine solution at 450K? carbon tetrachloride (CCl4) if 22 g of benzene is dissolved
(Un) (CBSE CFPQ,2024) in 122 g of carbon tetrachloride. (Ev) (NCERT Intext)
(a) Molarity < 5.33 M; Molality = 3.89 molal Ans. Given:
(b) Molarity < 5.33 M; Molality < 3.89 molal Mass of solute, benzene = 22g
(c) Molarity > 5.33 M; Molality = 3.89 molal Mass of solvent CCl4=122g
(d) Molarity = 5.33 M; Molality = 3.89 molal So, total mass of solution = 22g + 122g =144g
4. 4L of 0.02 M aqueous solution of NaCl was diluted by Mass % of benzene = (Mass of benzene/Total mass of
adding one litre of water. The molality of the resultant solution) × 100 (½ M)
solution is __________. (Un) (NCERT Exemplar) = 22/144 × 100 = 15.28 % (½ M)
(a) 0.004 (b) 0.008 Mass % of carbon tetrachloride
(c) 0.012 (d) 0.016 = (Mass of Carbon tetrachloride/Total mass of solution)
× 100 (½ M)
Assertion and Reason  (1 M) = 122/144 × 100 = 84.72 % (½ M)
Direction: The following questions consist of two statements 3. Calculate the moles and molarity of a solution containing
– Assertion (A) and Reason (R). Answer these questions by 5 g of NaOH in 450 mL solution. (An)
selecting the appropriate option given below: 5g
Ans. Moles of NaOH = = 0.125 mol (1 M)
(a) Both A and R are true, and R is the correct explanation of A. 40g mol−1
(b) Both A and R are true, but R is not the correct explanation
Volume of the solution in litres = 450mL/1000mL
of A.
(c) A is true, but R is false. n 0.125 mol × 1000 mL L−1
= Molarity = = 0.278M (1 M)
(d) A is false, but R is true. v 450 mL

25 Solutions P
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 4. A solution is prepared by dissolving 10 grams of solute X 2. Calculate (a) molality (b) molarity and (c) mole fraction
in 500 mL of solvent Y. The molar mass of solute X is 50 g of KI if the density of 20% (mass/mass) aqueous KI is
mol, and the density of solvent Y is 0.8 g/mL. Calculate 1.202g mL–1. (Ev) (NCERT Intext)
the molality of the solution. (Ev) Ans. 20% aq. KI solution ⇒ 20g of KI in 100g solution
Ans. (c) Mass of solute (m) = 10 g ∴ Mass of solvent = 100 − 20 = 80 g
Volume of solvent (V) = 500 mL = 0.5 L Moles of KI = Weight of KI/ Molar mass of KI = 20/166
Molar mass of solute (M) = 50 g/mol = 0.120 moles
Moles of solute (n) = mass of solute / molar mass of solute (i) Molality = no. of moles of KI / mass of solvent (kg)
= 10 g/50 g/mol = 0.2 mol = 0.120 / 0.080 = 1.5 mol kg−1 (1 M)
Mass of solvent (m) = volume of solvent × density (ii) Density of solution = 1.202g mL−1
= 500 mL × 0.8 g/mL = 400 g = 0.4 kg (1 M)
Volume of solution =100/1.202 = 83.2mL = 0.0832L
Molality (m) = moles of solute/mass of solvent in kg
∴ Molarity = 0.120 / 0.0832 =1.44 M (1 M)
= 0.2 mol/(0.4 kg) = 0.5 mol/kg = 0.5 m (1 M)
(iii) No. of moles of KI = 0.120
Short Answer Type Questions (3 M) nH = 80/18 = 4.44
2O
1. A solution of glucose in water is labelled as 10% w/w, xKI = 0.120/(0.120 + 4.44) = 0.120/4.560 = 0.0263
what would be the molality and mole fraction of each (1 M)
component in the solution? If the density of solution
is 1.2g mL–1, then what shall be the molarity of the Long Answer Type Questions (5 M)
solution? (Ev) (NCERT)
1. (a) D
 efine the term solution. How many types of solutions
Ans. 10% w/w solution of glucose in water means that 10 g of are formed? Write briefly about each type with an
glucose is present in 100 g of the solution and 90 g of water. example. (NCERT)
Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × (b) Give an example of a solid solution in which the solute
16 = 180 g mol–1 is a gas. (Un)
Then, number of moles of glucose = 10g/180g mol–1 = 0.056 mol Ans. (a) A solution is a homogeneous mixture composed of two
● Molality of solution = n/Wwater(kg) = 0.056 mol/0.09kg or more substances. In such a mixture, a solute is a substance
= 0.62 m (1 M) dissolved in another substance, known as a solvent. The
concentration of a solute in a solution can vary. (½ M)
Number of moles of water = 90g/18g mol–1 = 5 mol
Solutions can be formed with different phases of matter
Mole fraction of glucose = 0.056/(0.056 + 5) = 0.011 (solid, liquid, and gas). There are primarily three types of
(½ M) solutions based on the physical state of the solvent: (½ M)
And, mole fraction of water = 1 – mole fraction of glucose 1. G
 aseous Solutions: These are solutions where the
= 1 – 0.011 = 0.989 (½ M) solvent is a gas. An example is mixture of oxygen and
Density of the solution is 1.2g mL–1, then volume of the nitrogen gases. (1 M)
100 g solution can be given as: 2. Liquid
 Solutions: These are the most common types
= 100g/1.2g mL–1 of solutions, where the solvent is a liquid. They can
= 83.33 mL involve solutes of gas, liquid, or solid. An example is
a saline solution, where salt (sodium chloride) is the
= 83.33 × 10–3 L (½ M) solute dissolved in water, the solvent. (1 M)
n
● Molarity of the solution = = 0.056 mol/83.33 × 10–3 L 3. S
 olid Solutions: In these solutions, the solvent is solid.
V(L) An example is alloys like brass, where zinc is dissolved
= 0.67 M (½ M)
in copper. (1 M)
(b) Solution of hydrogen(solute) in palladium (solid) is an
example of solid solution with solute is a hydrogen gas.
Key Takeaways (1 M)
This question teaches students that the concepts of
concentration units such as weight/weight percentage,
molality, mole fraction, and molarity are crucial. The Nailing the Right Answer
formulas used are:
U To ensure students get full credit, they should begin by
Molality (m) = moles of solute / mass of solvent (kg)
clearly defining a solution as a homogeneous mixture
Mole fraction (χ) = moles of component / total moles of of two or more substances. Next, clearly explain the
all components three types of solutions with examples: gaseous, liquid,
Molarity (M) = moles of solute / volume of solution (L) and solid. Highlight important words by underlining.

P
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 2. This question contains four parts. Answer all of them. 60
Moles of H2O = = 3.33mol (½ M)
(a) A solution is prepared by mixing 250 mL of a 2 M 18g/mol
hydrochloric acid (HCl)solution with 500 mL of a 4 M Mole fraction of ethylene glycol
HCl solution. Calculate the molarity of the final moles of C 2 H 6O 2 0.645
solution. = = = 0.162
Moles of C 2 H 6O 2 +moles of H 2O 0.645 + 3.33
(b) How many moles of solute are present in 250 mL of (½ M)
a 0.5 M solution?
(c) Calculate molality of 3.5 g of ethanoic acid in 80 g
Hints & Explanations
benzene.
(d) Calculate the mole fraction of ethylene glycol (C2H6O2) Multiple Choice Questions
in a solution containing 40% C2H6O2 by mass. (Un)
1. (d) Solutions are homogeneous mixtures, meaning their
Ans. (a) Calculate moles for 2M solution composition and properties are uniform throughout the
moles = M × V mixture, not varying as stated in options a and c. The
2 × 250 physical state is determined by the solvent, not the solute
= = 0.5mol  (½ M) as stated in option b.
1000
Calculate moles for 4 molar solution. 2. (b) Oil does not dissolve in water; therefore, instead of
Moles = M × V forming a homogeneous solution, the mixture will be
heterogeneous with two separate layers.
4 × 500
= = 2mol (½ M)
1000 3. (a) Molarity is temperature dependent whereas molality
Now, calculate final molarity is: is not. Hence, as temperature increases, volume increases,
therefore, molarity decreases whereas, molality remains the
Total moles
Molarity = same.
Total volume
4. (d) Molarity of NaCl is 0.02 M
0.5 + 2
= = 3.3M (½ M) No. of moles of NaCl is
0.25 + 0.5
0.02 × 4 = 0.08 moles
(b) Moles (n) = Molarity (M) × volume
Final Volume of solution is 4 +1 =5 litres
250
= 0.5M × litre Mass of 5 L water is 5 kg
1000
(∴ 1 litre = 1000ml) Molality = No. of moles of solute/mass of solvent in kg
= 0.125 moles (1 M) = 0.08 moles / 5 kg = 0.016 m
(c) Molality (M)
Assertion and Reason
Moles of solute 3.5
= = 1. (d) Camphor dissolved in nitrogen gas is an example of
mass of solvent in kg 60 × 0.08kg
gaseous solution.
0.058
= = 0.725m. 2. (d) The molality is expressed in units of moles per 1000
0.08kg gm of solvent. As, mass is temperature independent, hence,
Molar mass of CH3COOH = 60 g/mol (1 M) the molality of the solution does not change with change in
(d) Solution contains 40g of ethylene glycol and 100g of temperature.
water. 3. (c) Percent composition by mass is a way to express the
Molar mass of C2H6O2 = 62 g/mol concentration of a solution. It is calculated by dividing the
40 g
mass of the solute by the total mass of the solution (not
Moles of C2H6O2 = = 0.645mol (½ M) solvent), multiplied by 100.
62 g / mol

27 Solutions P
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 2 SOLUBILITY
NCERT Definitions (Commonly asked in 1 mark)
U Solubility: Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a
specified temperature.
U Saturated Solution: A saturated solution is a solution in which no more solute can be dissolved at a same temperature and
pressure.
U Unsaturated Solution: An unsaturated solution is one in which more solute can be dissolved at the same temperature.

Important Facts

01 Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.

A principle which states that solute dissolves in a solvent if the intermolecular interactions are similar in
02 the two. ~Like dissolves like

The solubility of solids in liquids increases with a rise in temperature if the dissolution process is
03 endothermic, and if it is exothermic, the solubility should decrease,, while solubility of gases in liquids
decreases with temperature and increases with pressure.

04 A condition caused due to low concentrations of oxygen in the blood and tissues. ~Anoxia

Important Concepts

U Henry’s law: It states that at a constant temperature, the partial pressure of the gas in vapour phase (p) is directly proportional to
mole fraction of the gas (x) in the solution, i.e., p = KH χ where, KH is the Henry’s law constant.

Real Life Application Based Questions

1. Why is carbon dioxide used in soft drinks and how does temperature affect its solubility?
Ans. Carbon dioxide is used in soft drinks for carbonation, providing fizz and a tangy taste. The solubility of CO2 decreases with
increasing temperature, causing the drink to go flat faster when warm.
2 How does the solubility of oxygen in water affect aquatic life, especially in warmer conditions?
Ans. Oxygen’s solubility in water decreases with increasing temperature, which can lead to lower oxygen levels in warm water bodies.
This can stress or kill aquatic organisms that rely on dissolved oxygen for survival.
3. Why do divers use a tank filled with the air diluted with helium when diving at great depths?
Ans. Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased
pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually
decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries

P
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 and creates a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well as, the toxic
effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium.
4. How does the solubility of carbon dioxide in blood relate to respiratory function in humans?
Ans. The solubility of carbon dioxide in blood is crucial for transporting CO2 from body tissues to the lungs for exhalation. Changes in
blood pH can affect CO2 solubility, impacting respiration and acid-base balance.
5. Explain how the solubility principle is used in the decaffeination of coffee.
Ans. In decaffeination, solvents like supercritical CO2 dissolve caffeine from coffee beans. The caffeine-rich solvent is then separated,
leaving decaffeinated coffee. This process relies on the selective solubility of caffeine in the solvent.

Myth Buster

U Myth: Solubility always increases with temperature.

Fact: While this is often true, there are exceptions. Some substances like gas solubility decreases with increasing temperature.
An example that illustrates this myth and fact is the solubility of calcium sulphate (CaSO4). Unlike many solids whose solubility
increases with temperature, the solubility of calcium sulphate slightly decreases as temperature rises. This behaviour is
counterintuitive to the common belief that all solids dissolve better in warm liquids.
U Myth: Solubility is the same for all substances in a given solvent.
Fact: Solubility varies widely depending on the nature of the solute and solvent. Polar substances generally dissolve well in polar
solvents, while nonpolar substances dissolve in nonpolar solvents.
U Myth: All solutes dissolve equally well in all solvents.
Fact: The solubility of a solute in a solvent depends on the nature of both. Generally, “like dissolves like,” meaning polar solutes
dissolve well in polar solvents and non-polar solutes dissolve well in non-polar solvents.
U Myth: Adding more solvent always increases solubility.
Fact: Adding more solvent to a saturated solution does not increase solubility; it only dilutes the solution. Solubility is an intrinsic
property that depends on temperature and pressure, not solvent quantity.

29 Solutions P
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 competency BASED SOlved Examples
1. Assertion (A): The solubility of most solid solutes in liquid
Multiple Choice Questions (1 M) solvents increases with an increase in temperature.
Reason (R): Increasing temperature increases solvent
1. As per Henry’s law, KH = p/x; where p is the partial
molecule kinetic energy, which helps to break the solute-
pressure, x is the mole fraction of the gas, and KH is the
solvent intermolecular forces more effectively. (Re)
Henry’s law constant. If, the concentration of N2 gas in
water at constant pressure increases quadratically, how 2. Assertion (A): The solubility of gases in liquids increases
will the value of KH change? (Un) (CBSE CFPQ, 2024) with an increase in temperature.
Reason (R): Higher temperatures decrease the intermolecular
(a) Increases linearly
forces between gas molecules and the solvent, leading to a
(b) Decreases quadratically lower solubility of gases in liquids. (Re)
(c) Decreases linearly 3. Assertion (A): According to Henry’s Law, the solubility of a
(d) Remains the same gas in a liquid is directly proportional to the partial pressure
of the gas above the surface of the liquid.
2. Low concentration of oxygen in the blood and tissues of
people living at high altitude is due to: (Un) (CBSE, 2023) Reason (R): Increasing the pressure of the gas over the
liquid decreases the number of gas molecules that come into
(a) high atmospheric pressure contact with the liquid, thus decreasing the gas’s solubility.
(b) low temperature (Un)
(c) low atmospheric pressure
(d) both low temperature and high atmospheric pressure Subjective Questions
3. The table below shows the KH values for some gases at
293 K and at the same pressure. (An) (CBSE APQ, 2023) Very Short Answer Type Questions  (2 M)
1. Aquatic animals feel more comfortable in cold water than
KH values 144.97 69.16 76.48 34.86
warm water as the solubility of oxygen in cold water is
(kbar)
more than that in warm water. The graph below shows
Gas Helium Hydrogen Nitrogen Oxygen the solubility of oxygen in the water as a function of
In which of the following are the gases arranged in their pressure at different temperatures T1 and T2.
decreasing order of solubility (from left to right)? (Ap) (CBSE CFPQ, 2024)
(a) Helium > Nitrogen > Hydrogen > Oxygen A T1
(b) Hydrogen > Helium > Nitrogen > Oxygen
(c) Nitrogen > Hydrogen > Oxygen > Helium
(d) Oxygen > Hydrogen > Nitrogen > Helium
T2
4. The value of Henry’s constant KH is ________.
(Un) (NCERT Exemplar)
Partial pressure

(a) greater for gases with higher solubility.

(b) greater for gases with lower solubility.
(c) constant for all gases.
(d) not related to the solubility of gases. 60° 30°
O X (mole fraction)
Assertion and Reason (1 M)
(i) Based on the above graph, what is the ratio of KH at
Direction: The following questions consist of two statements T1 and T2?
– Assertion (A) and Reason (R). Answer these questions by
(ii) Between T1 and T2, which one is greater?
selecting the appropriate option given below:
Ans. (i) Ratio of KH (T1) : KH (T2)
(a) Both A and R are true, and R is the correct explanation of A.
As per Henry’s law, p = KHχ. So the slope of the p vs x
(b) Both A and R are true, but R is not the correct explanation
curve will be the Henry law constant.
of A.
(c) A is true, but R is false. KH : KH = tan60/tan30 = 3 /(1/ 3 ) = 3 (1½ M)
1 2
(d) A is false, but R is true. (ii) T1 > T2 (½ M)
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 2. What is Henry’s law? Give one application of it. n mol n
(Re) (CBSE, 2023) x(Nitrogen)= = = 1.29 × 10−5 (½ M)
n mol + 55.5 mol 55.5
Ans. It states that the partial pressure of the gas in vapour phase (n in denominator is neglected as it is < < 55.5)
(p) is proportional to the mole fraction of the gas (χ) in the Thus n = 1.29 ×10–5 × 55.5 mol = 7.16 × 10–4 mol
solution. The formula of Henry’s law is p = KH χ (1 M) 7.16 × 10−4 mol × 1000mmol
= = 0.716 mmol (1 M)
Where, 1 mol
p = Partial pressure of the gas in the atmosphere above the Short Answer Type Questions (3 M)
liquid
1. (a) The Henry’s law constant for nitrogen gas (N 2)
χ = mole fraction of the gas
dissolved in water at a certain temperature is 6.8 × 10–4
KH = Henry’s law constant of the gas M/atm. Calculate the solubility of nitrogen gas in water
Application: It is used to increase the solubility of CO2 in when the partial pressure of N2 above the solution is
soft drinks and soda water and to add flavour to the drinks. 2.5 atm.
(1 M)
(b) Define the solubility rule “like dissolves like”. (Un)
3. H2S, a toxic gas with a rotten egg like smell, is used for Ans. (a) Given: Henry’s law constant (KH) = 6.8 × 10–4 M/atm
the qualitative analysis. If the solubility of H2S in water
Partial pressure of N2 (P) = 2.5 atm
at STP is 0.195 m, calculate Henry’s law constant.
(Ev) (NCERT Intext) The formula to calculate solubility is:

Ans. 0.195 mol of H2S is dissolved in 1000 g of water. Solubility= KH × P (1 M)

Moles of water = 1000 g / 18 g mol–1 = 55.56 mol (½ M) Substitute the values in above equation:

∴ Mole fraction of H2S, χ = Moles of H2S / (Moles of H2S Solubility of N2 = (6.8 × 10–4 M/atm) × (2.5 atm)
+ Moles of water) On solving,
= 0.195 / (0.195 + 55.56) Solubility = 1.7 × 10–3 M
= 0.0035 (½ M) Therefore, the solubility of nitrogen gas in water is
At STP, pressure (p) = 1 bar 1.7 × 10–3 M (1 M)

According to Henry’s law: p= KHχ (½ M) (b) According to this rule, a solute dissolves in a solvent if
the intermolecular interactions are similar in the two.(1 M)
Where, KH is the Henry’s constant. P is the partial pressure
of the gas and χ is the mole fraction 2. (a) Explain the following phenomena with the help of
Henry’s law.
⇒ KH = p/χ
(i) A painful condition known as bends.
= 1 / 0.0035 bar
(ii) Feeling of weakness and discomfort in breathing
= 285 bar (½ M)
at high altitude.
(b) Why soda water bottle kept at room temperature
fizzes on opening? (Ap) (NCERT Exemplar)
Nailing the Right Answer
Ans. (a) Henry’s Law Explanations:
U To ensure students get full credit, start by clearly
(i) Bends: Divers breathe compressed air containing
stating the formula used and define each variable.
nitrogen at high pressures underwater.
Then, plug in the given values accurately, ensuring
units are consistent. Simplify the calculation step-by- 
Upon rapid ascent, pressure decreases quickly, causing
step, showing all work. Finally, present the result with nitrogen to form bubbles in the blood. These bubbles
the correct units. can cause severe pain and other symptoms, known as
bends. (1 M)
4. If N2 gas is bubbled through water at 293 K, how many (ii) High Altitude Discomfort: At high altitudes,
millimoles of N2 gas would dissolve in 1 litre of water? atmospheric pressure is lower. According to Henry’s
Assume that N2 exerts a partial pressure of 0.987 bar. Given Law, lower pressure reduces the amount of oxygen
that Henry’s law constant for N2 at 293 K is 76.48 kbar. dissolved in the blood. This reduced oxygen availability
(Ev) leads to hypoxia, causing feelings of weakness and
Ans. The mole fraction of the gas in the solution is calculated by difficulty in breathing. (1 M)
applying Henry’s law. Thus: (b) Soda Water Fizzing: Henry’s Law explains that carbon
p (nitrogen) 0.987 bar dioxide is dissolved in the soda under high pressure.
x(Nitrogen)= = = 1.29 × 10−5  (½ M)
KH 76480 bar 
When the bottle is opened, the pressure is released. The
As 1 litre of water contains 55.5 mol of it, therefore if n sudden decrease in pressure causes carbon dioxide to
represents number of moles of N2 in solution. escape from the liquid rapidly, resulting in fizzing.(1 M)

31 Solutions P
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 Long Answer Type Questions (5 M)
1. (a) The partial pressure of ethane over a solution
Mistakes 101 : What not to do!
containing 6.56 × 10–3 g of ethane is 1 bar. If the solution
contains 5.00 × 10–2 g of ethane, then what shall be the Students might make mistake in:
partial pressure of the gas? (NCERT) U Partial pressure calculation: Misapplying the direct
(b) Why do gases always tend to be less soluble in liquids proportionality of pressure to the amount of gas.
as the temperature is raised? (NCERT) U Gas solubility: Ignoring the inverse relationship
between temperature and gas solubility.
(c) What is the significance of Henry’s Law constant
U Henry’s Law constant: Confusing it with other
KH? (NCERT Exemplar)
constants, not emphasizing its role in gas solubility.
(d) Explain why climbers experience symptoms of U Anoxia: Failing to link it to lower oxygen partial
anoxia at high altitudes. (Un) pressure and reduced oxygen solubility at high altitudes.
Ans. (a) According to Henry’s law, the solubility of gas in a
liquid is directly proportional to the pressure of the gas.
Hints & Explanations
χ = KH × P
Mole fraction of ethane will be directly proportional to its Multiple Choice Questions
mass as.
1. (d) The value of KH remains the same as it is a function of
W1 P2 = W2 P1 (1 M) the nature of the gas.
W1 is the mass of ethane in the first solution and P1 is the 2. (c) Low concentration of oxygen in the blood and tissues of
partial pressure of ethane in the first solution. W2 is the people living at high altitude is due to low atmospheric pressure.
mass of ethane in the second solution and P2 is the partial 3. (d) Higher the value of KH, lower is the solubility. Hence,
pressure of ethane in the second solution. the correct order of solubility is: Oxygen > Hydrogen >
6.56 × 10−3 × P2 = 5.00 × 10−2 × 1 (½ M) Nitrogen > Helium (1 M)
P2 = 7.6 atm (½ M) 4. (b) The higher the value of KH, the lower is the solubility
of gas in the liquid.
Partial pressure of gas will be 7.6 bar.
(b) When gases are dissolved in water, it is accompanied by Assertion and Reason
a release of heat energy, i.e., process is exothermic. When 1. (a) The solubility of solids in liquids often increases with
the temperature is increased, according to Le-Chatelier’s temperature because the increased kinetic energy of the
Principle, the equilibrium shifts in backward direction, and solvent molecules aids in overcoming the lattice energy of
thus gases becomes less soluble in liquids. (1 M) the solid solute, facilitating its dissolution.
(c) Higher the value of Henry’s law constant KH, the lower 2. (d) As temperature increases, the increased kinetic energy
is the solubility of the gas in the liquid. (1 M) of the gas molecules allows them to escape more easily from
(d) At high altitudes the partial pressure of oxygen is the solvent, thereby reducing their solubility in the liquid.
less than that at the ground level. This leads to low 3. (c) Henry’s Law clearly states that the solubility of a gas
concentrations of oxygen in the blood and tissues of people in a liquid is proportional to the partial pressure of the gas
living at high altitudes or climbers. Low blood oxygen present above the surface of liquid or solution. Increasing
causes climbers to become weak and unable to think clearly, pressure effectively increases solubility by forcing more
symptoms of a condition known as anoxia. (1 M) gas molecules into the solvent.

P
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 VAPOUR PRESSURE OF LIQUID SOLUTIONS &
3 IDEAL AND NON-IDEAL SOLUTIONS
NCERT Definitions (Commonly asked in 1 mark)

U Binary Solution: A solution containing two components, for instance, liquid-liquid or solid-liquid mixtures.
U Partial Vapour Pressure: The pressure exerted by a single component of a mixture in the vapour phase.
U Ideal Solution: A solution that obeys Raoult’s law over the entire range of concentration.
U Non-Ideal Solution: A solution that does not obey Raoult’s law over the entire range of concentration.
U Azeotrope: A mixture of two liquids that has the same composition in both liquid and vapour phase and boils at a constant
temperature.

Important Facts

A solution which shows a large positive deviation from Raoult’s law.

01
~Minimum Boiling Azeotrope

An example of azeotrope that has the approximate composition, 68% nitric acid and 32% water by mass,
02 with a boiling point of 393.5 K ~Maximum Boiling Azeotrope

03 Positive deviations from Raoult’s Law indicates weaker interactions between different molecules.

04 Negative deviations from Raoult’s Law indicates stronger interactions between different molecules.

Important Concepts
U Raoult’s Law: States that the partial vapour pressure of each component of a solution is directly proportional to its mole fraction
present in the solution that is, p = po x1 where po is the vapour pressure of pure component.
U Dalton’s Law of Partial Pressures: The total pressure over the solution phase in the container will be the sum of the partial
pressures of the components of the solution, that is, ptotal = p1 + p2.
U According to Raoult’s law
p1 = p10 x1 and p2 = p20 x2
Where, p10 and p20 represents the vapour pressure of pure component 1 and 2 respectively.
x1 and x2 are the mole fraction of component 1 and 2 respectively.
p1 and p2 represents the partial vapour pressure of each component of the solution.
Substituting the values of p1 and p2, we get
ptotal = x1 p10 + x2 p20 = (1 – x2) p10 + x2 p20 = p10 + (p20 – p10) x2
If y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then, using Dalton’s law of
partial pressure:
p1 = y1 ptotal
p2 = y2 ptotal
In general, pi = yi ptotal

33 Solutions P
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 Difference Between

Ideal Solutions vs. Non-Ideal Solutions

Aspect Ideal Solutions Non-Ideal Solutions
Obey Raoult’s Law over the entire range of Do not obey Raoult’s Law over the entire range of
Raoult’s Law
concentrations. concentrations.
Change in enthalpy/ No change in enthalpy or volume upon mixing (Dmix Mixing may involve changes in enthalpy and
volume H = 0, Dmix V = 0). volume.
Intermolecular forces between unlike molecules
Intermolecular Intermolecular forces between unlike molecules are
differ significantly from those between like
forces similar to those between like molecules.
molecules.
Represent ideal mixing behaviour without energy Show either positive or negative deviation based on
Mixing behaviour
change. the nature of intermolecular interactions.
Ethanol and acetone mixture (positive deviation)
Example Benzene and toluene mixture. or chloroform and acetone mixture (negative
deviation).

Positive Deviation from Raoult’s Law vs. Negative Deviation from Raoult’s Law

Aspect Positive Deviation Negative Deviation

Occurs when A-B interactions are weaker than Occurs when A-B interactions are stronger than
A-B interactions
A-A or B-B interactions. A-A or B-B interactions.
Results in a higher vapour pressure than predicted Results in a lower vapour pressure than predicted
Vapour pressure
by Raoult’s Law. by Raoult’s Law.
Enthalpy of mixing Enthalpy of mixing is positive (i.e; DHmix > 0) Enthalpy of mixing is negative (i.e; DHmix < 0)
Volume of mixing Volume of mixing is positive (i.e; DVmix > 0) Volume of mixing is negative (i.e; DVmix < 0)
Often leads to the formation of minimum boiling Often leads to the formation of maximum boiling
Azeotropes
azeotropes. azeotropes.
Example Ethanol and acetone. Phenol and aniline.

Minimum Boiling Azeotropes vs. Maximum Boiling Azeotropes

Aspect Minimum Boiling Azeotropes Maximum Boiling Azeotropes

Composition remains constant at boiling point, but

Composition remains constant at boiling point, but
Composition boils at a lower temperature than either of the pure
boils at a higher temperature than either component.
components.
Deviation from Formed by solutions showing large positive Formed by solutions showing large negative
Raoult’s Law deviations from Raoult’s Law. deviations from Raoult’s Law.
Example Ethanol-water mixture. Nitric acid-water mixture.

Real Life Application Based Questions

1. Why is it difficult to separate ethanol and water by simple distillation?

Ans. Ethanol and water form an azeotrope with a constant boiling point. This makes simple distillation ineffective, as the vapor
composition remains the same as the liquid.
P
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 2. How does the positive deviation from Raoult’s Law affect the formulation of perfumes?
Ans. Positive deviation indicates weaker interactions between different molecules, enhancing the evaporation rate of volatile compounds
in perfumes, which contributes to their fragrance release.
3. Explain the significance of negative deviation from Raoult’s Law in the manufacture of chloroform-acetone mixtures.
Ans. Negative deviation indicates stronger intermolecular forces, which can lead to a decrease in vapor pressure. This property is crucial
for creating stable mixtures with specific boiling points needed in industrial applications In the chemical industry, the chloroform-
acetone mixture is used as a solvent system where the controlled volatility is essential for reactions and extractions. Understanding
and utilizing the negative deviation allows chemists to optimize solvent properties for specific applications, enhancing efficiency
and safety..

Myth Buster

U Myth: All liquids have the same vapour pressure at a given temperature.
Fact: Vapour pressure varies among different liquids based on their intermolecular forces. Volatile liquids have higher vapor
pressures than less volatile ones.
U Myth: Increasing the amount of liquid increases its vapour pressure.
Fact: Vapour pressure depends on temperature, not the quantity of liquid. It is the pressure exerted by the vapour in equilibrium
with its liquid phase.
U Myth: Non-volatile solutes do not affect the vapour pressure of a solvent.
Fact: Non-volatile solutes lower the vapour pressure of a solvent by occupying surface area and reducing the number of solvent
molecules escaping into the vapour phase.
U Myth: Raoult’s Law applies to all types of solutions.
Fact: Raoult’s Law applies specifically to ideal solutions. Non-ideal solutions require modifications to account for deviations
from ideal behaviour.
U Myth: Enthalpy of mixing is always zero.
Fact: Enthalpy of mixing is zero only for ideal solutions. Non-ideal solutions exhibit either exothermic or endothermic enthalpy
changes
U Myth: Azeotropes can be easily separated by distillation.
Fact: Azeotropes have constant boiling points and compositions, making them inseparable by simple distillation. Special
techniques are required.

Mnemonics

U CAN: Chloroform and Acetone mixture (Negative deviation).

C A N

Chloroform Acetone Negative deviation

U EAP: Ethanol and Acetone mixture (Positive deviation)
E A P

Ethanol Acetone Positive Deviation

U MinE: Minimum Boiling Azeotrope Example: Ethanol-water mixture.
Min E

Minimum boiling azeotrope Ethanol + water

U MaN: Maximum Boiling Azeotrope Example: Nitric acid-water mixture.
Ma N

Maximum boiling azeotrope Nitric acid + water

35 Solutions P
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 Competency BASED Solved Examples
2. Assertion (A): In case of positive deviation from Raoult’s
Multiple Choice Questions (1 M) law, A-B interactions are weaker than those between A-A
or B-B.
1. In which case Raoult’s law is not applicable? (Un)
Reason (R): In case of negative deviations from Raoult’s
(a) 1 m NaCl (b) 1 m urea
law, the intermolecular attractive forces between A-A and
(c) 1 m glucose (d) 1 m sucrose
B-B are weaker than those between A-B. (Re)
2. Which one of the following pairs will form an ideal
solution? (Re) (CBSE, 2022 Term-1) 3. Assertion (A): Azeotropes signify a deviation from Raoult’s
Law, demonstrating the limit of non-ideal solution behaviour.
(a) Chloroform and acetone (b) Ethanol and acetone
(c) n-hexane and n-heptane (d) Phenol and aniline Reason (R): Azeotropes are mixtures that boils at a different
temperature and retain the same composition in both liquid
3. On the basis of information given below, mark the correct
and vapour phases, due to unique intermolecular interactions
option.
that prevent the application of Raoult’s Law. (Re)
Information:
(i) In bromoethane and chloroethane mixture,
intermolecular interactions of A-A and B-B type are Subjective Questions
nearly same as A-B type interactions.
(ii) In ethanol and acetone mixture, A-A or B-B type Very Short Answer Type Questions  (2 M)
intermolecular interactions are stronger than A-B 1. State
 Raoult’s law for a solution containing volatile
type interactions. components. What is the similarity between Raoult’s law
(iii) In chloroform and acetone mixture, A-A or B-B type and Henry’s law? (Re) (CBSE, 2020)
intermolecular interactions are weaker than A-B
Ans. Raoult’s law states that for a solution of volatile liquids, the
type interactions.
partial vapour pressure of each component in the solution is
(a) Solution (ii) and (iii) will follow Raoult’s law.
directly proportional to its mole fraction. (½ M)
(b) Solution (i) will follow Raoult’s law.
Thus, if there is a solution of two liquid components (1) and
(c) Solution (ii) will show negative deviation from Raoult’s
(2), then for component (1), p1 α X1
law.
(d) Solution (iii) will show positive deviation from Raoult’s And, For component (2), p2 α X2
law. (Re) (NCERT Exemplar) where p 1 ,p 2 = Partial vapour pressure of two volatile
4. Considering the formation, breaking and strength of components (1) and (2) of the solution and X1,X2 are mole
hydrogen bond, predict which of the following mixtures fractions of the components (1) and (2), respectively.(½ M)
will show a positive deviation from Raoult’s law? Henry’s law states that,
(Re) (NCERT Exemplar) p = KH X (½ M)
(a) Methanol and acetone. (b) Chloroform and acetone. In both the laws, the partial pressure is directly proportional
(c) Nitric acid and water. (d) Phenol and aniline. to the mole fraction of the components. therefore, on
comparing both laws it has been observed that Raoult’s law
becomes a special case of Henry’s law in which KH becomes
Assertion and Reason (1 M) equal to poi . (½ M)
Direction: The following questions consist of two statements Topper’s Explanation
– Assertion (A) and Reason (R). Answer these questions by
selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation
of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
1. Assertion (A): Vapour pressure of a solution decreases when
a non-volatile solute is dissolved in a solvent.
Reason (R): Dissolving a non-volatile solute in a solvent
decreases the number of solvent molecules at the surface,
reducing the rate of evaporation. (Un)
P
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 2. The vapour pressure of pure liquid X and pure liquid The sign of ΔmixH (enthalpy change of mixing) indicates
Y at 25°C are 120 mm Hg and 160 mm Hg respectively. the nature of these deviations: positive ΔmixH for positive
If equal moles of X and Y are mixed to form an ideal deviations (endothermic process) and negative ΔmixH for
solution, calculate the vapour pressure of the solution. negative deviations (exothermic process), reflecting the
(Ev) (CBSE, 2023) energy required or released when mixing solute and solvent.
0 0 (1 M)
Ans. PTotal = PX χ X + PY χ Y ,  (1 M)
2. The vapour pressure of pure liquids A and B are 450
Here, χ X =χ Y =0.5 and 700 mm Hg respectively, at 350 K. Find out the
So, PTotal = (120 × 0.5 ) + (160 × 0.5 ) = 140 mm Hg  (1 M) composition of the liquid mixture if total vapour pressure
is 600 mm Hg. Also find the composition of the vapour
3. The images below show the evaporation of the solvent on
phase. (Ev) (NCERT Intext)
account of the presence of non-volatile solutes. In each
of the three cases, the solvent taken is of the same type. Ans. Composition of Liquid mixture:
The solvent is volatile and its quantity is the same in all
Vapour pressure of pure liquid A (p0A) = 450 mm Hg
three cases. (Ap) (CBSE CFPQ, 2024)
Vapour pressure of pure liquid B (p0B) = 700mm Hg
Total vapour pressure of solution (P) = 600 mm Hg
According to Raoult’s law:
Solvent Solvent Solvent
+ + + p = p0AχA+p0 BχB (½ M)
Solute Solute Solute
p = p0A.χA + p0B (1 − χA)
P Q R
→Molecules or particles of solvent vapour p = 450.χA + 700 (1 − χA)
Which of the above three solutions has the least amount 600 = 450 χA + 700 − 700χA
of solute in it? How did you reach that conclusion?
600 − 700 = −250 χA
Ans. Solution Q has the least amount of solute in it

Reason: The evaporation rate of molecules of a given χA = 0.40
solvent is the highest in Q as there is more number of particles So, mole fraction of B = 0.60 (½ M)
outside the solution responsible for vapour pressure. (1 M) Composition of vapour phase:
The higher the vapour pressure, the higher will be mole
pA = p0AχA
fraction of the solvent and the lower will be the amount of
solute. (1 M) = 450 × 0.40 = 180 mm Hg (½ M)
pB = p0 BχB
= 700 × 0.60
Nailing the Right Answer = 420 mm Hg (½ M)
U While writing the answer, the student should clearly  pA = yA ptotal
state that the solution with the least amount of solute pA 180
is the one with the most significant evaporation, as ⇒ yA = = = 0.3 (½ M)
p total 600
the presence of non-volatile solute decreases vapour
pB = yB ptotal
pressure, reducing evaporation. Students should
accurately explain that less solute results in higher pB 420
⇒ yB = = = 0.70 (½ M)
vapour pressure and thus more evaporation, and they p total 600
should reference the images to support their explanation. ∴ Mole fraction of A in the vapour phase (yA) = 0.3

Short Answer Type Questions (3 M) Mole fraction of B in the vapour phase (yB) = 0.70
1. What is meant by positive and negative deviations from
Raoult’s law and how is the sign of DmixH related to Long Answer Type Questions (5 M)
positive and negative deviations from Raoult’s law?
(Re) (NCERT) 1. (a) Differentiate between Ideal solution and Non-ideal
Ans. Positive deviations from Raoult’s law occur when the solution. (CBSE, 2023)
solute-solvent interactions are weaker than the solvent- (b) What type of deviation from Raoult’s Law is
solvent and solute-solute interactions, leading to a higher expected when phenol and aniline are mixed with each
vapour pressure than predicted. (1 M) other? What change in the net volume of the mixture is
expected? Graphically represent the deviation.
Negative deviations happen when solute-solvent
interactions are stronger than the solute-solute and solvent- (CBSE SQP, 2023)
solvent interactions, resulting in lower vapour pressure. (c) Why is the vapour pressure of an aqueous solution of
(1 M) glucose lower than of water? (Un) (NCERT Exemplar)

37 Solutions P
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Ans. (a) (ii) Using Raoult’s law, Ptotal = PoA.XA + PoB. XB
⇒ 760 = 526 XA + 11250 XB … (i) (1 M)
Ideal Solutions Non-Ideal Solutions
Also
Obey Raoult’s Law Do not obey Raoult’s ⇒ X A + XB = 1 … (ii) (1 M)
over the entire range of Law over the entire Solving equ. (i) and (ii), we get XA = 0.98; XB = 0.02 (1 M)
concentrations. range of concentrations. (b) Nitric acid and water is an example of maximum boiling
No change in enthalpy or Mixing may involve azeotrope. This azeotrope has the approximate composition,
volume upon mixing (Δmix changes in enthalpy and 68% nitric acid and 32% water by mass, with a boiling point
H = 0, Δmix V = 0). volume. of 393.5 K (1 M)
(2 M) Hints & Explanations
(b) 
Negative Deviation is expected when phenol and
aniline are mixed with each other. The net volume Multiple Choice Questions
of the mixture will decrease, ∆V< 0 due to stronger
intermolecular interactions. (1 M) 1. (a) If the total number of particles of solute changes in
the solution due to association or dissociation, Raoult’s
law is not applicable. Among the given compounds, NaCl
undergoes dissociation and forms Na+ and Cl– ions. Thus,
Raoult’s law is not applicable to NaCl.
2. (c) n-hexane and n-heptane follow Raoult’s law over the
entire range of concentration. The new intermolecular
interactions formed between hexane and heptane in solution
are similar to one present in pure hexane and heptane.
Hence, they can form nearly ideal solutions.
3. (b) Solution of Bromoethane and chloroethane is the
example of an ideal solution. Hence, the intermolecular
attractive forces between the A-A and B-B are nearly equal
(1 M) to those between A-B.
(c) The vapour pressure of an aqueous glucose solution is 4. (a) Methanol and acetone will show a positive deviation
lower than that of pure water because glucose molecules from Raoult’s law. This is because the hydrogen
disperse among water molecules, hindering some water bonding between methanol molecules is stronger than
from evaporating.  (1 M) the intermolecular forces between methanol and acetone.
2. (a) The vapour pressure of compound A at 90°C is 526 mm Therefore, when mixed, the weaker methanol-acetone
Hg and that of compound B is 11250 mm of Hg. interactions lead to an increase in vapour pressure compared
(i) What will be the total concentration (in terms of to what Raoult’s law predicts, indicating a positive deviation.
mole fraction) of the boiling mixture of A and B at
90° C if the two liquids are completely miscible with Assertion and Reason
each other? 1. (a) The presence of a non-volatile solute reduces the
(ii) Using (i), calculate XA and XB. (Round off to two solvent’s vapour pressure because it decreases the fraction
decimal places) of solvent molecules able to escape into the vapour phase,
(Take Ptotal = 760 mm Hg) (CBSE CFPQ,2024) which aligns with the reason provided.
(b) Give an example of maximum boiling azeotrope with 2. (b) Both A and R are true, but R is not the correct explanation
its composition.  (An) of A.
Ans. (a) (i) For binary solution, XA + XB = 1; where A and B are 3. (c) Azeotropes are binary mixtures having the same
any two liquids/compounds that are mixed. composition in liquid and vapour phase and boil at a
So, the total concentration of A and B = 1 (1 M) constant temperature.

P
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 COLLIGATIVE PROPERTIES OF SOLUTIONS AND
4 ABNORMAL MOLAR MASS
NCERT Definitions (Commonly asked in 1 mark)

U Colligative Properties: Properties which depend on the number of solute particles irrespective of their nature relative to the
total number of particles present in the solution.
U Osmotic Pressure: The pressure required to prevent the flow of solvent into a solution through a semipermeable membrane.
U Semipermeable Membrane (SPM): A membrane that allows the passage of solvent molecules but blocks solute particles.
U Osmosis: The movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration
to a region of higher solute concentration.
U Reverse Osmosis: The process by which solvent molecules are forced to move from a region of higher solute concentration to a
region of lower solute concentration by applying pressure greater than the osmotic pressure.
U Isotonic Solutions: Two solutions having the same osmotic pressure at a given temperature.
U Hypotonic Solutions: If the salt concentration is less than 0.9% (mass/volume), the solution is said to be hypotonic.
U Hypertonic Solution: If the salt concentration is more than 0.9% (mass/volume), the solution is said to be hypertonic.
U Van’t Hoff Factor (i): The van’t Hoff factor (i) is a dimensionless quantity that represents the number of particles into which
a solute dissociates or associates in a solution.

Important Facts

01 A membrane through which only solvent molecules can pass through. ~Semipermeable Membrane

02 A colligative property used for water purification. ~Reverse Osmosis

03 Cryoscopic and ebullioscopic constants are unique for each solvent.

04 A proportionality constant used in calculating freezing point depression. ~Cryoscopic Constant

05 A proportionality constant used in calculating boiling point elevation. ~Ebullioscopic Constant

39 Solutions P
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 Important Formulae

Relative lowering of Vapour Pressure: (where, n1 and n2

∆p1 p10 − p1 0
n2  n2 
are the number of moles of solvent and solute respectively,= 0
= 0 = x 2 or p1 − p1 =  since x 2 
w1 and w2 are the masses and M1 and M2 are the molar p1 p1 0 n1 + n 2  n1 + n 2 
p1
masses of the solvent and solute respectively. p1o is the vapour
pressure of pure solvent and p1 is the vapour pressure of the
solution. x2 is the mole fraction of solute) p10 − p1 n 2 p10 − p1 w 2 × M1
= or =
For dilute solutions, n2 < < n1 p10 n1 p10 M 2 × w1

Elevation of Boiling Point: (where, DTb is the elevation of DTb = Kb m

Boiling point, Kb is the Boiling Point Elevation Constant or
K × w 2 × 1000
Molal Elevation Constant (Ebullioscopic Constant), m is ∆Tb =b
the molality, w2 is the grams of solute, M2 is molar mass of M 2 × w1
solute, W1 is the grams of solvent.)

Depression of Freezing Point: (where, DTf is the depression ∆Tf = Kf m

of freezing point, Kf is the Molal depression constant or K f × w2 × 1000
Cryoscopic constant) ∆T f =
M 2 × w1

Osmotic Pressure: (where,

Π represents osmotic pressure Π = C R T = (n2/V) R T
R is Gas constant
C is molar concentration of solution Π = w 2 R T / M2 × V
T is temperature in Kelvins
V is volume of a solution in litres containing n2 moles of
solute.
W2 is grams of solute,
M2 is the molar mass of solute present in the solution)

Normal molar mass

i=
Abnormal molar mass
Observed colligative property
Van’t Hoff factor, i: =
Calculated colligative property
Total number of moles of particles after association/dissociation
i=
Number of moles of particles before association/dissociation

Inclusion of van’t Hoff factor modifies the equations for

colligative properties as follows:
p1o − p1 n2
Relative lowering of vapour pressure of solvent = i.
p1o n1
Elevation of Boiling point,
∆Tb = iKbm
Depression of Freezing point ∆Tf = iKf m
Osmotic pressure of solution, Π = C R T = i(n2 R T) / V

Real Life Application Based Questions

1. How is the concept of freezing point depression used in making ice cream?
Ans. Salt is added to the ice surrounding the ice cream mixture. This lowers the freezing point of the ice, causing it to absorb more heat
from the mixture, which helps freeze the ice cream faster.

P
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 2. Why is it important for intravenous (IV) solutions to have the correct osmotic pressure?
Ans. IV solutions must be isotonic with blood plasma to prevent red blood cells from shrinking (crenation) or swelling and bursting
(hemolysis) due to osmotic pressure differences.
3. How does the boiling point elevation principle apply to car radiators in cold climates?
Ans. Antifreeze lowers the freezing point and raises the boiling point of the radiator fluid, preventing it from freezing in winter and
boiling in summer, ensuring the engine runs efficiently
4. Explain the role of osmotic pressure in desalination processes using reverse osmosis.
Ans. In reverse osmosis, pressure greater than the osmotic pressure is applied to saltwater, forcing water molecules to move through a
semipermeable membrane, leaving the salt behind, thus purifying the water.
5. Why is it important to use a saline solution for intravenous (IV) injections instead of pure water?
Ans. A saline solution matches the ionic concentration of blood plasma, preventing cell damage. Pure water can cause cells to swell
and burst due to osmotic pressure differences.

Myth Buster

U Myth: Colligative properties depend on the chemical nature of the solute.

Fact: Colligative properties depend only on the number of solute particles, not their chemical nature.
U Myth: All solutes affect boiling point elevation and freezing point depression equally.
Fact: The effect on colligative properties varies with the number of particles the solute dissociates into, which is accounted by
the Van’t Hoff factor.
U Myth: Only non-electrolytes exhibit colligative properties.
Fact: Both electrolytes and non-electrolytes exhibit colligative properties, but electrolytes have a greater effect due to their
dissociation into multiple ions.
U Myth: Abnormal molar masses are errors in measurement.
Fact: Abnormal molar masses occur due to the association or dissociation of solute molecules, not measurement errors. These
can be corrected using the Van’t Hoff factor.

41 Solutions P
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 competency BASED SOlved Examples
1. Assertion (A): 0.02 m solutions of urea and sucrose will
Multiple Choice Questions (1 M) freeze at same temperature.
Reason (R): Freezing point of solution is inversely related
1. Colligative properties depend on _________ . to the concentration of solution. (Un) (CBSE, 2023)
(Re) (NCERT Exemplar)
2. Assertion (A): 0.1 M solution of KCl has greater osmotic
(a) the nature of the solute particles dissolved in solution. pressure than 0.1 M solution of glucose at same temperature.
(b) the number of solute particles in solution.
Reason (R): In solution, KCl dissociates to produce more

(c) the physical properties of the solute particles dissolved
in solution number of particles. (Un) (CBSE, 2020)
(d) the nature of solvent particles.
2. In comparison to a 0.01 M solution of glucose, the Subjective Questions
depression in freezing point of a 0.01 M MgCl2 solution
is _________ . (Un) (NCERT Exemplar)
Very Short Answer Type Questions  (2 M)
(a) the same (b) about twice
(c) about three times (d) about six times 1. (a) What are colligative properties? Give an example.
3. An unripe mango placed in a concentrated salt solution (b) How does sprinkling of salt help in clearing the snow
to prepare pickle, shrivels because ________ . covered roads in hilly areas? Explain the phenomenon
(Un) (NCERT Exemplar) involved in the process. (Re) (NCERT Exemplar)
(a) it gains water due to osmosis. Ans. (a) Colligative properties are those properties of solutions
(b) it loses water due to reverse osmosis. that depend on the number of solute particles in a given
(c) it gains water due to reverse osmosis. amount of solvent and not on the nature of the solute
(d) it loses water due to osmosis. particles. (½ M)
4. Which of the following statements is false?
For Example: Relative lowering of vapour pressure.(½ M)
(Re) (NCERT Exemplar)
( b) Sprinkling salt on snow-covered roads lowers the freezing
(a) Two different solutions of sucrose of same molality
point of water, a process known as freezing point depression.
prepared in different solvents will have the same
depression in freezing point. This causes the snow to melt at lower temperatures than
(b) The osmotic pressure of a solution is given by the usual, preventing it from freezing into ice.
equation Π = CRT (where C is the molarity of the The salt disrupts water molecules’ ability to form solid ice,
solution). effectively clearing the roads and making them safer for
(c) Decreasing order of osmotic pressure for 0.01 M aqueous travel in hilly areas. (1 M)
solutions of barium chloride, potassium chloride, acetic
acid and sucrose is BaCl2 > KCl > CH3COOH > sucrose. 2. Calculate the freezing point of a solution containing 60 g
(d) According to Raoult’s law, the vapour pressure exerted of glucose (Molar mass = 180 g mol–1) in 250 g of water.
by a volatile component of a solution is directly (Kf of water = 1.86 K kg mol–1) (Ev) (CBSE, 2018)
proportional to its mole fraction in the solution.
Ans. WB = 250 g, WA = 60 g, MA = 180 g mol–1
Kf = 1.86 K kg mol–1
Assertion and Reason (1 M) WA  1000
Direction: The following questions consist of two statements DTf = Kf m  K f  (1 M)
M A  WB
– Assertion (A) and Reason (R). Answer these questions by
selecting the appropriate option given below: 60  1000 1.86  600 1116
 1.86     2  48 K
(a) Both A and R are true, and R is the correct explanation of 180  250 18  25 450
A. DTf = Tsolvent – Tsolution

(b) Both A and R are true, but R is not the correct explanation Tsolution = Tsolvent – DTf
of A.
(c) A is true, but R is false. = 273 . 15 K – 2 . 48 K
(d) A is false, but R is true. = 270 . 67 K (1 M)
P
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Topper’s Explanation Molality is the number of moles of solute in 1 kg of solvent.
0.25
Molality = = 0.50m (½ M)
500
1000
Now, depression in freezing point is given by
∆Tf = K f × m = 1.86 × 0.50 = 0.93 K  (½ M)
Now the van’t Hoff factor is given by the ratio of observed
freezing point depression to the calculated freezing point
depression.
1
=i = 1.0753  (½ M)
0.93
Let C be the initial concentration and a be the degree of
dissociation.
CH2FCOOH→CH3FCOO– + H+
C (1 – a)    Ca   Ca
The total number of moles = C(1 – a) + Ca + Ca = C(1
3. Calculate the osmotic pressure in pascals exerted by a +a)
solution prepared by dissolving 1.0 g of polymer of molar
mass 185,000 in 450 mL of water at 37°C. C(1 + α)
So, i = = 1 + α = 1.0753  (1 M)
(Ev) (NCERT Intext) C
Ans. Given: Hence, a = 0.0753
2. A 10% solution (by mass) of sucrose in water has
Mass of polymer = 1.0g
freezing point of 269.15 K. Calculate the freezing point
Molar mass = 185, 000 of 10% glucose in water, if freezing point of pure water
V = 450ml = 0.45 L is 273.15 K.
T = 37 + 273 = 310 K Given: (Molar mass of sucrose = 342 g mol–1)
n (Molar mass of glucose = 180 g mol–1) (Ev)
Osmotic Pressure, Π = RT (1 M)
v Ans. ∆Tf = k f × m  (½ M)
1 1
=Π × × 8.314 × 103 × 310K Tf 273.15 − 269 ⋅=
∆= 15 4K (½ M)
185000 0.45
10 1000
Π = 30.95 Pa (1 M) 4 = kf × ×
90 342
4
kf
= = 12.31  (½ M)
0.325
Key Takeaways For glucose
This question teaches students that the concept of osmotic ∆Tf = K f × m
pressure is essential in understanding colligative properties.
It demonstrates how to calculate osmotic pressure using 10 × 1000
12.3 ×
= 7.6
=
the formula Π = n/VRT where n is the number of moles of 180 × 90
solute, V is the volume of the solution in liters, R is the gas ∆Tf =
7.6 K (½ M)
constant, and T is the temperature in Kelvin.
7.6 = T(solvent) – T(solution) (½ M)
T(solution) = 273.15 – 7.6
Short Answer Type Questions (3 M)
= 265.55 K (½ M)
1. When 19.5 g of F – CH2 – COOH (Molar mass = 78 3. Give reasons for the following: (Ap) (CBSE 2018)
g mol–1), is dissolved in 500 g of water, the depression
in freezing point is observed to be 1°C. Calculate the (a) Measurement of osmotic pressure method is
preferred for the determination of molar masses of
degree of dissociation of F – CH2 – COOH.
macromolecules such as proteins and polymers.
[Given: Kf for water = 1.86 K kg mol–1]
(b) Aquatic animals are more comfortable in cold
(Ev) (CBSE, 2023) water than in warm water.
Ans. Number of moles of fluoroacetic acid = 19.5/78 = 0.25 mol (c) Elevation of boiling point of 1 M KCl solution is
(½ M) nearly double than that of 1 M sugar solution.

43 Solutions P
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Ans. (a) 
The osmotic pressure method records pressure at (iii) ∆Tb = iK b m (½ M)
standard room temperatures and utilises the solution’s Here, i = 3
molarity rather than molality. This approach is ideal
w B ×1000
for large molecules such as proteins, which degrade Now, m = (½ M)
MB × w A
at elevated temperatures, and for polymers that don’t
easily dissolve. Because of their large size, osmotic 3 × 0.512 × 1000 × 10
So, ∆Tb = (½ M)
pressure is the only colligative property that can be 95 × 200
measured with any significant effect, making this ∆Tb = 0.81 K or o C (½ M)
technique the preferred choice. (1 M)
(b) Aquatic animals are more comfortable in cold water
than in warm water primarily due to the higher
solubility of oxygen in colder water. Cold water Mistakes 101 : What not to do!
can hold more dissolved oxygen than warm water, Incorrect Dissociation Assumption: Do not assume that
providing a better environment for the respiratory NaCl and glucose dissociate similarly. NaCl is ionic and
needs of aquatic animals. (1 M) dissociates into ions, while glucose does not dissociate.
(c) The boiling point elevation of a 1M KCl solution Incorrect Mole Fraction Calculation: Ensure the correct
is nearly twice that of a 1 M sugar solution due to calculation of mole fraction by using the total number of
dissociation. KCl, an ionic compound, dissociates into moles in the solution (solute + solvent).
K+ and Cl– ions, providing 2 moles of particles per litre, Wrong van’t Hoff Factor: For MgCl2, the van’t Hoff
elevating the boiling point more than sugar, a covalent factor i is 3 because it dissociates into three ions (Mg²⁺
compound that doesn’t dissociate and provides only 1 and 2Cl⁻). Using an incorrect value will lead to wrong
mole of particles per litre. (1 M) calculations.
Long Answer Type Questions Neglecting Unit Conversion: Always convert masses to
(5 M)
kilograms when calculating molality.
1. (i) Why is boiling point of 1M NaCl solution more than
that of 1 M glucose solution?
2. (i) Why is the value of Van’t Hoff factor for ethanoic
(ii) A non-volatile solute ‘X’ (molar mass = 50 g mol–1) when
dissolved in 78 g of benzene reduced its vapour pressure to acid in benzene close to 0.5?
90%. Calculate the mass of X dissolved in the solution. (ii) 
Determine the osmotic pressure of a solution
(iii) Calculate the boiling point elevation for a solution prepared by dissolving 2.32 × 10–2 g of K2SO4 in
prepared by adding 10 g of MgCl2 to 200 g of water 2 L of solution at 25°C, assuming that K2SO4 is
assuming MgCl2 is completely dissociated.
completely dissociated.
(Kb for Water = 0.512 K kg mol–1, Molar mass MgCl2 =
95 g mol–1) (An) (CBSE, 2023) (iii) When 25.6 g of Sulphur was dissolved in 1000 g
Ans. (i) NaCl is an ionic compound that has the van’t Hoff of benzene, the freezing point lowered by 0.512 K.
Calculate the formula of Sulphur (Sx).
factor 2, whereas glucose is a covalent compound that does
not undergo dissociation and has the van’t Hoff factor 1. (  Kf for benzene = 5.12 K kg mol–1, Atomic mass of
Elevation in boiling point is directly proportional to Sulphur = 32 g mol–1) (Ev) (CBSE, 2023)
molality and van’t Hoff factor. As a result, the boiling point Ans. (i) In benzene, two molecules of ethanoic acid associate to
of 1 M NaCl is higher than 1 M glucose solution. (1 M) form a dimer.
P o − Ps 2CH3 COOH (CH3 COOH)2
(ii) Relative lowering of vapour pressure is = χ2
Po So, the van’t Hoff factor,
P o − Ps nB
= (½ M) Number of moles of particles after dissociation or association
P o
nA + nB i=
Number of moles of particles before dissociation or association
w A 78 = wB wB
Here, =
nA = , nB = (½ M) (½ M)
M A 78 M B 50
1
i= = 0.5
P o − 0.9P o w / 50 2 (½ M)
= (½ M)
Po w / 50 + 1
(ii) Mass of K2SO4 = 2.32 × g 10–2
w  w Temperature, T = 25°C = 25 + 273 = 298 K
0.1 + 1 =
 50  50 Molar mass of K2SO4 = (2 × 39) + 32 + 4 × 16 = 174 g/mol
w w w w 9w As K2SO4 dissociates completely
+ 1= ⇒ − 1⇒
= =1
50 5 5 50 50
K 2SO 4 → 2K + + SO 24 −
w = 5.55 g (½ M)
P
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 After dissociation, the total number of ions produced = 3 2. (c) The depression in freezing point is about three times
(½ M) for a 0.01 M MgCl2 solution compared to a 0.01 M glucose
Here, i = 3, Π = iCRT (½ M) solution because MgCl2 dissociates into three ions (Mg2+
−2
and 2Cl–1), increasing the solution’s particle concentration.
=So, Π 3 × 2.32 × 10 × 0.082 × 298  (½ M) 3. (d) An unripe mango placed in a concentrated salt solution
174 × 2
to prepare pickle shrivels because it loses water due to
Π = 4.8 × 10−3 atm 
= 0.0048 (½ M) osmosis. The concentrated salt solution is the hypertonic
solution.
(iii) Kf for benzene = 5.12 K kg mol–1
4. (a) Two different solutions of sucrose with the same
Mass of solute, WB = 25.6 g
molality prepared in different solvents may not have the
WA = 1000 g same depression in freezing point because the freezing
Atomic mass of sulphur = 32 g/mol point depression also depends on the specific properties of
∆Tf =
0.512 K the solvent, such as its Kf value (cryoscopic constant).

Assertion and Reason

Now, M B = K f × WB ×1000 (½ M)
∆Tf × WA 1. (b) The value of ‘i’ is same for both urea and sucrose,
therefore both solutions will freeze at the same temperature.
5.12 × 25.6 ×1000
MB = ΔTf = Kfm
0.512 ×1000
(T0f – Tf) = Kf m
MB = 256 (½ M)
On increasing m, ΔTf will increase and hence freezing
Now, Sx =x × 32 =256 (½ M) point will decrease.
So, x = 8 2. (a) In an aqueous solution, KCl dissociates completely
Thus, the correct formula of sulphur is S8. (½ M) into K⁺ and Cl– ions. Therefore, a 0.1 M solution of KCl
produces 0.1 M K⁺ and 0.1 M Cl–, resulting in a total of
Hints & Explanations 0.2 M particles. In contrast, glucose does not dissociate
and remains as individual glucose molecules, so a 0.1 M
Multiple Choice Questions solution of glucose produces only 0.1 M particles.. Thus,
1. (b) Colligative properties depend on the number of solute KCl dissociates into more particles,and hence it results in
particles irrespective of their nature relative to the total a higher osmotic pressure compared to glucose, which does
number of particles present in the solution. not dissociate.

45 Solutions P
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 miscellaneous Exercise
5. KH value for Ar(g), CO2(g), HCHO (g) and CH4 (g) are
Multiple Choice Questions (1 M) 40.39, 1.67, 1.83 × 10–5 and 0.413 respectively.
Arrange these gases in the order of their increasing solubility.
1. Which of the following units is useful in relating concentration (Ap) (NCERT Exemplar)
of solution with its vapour pressure?
(a) HCHO < CH4 < CO2 < Ar
(Re) (NCERT Exemplar)
(b) HCHO < CO2 < CH4 < Ar
(a) mole fraction (b) parts per million
(c) Ar < CO2 < CH4 < HCHO
(c) mass percentage (d) molality
(d) Ar < CH4 < CO2 < HCHO
2. Value of Henry’s constant KH __________ .
(Un) (NCERT Exemplar) 6. The unit of ebullioscopic constant is ________ .
(Re) (NCERT Exemplar)
(a) increases with increase in temperature.
(a) K kg mol–1 or K (molality)–1
(b) decreases with increase in temperature.
(b) mol kg K–1 or K–1 (molality)
(c) remains constant.
(d) first increases then decreases. (c) kg mol–1 K–1 or K–1 (molality)–1
(d) K mol kg–1 or K (molality)
3. 93 g of ethylene glycol is added with 1 kg of water to change
7. Two beakers of capacity 500 mL were taken. One of these
the freezing point of the solution. If the freezing point of
beakers, labelled as “A”, was filled with 400 mL water
water is 273 K at 1 bar, and Kf of water is 2 K kg/mol,
whereas the beaker labelled “B” was filled with 400 mL
which of the following graphs represents the depression in of 2 M solution of NaCl. At the same temperature both the
the freezing point of the water-ethylene glycol solution? beakers were placed in closed containers of same material
(Molar mass of ethylene glycol is 62 g/mol). and same capacity as shown in Fig.
(Un) (CBSE CFPQ, 2024) (Ap) (NCERT Exemplar)
At a given temperature, which of the following statement
1 Water 1 is correct about the vapour pressure of pure water and that
Ice Ice Water of NaCl solution.
VP (in bar)

VP (in bar)

Water + Ethylene glycol

Water + Ethylene glycol (a) vapour pressure in container (A) is more than that in
container (B).
270 273 271 273 (b) vapour pressure in container (A) is less than that in
T (in kelvin) T (in kelvin) container (B).
A B (c) vapour pressure is equal in both the containers.
(d) vapour pressure in container (B) is twice the vapour
1
Water
1 pressure in container (A).
Ice Ice
VP (in bar)

VP (in bar)

Water
Water + Ethylene glycol

Water + Ethylene glycol

270 273 271 273

T (in kelvin) T (in kelvin)
C D

(a) A (b) B 8. The mixture which shows positive deviation from Raoult’s
(c) C (d) D law is: (Re)
(a) Benzene + Toluene
4. Which of the following should be done to change 100 ml of
0.1M KCl solution to 0.2M? (Un) (CBSE CFPQ, 2024) (b) Acetone + Chloroform
(c) Chloroethane + Bromoethane
(i) Reduce volume of solution to half by evaporation
(d) Ethanol + Acetone
(ii) Add 50 ml water
9. A mixture of acetone and chloroform forms a maximum
(iii) Add 0.1 mol KCl
boiling azeotrope at a specific composition. Which of these
(iv) Add 0.01 mol KCl is CORRECT for the mixture? (Un) (CBSE CFPQ, 2024)
(a) (i) and (iii) (b) (i) and (iv) (a) Change in volume on mixing will be positive.
(c) (ii) and (iv) (d) Any of (i), (ii), (iii) and (iv) (b) Change in enthalpy on mixing will be positive.
P
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 (c) Interaction between unlike molecules is stronger than 14. On the basis of information given below mark the correct
that between like molecules in the mixture. option.
(d) The proportion of acetone and chloroform in the mixture Information: On adding acetone to methanol some of the
in the liquid phase is not the same as in the vapour phase hydrogen bonds between methanol molecules break.
10. Consider the Fig. 2.1 and mark the correct option. (Re) (NCERT Exemplar)
(Un) (NCERT Exemplar) (a) At specific composition, methanol-acetone mixture will
(a) water will move from side (A) to side (B) if a pressure form minimum boiling azeotrope and will show positive
lower than osmotic pressure is applied on piston (B). deviation from Raoult’s law.
(b) water will move from side (B) to side (A) if a pressure (b) At specific composition, methanol-acetone mixture
greater than osmotic pressure is applied on piston (B). forms maximum boiling azeotrope and will show
(c) water will move from side (B) to side (A) if a pressure positive deviation from Raoult’s law.
equal to osmotic pressure is applied on piston (B). (c) At specific composition, methanol-acetone mixture will
(d) water will move from side (A) to side (B) if pressure form minimum boiling azeotrope and will show negative
equal to osmotic pressure is applied on piston (A). deviation from Raoult’s law.
(d) At specific composition, methanol-acetone mixture
will form maximum boiling azeotrope and will show
SPM negative deviation from Raoult’s law.
15. An azeotropic mixture of two liquids will have a boiling
point lower than either of the two liquids when it
(Re) (CBSE, 2023 & 2022 Term-I)
(a) shows a negative deviation from Raoult’s law
(b) forms an ideal solution
(c) shows a positive deviation from Raoult’s law
Fig.; 2.1 (d) is saturated
16. Van’t Hoff factor of Ca(NO3)2 is: (Un)
11. In the following diagram point, ‘X’ represents.
(a) One (b) Two
(Un) (CBSE, 2022 Term-I)
(c) Three (d) Four
17. Which one of the following is incorrect for ideal solution?

Vapour (a) ∆P = Pobs – Pcalculated by Raoult’s law = 0 (Un)

Pressure X (b) ∆Smix = 0
(c) ∆Hmix = 0

Temperature (d) ∆Umix = 0

18. In a laboratory experiment, a student is investigating osmotic
(a) Boiling point of solution
pressure. They have set up a special apparatus and obtained
(b) Freezing point of solvent some data. The student now needs to express the osmotic
(c) Boiling point of solvent pressure in the correct unit. What unit should the student use
(d) Freezing point of solution to express osmotic pressure? (Re)
12. Elevation in boiling point of 1 M aqueous glucose solution (a) Mol L–1 (b) Calorie
(ρ = 1.2 g/mL) is-
(c) Kg mol–1 (d) Atm
(a) Kb (b) 1.2 Kb 19. A solution of urea contain 8.6 g/litre (mol. wt. 60.0). It is
(c) 1.02 Kb (d) 0.98 Kb isotonic with a 5% solution of a non-volatile solute. The
13. 342.3 g of sucrose is dissolved in 1 kg of water in a pot molecular weight of the solute will be (Un)
to form a solution. The boiling point of water (solvent) is (a) 348.9 (b) 34.89
373.15 K. Which of the following is likely to be the boiling
(c) 3489 (d) 861.2
point of the solution? (An) (CBSE CFPQ, 2024)
(Molar mass of sucrose= 342.3 g/mol; Atmospheric 20. The values of Van’t Hoff factors for KCl, NaCl and K2 SO4,
pressure= 1.013 bar; Kb = 0.52 K kg mol–1) respectively, are _______ . (Un) (NCERT Exemplar)
(a) 373 K (b) 373.15 K (a) 2, 2 and 2 (b) 2, 2 and 3
(c) 373.67 K (d) 372.63 K (c) 1, 1 and 2 (d) 1, 1 and 1

47 Solutions P
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 (b) Potassium chloride solution freezes at a lower
Assertion and Reason  (1 M) temperature than water.
Direction: The following questions consist of two statements 3. Rakesh took 20 g of solute A to prepare a 50 ml solution. This
– Assertion (A) and Reason (R). Answer these questions by solution is isotonic to another solution of the same volume
selecting the appropriate option given below: with a weight of 40 g of a different solute B.
(i) If both the solution is prepared at the same temperature,
(a) Both A and R are true, and R is the correct explanation of A. then, what is the ratio of molecular mass of solute A to
(b) Both A and R are true, but R is not the correct explanation that of B?
of A. (ii) If the two solutions are placed at different temperatures,
(c) A is true, but R is false. keeping all other variables constant, and separated by
(d) A is false, but R is true. SPM, will the osmosis happen, and why?
(Ap) (CBSE CFPQ, 2024)
1. Assertion (A): Molarity of a solution in liquid state changes
4. The depression in the freezing point of water observed for
with temperature.
the same amount of acetic acid, trichloroacetic acid and
Reason (R): The volume of a solution changes with change trifluoroacetic acid increases in the order given above.
in temperature. (Un) Explain briefly. (Un) (NCERT)
2. Assertion (A): When NaCl is added to water, a depression 5. What happens when
in freezing point is observed. (i) a pressure greater than osmotic pressure is applied on the
Reason (R): The lowering of vapour pressure of a solution solution side separated from solvent by a semipermeable
causes depression in the freezing point. (Un) membrane?
(ii) acetone is added to pure ethanol? (Un) (CBSE, 2020)
3. Assertion (A): When methyl alcohol is added to water,
boiling point of water increases. 6. State Henry’s law. Calculate the solubility of CO2 in water
at 298 K under 760 mm Hg.
Reason (R): When a volatile solute is added to a volatile
solvent, vapour pressure of solvent increases. (Un) (KH for CO2 in water at 298 K is 1.25 × 106 mm Hg
(Ev) (CBSE, 2020)
4. Assertion (A): The freezing point is the temperature at which
a solid dissolves in the solution. 7. ‘Colligative properties help in determining the molar masses
of the solutes.’ The method based on which colligative
Reason (R): Depression of freezing point for dilute solution
property is preferred over others for determining molar
is directly proportional to molality, m of the solution. (Re)
masses of biomolecules and why? (Un) (CBSE APQ, 2023)
5. Assertion (A): When a solution is separated from the 8. (a) Write the colligative property which is used to find the
pure solvent by a semipermeable membrane, the solvent molecular mass of macromolecules.
molecules pass through it from pure solvent side to the
(b) In non-ideal solution, what type of deviation shows the
solution side.
formation of minimum boiling azeotropes?
Reason (R): Diffusion of solvent occurs from a region of
(Re) (CBSE, 2016)
high concentration solution to a region of low concentration
solution. (Re) (NCERT Exemplar) 9. A 5% solution of Na2SO4.10H2O (M.W = 322) is isotonic
with 2% solution of non-electrolytic, non volatile substance
6. Assertion (A): Osmotic pressure is a colligative property.
X. Find out the molecular weight of X. (Ev)
Reason (R): Two solutions having the same osmotic
pressure at a given temperature are called isotonic solutions. Short Answer Type Questions (3 M)
(Re)
1. A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water
7. Assertion (A): A raw mango placed in a saline solution loses has a freezing point of 271.15 K. Calculate the freezing point
water and shrivel into pickle. of 5% glucose (M = 180 g mol–1) in water.
Reason (R): Through the process of reverse osmosis, raw (Ev) (CBSE, 2019)
mango shrivel into pickle. (Ap) (CBSE, 2022-Term-1) (Given: Freezing point of pure water = 273.15 K)
2. The air is a mixture of a number of gases. The major
Subjective Questions components are oxygen and nitrogen with approximate
proportion of 20% is to 79% by volume at 298 K. The water
Very Short Answer Type Questions  is in equilibrium with air at a pressure of 10 atm. At 298 K,
(2 M) if the Henry’s law constants for oxygen and nitrogen are
1. Define the following terms: (Re) 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate
(i) Ideal solution the composition of these gases in water. (Ev) (NCERT)
(ii) Molarity (M) 3. The freezing point of a solution containing 5 g of benzoic
acid (M = 122 g mol–1) in 35 g of benzene is depressed by
2. Give reasons: (Ap) (CBSE, 2019) 2.94 K. What is the percentage association of benzoic acid
(a) A decrease in temperature is observed on mixing ethanol if it forms a dimer in solution?
and acetone. (Kf for benzene = 4.9 K kg mol–1) (Ev) (CBSE, 2020)
P
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 4. A 0.01 m aqueous solution of AlCl3 freezes at –0.068°C. (c) 
On addition of solute to solvent during dissolution,
Calculate the percentage of dissociation. [Given: Kf for some solute particles separate out from the solution
Water = 1.86 K kg mol–1 ] (Un) (CBSE, 2020) due to crystallisation. At equilibrium, what happens to
5. Answer the following questions: the concentration of solute in the solution at a given
temperature and pressure? (Ev)
(a) 
Give an example of the liquids which when mixed result
in an endothermic process, What will be the change in 5. The following table contains osmotic pressure data for three
volume when the liquids are mixed? compounds dissolved in various solvents.
(An) (CBSE APQ, 2023)
(b) 
At 300K, what is the relation between the osmotic
pressure of two equimolar solutions, one whose Van’t Compounds Concentration, C Osmotic pressure
Hoff factor is 2 and for the other is ½? (g/L) (atm)
(c) 
Which of the two aqueous solutions has a higher melting Cellulose 12.5 0.0021
point: 2molal glucose solution or 3 molal sucrose
Protein 28.5 0.0026
solution? Why? (Ap) (CBSE APQ, 2023)
Haemoglobin 5 0.0018
Long Answer Type Questions (5 M) (R = 0.083 L bar mol-1 K-1)
1. (a) A
 mongst the following compounds, identify which are (a) If the concentration of protein is doubled keeping all
insoluble, partially soluble and highly soluble in water? other variables constant, what will be the osmotic
pressure of the new solution?
(Un) (NCERT)
(b) When one litre of cellulose solution was heated to 315 K,
(i) phenol (ii) toluene its osmotic pressure changed to 0.00248 atm. What is
(iii) formic acid (iv) ethylene glycol the molecular mass of the cellulose in the solution?
(v) chloroform (vi) pentanol. (c) A solution of 10 g of protein in a litre of solvent was
(b) The boiling point of benzene is 353.23 K. When 1.80 found to be isotonic to the haemoglobin solution given
g of a non-volatile solute was dissolved in 90 g of benzene, above in the table, at the same temperature. If the
molecular weight of the protein is 130,000 g/mol, what
the boiling point is raised to 354.11 K. Calculate the molar
is the molecular weight of haemoglobin.
mass of the solute. Kb for benzene is 2.53 K kg mol–1
OR
2. (a) What is the effect of temperature on the solubility of The relation between the osmotic pressure of three solutions
glucose in water? A, B, and C is:
(b) Ibrahim collected a 10mL each of fresh water and ocean πB < πC
water. He observed that one sample labeled “P” froze at 0°C πC > πA
while the other “Q” at –1.3°C. Ibrahim forgot which of the πA> πB
two, “P” or “Q” was ocean water. Help him identify which
The three solutions have the same molarity and are at the
container contains ocean water, giving rationalization for
same temperature.
your answer.
(a) For which of the solutions is the value of ‘i’ expected
(c) Calculate Van’t Hoff factor for an aqueous solution of to be the greatest? Give a reason.
K3[Fe(CN)6] if the degree of dissociation (α) is 0.852. What
(b) Which of the solutions is MOST LIKELY to be glucose,
will be boiling point of this solution if its concentration is 1
potassium sulphate, and sodium chloride?
molal? (Kb = 0.52 K kg/mol) (Ev) (CBSE SQP, 2023)
(c) Which of the solutions is expected to give a vapour
3. (a) Depict the elevation in boiling point colligative property
pressure-mole fraction graph similar to that of an
using a diagram. acetone-chloroform mixture? Give reason.
(b) Calculate the osmotic pressure of a 0.5L solution made
of 18g glucose, 6g urea and 34.2g sucrose at 300 K?
Case Based Questions
(c) Differentiate between osmosis and reverse osmosis.
(Un) Case Based-I: The concentration of a solute plays a crucial role
in the study of chemical reactions as it governs the frequency of
4. (a) 
T he vapour pressure of pure water at a certain
molecular collisions within a solution, indirectly influencing the
temperature is 23.80 mm Hg. If 1 mole of a non-volatile
reaction rate and equilibrium conditions. Expressing the amount
non-electrolytic solute is dissolved in 100g water,
of solute in a solution can be achieved through various methods,
Calculate the resultant vapour pressure of the solution. providing valuable insights into the composition of the solution.
(CBSE SQP, 2023) Some common ways to express concentration include molarity,
30 g of urea is dissolved in 846 g of water. Calculate
(b)  molality, parts per million, mass percentage, and volume
the vapour pressure of water for this solution if vapour percentage, each serving distinct purposes in different scenarios.
pressure of pure water at 298 K is 23.8 mm Hg. (Ap)

49 Solutions P
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 (a) Which method of expressing concentration is commonly not appreciate that Henry’s constant is not a true constant but has a
used in laboratory experiments and industrial applications? significant non-linear temperature dependence. Figure 1 illustrates
(1 M) some typical behavior of Henry’s constant vs temperature for
OR several solutes in water. The Henry’s constant typically increases

When is molality preferred over molarity for expressing with temperature at low temperatures, reaches a maximum, and
concentration? (1 M) then decreases at higher temperatures. The temperature at which
(b) In an experiment, a chemist prepares a solution by dissolving the maximum occurs depends on the specific solute-solvent pair.
2 grams of salt in 250 grams of water. What is the mass Clearly, the use of a Henry’s constant that was derived at 25°C
percentage of the solute in the solution? (1 M) at a different temperature could lead to serious errors during
manufacturing. Even a variation as small as 10 K can cause the
(c) (i)  A chemist wants to prepare a solution of hydrochloric
Henry’s constant to change by a factor of two, which could have
acid (HCl) in water with a mass percentage of 25%.
a serious impact on many process designs.
The chemist has 200 grams of water. How much
hydrochloric acid (in grams) should the chemist add (Ev) (CBSE APQ, 2023)
to achieve the desired mass percentage in the final
solution? (1 M)
(ii) The concentration of lead (Pb) in a water sample is found
to be 12 parts per million (ppm). If the total volume of
the water sample is 500 liters, calculate the mass of lead
present in the sample. (1 M)
(Given: Molar Mass of Pb = 207.2 g/mol)
Case Based-II: An ideal solution is one that follows Raoult’s law
exactly across the whole concentration range. Non-ideal solutions
are those in which the vapour pressure is either higher or lower
than expected by Raoult’s law. Non-ideal solutions can deviate
from Raoult’s law by showing either a positive or negative
deviation, depending on whether the A-B interactions in solution Temperature dependence of Henry’s constant
are greater or weaker than the A-A and B-B interactions. (Re) for several solutes in water.
(a) Provide an example of a solution exhibiting negative
(a) Which of the following is not a true constant like Henry’s
deviations. (1 M)
constant (1 M)
(b) What is a non-ideal solution? (1 M)
(c) (i) Why do some solutions show positive deviations from (i) speed of light (c) (ii) Planck’s constant (h)
Raoult’s law? (iii) rate constant (k) (iv) gas constant (R)
(ii) How does positive deviations impact vapour pressure? (b) How will the solubility of n-octane vary with a change in
(2 × 1 = 2) temperature, provided the pressure is kept constant? (1 M)
OR OR

Contrast positive and negative deviations from Raoult’s law
Rishita works in an aerated drinks factory. To increase the
and discuss the key factors that determine the direction of fizz in the drink, she proposes to bottle the aerated drinks at
the deviation in a given solution. (2 M) 40°C instead of 20°C. Do you support her proposal? Why
Case Based-III: Temperature dependence of Henry’s law constant or why not?  (1 M)
Henry’s law and Henry’s law constant are widely used in chemical (c) Find the ratio of solubility of toluene in water at 20°C and
and environmental engineering. Unfortunately, many people do 60°C. (2 M)

P
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 Answer keys

Multiple Choice Questions

1. (a) 2. (a) 3. (c) 4. (b) 5. (c) 6. (a) 7. (a) 8. (d) 9. (c) 10. (b)
11. (d) 12. (d) 13. (c) 14. (a) 15. (c) 16. (c) 17. (b) 18. (d) 19. (a) 20. (b)

Assertion and Reason

1. (a) 2. (a) 3. (d) 4. (d) 5. (c) 6. (b) 7. (c)

Hints & Explanations

(iv) Add 0.01 mol KCl: Adding 0.01 mol KCl to the
Multiple Choice Questions existing 0.01 mol KCl:
Total moles = 0.01 moles + 0.01 moles = 0.02 moles
1. (a) Mole fraction is used in Raoult’s Law to relate the
concentration of a solution to its vapour pressure. The volume remains the same at 0.1 L, so the concentration
will be:
2. (a) Value of Henry’s constant increases with increase in
0.02 moles
temperature. New concentration = = 0.2M
0.1L
3. (c) Number of moles of Ethylene glycol = 93/62 =1.5 mol This method will also work.
Molality of ethylene glycol = 1.5/1 = 1.5 m 5. (c) The higher the value of KH, the lower is the solubility
ΔTf = Kf x m = 2 K kg/mol × 1.5 mol/kg = 3 K of the gas in the liquid. Hence, the order of increasing
Hence, the freezing point of the solution = 270 K solubility of the gases will be Ar < CO2 < CH4 < HCHO.
From the provided graphs: 6. (a) K kg mol–1 or K (molality)–1
z Graph A shows a depression to around 270 K, but the 7. (a) Since NaCl is a non-volatile solute, it will reduce
representation line for the solution is wrong. vapour pressure of the solution. Hence, vapour pressure of
z Graph B shows a depression to around 272 K. pure water in container A is more than in container B.
z Graph C shows a depression to 270 K. 8. (d) Ethanol-acetone mixture shows positive deviation due
z Graph D shows a depression to around 271 K but the to weaker ethanol-acetone interaction in comparison to
representation line for both the solution and solvent is ethanol-ethanol or acetone-acetone interaction.
wrong.
9. (c) Interaction between unlike molecules is stronger than that
4. (b) (i) Reduce the volume of solution to half by evaporation: between like molecules in the mixture.
If we reduce the volume to half, the new volume will be:
10. (b) Water will move from side B (concentrated sodium
100 ml/2 = 50 ml = 0.05 L chloride solution) to side A (fresh water) if a pressure
The concentration will be: greater than osmotic pressure is applied on piston B.
moles of solute 0.01moles 11. (d) Here, Tfo is the freezing temperature of solvent and Tf is
New concentration = = = 0.2 M
New volume 0.05L the freezing temperature of solution, thus X represents the
This method will work. freezing point of solution.

51 Solutions P
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 lven
t Assertion and Reason
id so
Iiqu
tion
solu n

Vapour pressure
e nt 1. (a) Molarity = × 100 .
olv
frozen s v
As volume change with temp. so molarity also change with
∆Tf
temp.
Tf Tf0
Temperature/K 2. (a) Addition of non-volatile solute leads to decrease in
12. (d) 1 M aqueous glucose solution means 1 mole of glucose vapour pressure of a solution which may causes depression
is present per litre of solution. in freezing point.
So, no. of moles of solute = 1 3. (d) When methyl alcohol (volatile) is added to water,
∴ Weight of solute = 180 g the boiling point of water decreases because vapour
pressure increases when volatile solute is added to
ρ of solution = 1.2 g/mL
volatile solvent.
So, 1 mL of solution weighs = 1.2 g
4. (d) The freezing point of a substance may be defined as the
1000 mL of solution weighs = 1.2 × 1000 = 1200 g temperature at which the vapour pressure of the substance
∴ Wt. of solvent = 1200 – 180 = 1020 g in its liquid phase is equal to its vapour pressure in the solid
ΔTb = Kbm phase.
1 1000 5. (c) Solvent molecules pass through the semipermeable
= Kb × ⇒ Kb ×
1020 / 1000 1020 membrane from the region of low concentration solution to
the region of high concentration solution.
ΔTb = 0.98 Kb
6. (b) Both A and R are true, but R is not the correct
13. (c) ∆Tb = Kb × m = 0.52 × (342.3/342.3 х1) = 0.52 K
explanation of A.
Also, ΔTb = Tb(solution) – Tob(solvent) = 0.52
7. (c) A is true but R is false. When a raw mango is placed in
Hence, Tb(solution) = 0.52 + 373.15 = 373.67 K a saline solution to prepare pickles then the mango loses
14. (a) At specific composition, methanol-acetone mixture will water due to osmosis and gets shrivel.
show positive deviation from Raoult’s law as it has lesser
interactions than methanol- methanol and acetone-acetone
Subjective Questions
interactions. Hence it forms a minimum boiling azeotrope.
15. (c) An azeotropic mixture of two liquids will have a Very Short Answer Type Questions
boiling point lower than either of the two liquids when it
1. (i) A n ideal solution is a solution that follows Raoult’s
shows a positive deviation from Raoult’s law.
Law at all concentrations for both solute and solvent.
16. (c) Ca ( NO3 )2 
 Ca 2+ + 2NO-3 (1 M)
(ii) Molarity (M) is a measure of the concentration of a
Its van’t hoff factor is 3 because it furnishes three ions per solute in a solution.
formula unit. It is defined as the number of moles of solute dissolved
17. (b) In case of an ideal solution ∆Hmix = 0 in one litre of solution,
moles of solute
∆Umix = 0 but ∆Smix ≠ 0 M= (1 M)
Volume of solution (in l)
18. (d) The osmotic pressure is expressed in the unit of 2. (a) When ethanol and acetone are mixed, hydrogen bonds
atmosphere (atm). form between ethanol’s hydroxyl groups and acetone’s
19. (a) If two non-electrolytic solution are isotonic, carbonyl groups, releasing energy. This exothermic
interaction results in a decrease in temperature as the
Then, C1 = C2 system stabilizes. (1 M)
8.6 5 × 1000 (b) Potassium chloride has low freezing point because of
=
Therefore, = \ M.wt. = 348.9 the colligative property which states that when a solute such
60 × 1 M.wt. × 100
as KCl is added to solvent such as water then according to
Hence, molecular weight of the solute = 348.9 g colligative property there is a lowering in freezing point.
20. (b) Van’t Hoff factors for ionic compounds are equal to the Freezing point depends on the inter-molecular force of
number of ions produced when the compound dissolves in attraction between the liquid. When solute KCl is added to
solvent water then it decreases the inter-molecular force of
solution. KCl and NaCl both dissociate into 2 ions each,
attraction between the molecules, hence the freezing point
while K2SO4 dissociates into 3 ions (2K + and1SO 24 − ) . decreases (1 M)
P
W CBSE Class – XII CHEMISTRY 52
3. (i) For any solution, osmotic pressure is given by π = (w/ It is preferred for biomolecules as the pressure measurement
MV) × RT; w = weight of the solute, V = volume of solution, is done around room temperature and biomolecules are
R = gas constant and T is temperature generally not stable at higher temperatures. (1 M)
For two solutions to be isotonic, π1 = π2 (½ M) 8. (a) The colligative property known as osmotic pressure
is utilized for determining the molecular mass of
⇒ 20/M1 = 40/M2
macromolecules. (1 M)
⇒ M1/M2 = 1/2 (½ M) (b) In non-ideal solutions, a positive deviation indicates the
(ii) Yes, because at different temperatures the solutions are formation of minimum boiling azeotropes. (1 M)
no longer isotonic and hence there will be movement of Topper’s Explanation
particles through osmosis. (1 M)
4. Within the trio of H, Cl, and F, H exhibits the lowest
electronegativity, in contrast, F ranks as the highest. This
makes F superior in attracting electrons compared to Cl and
H, enabling trifluoroacetic acid to readily release H+ ions,
thereby enhancing its ionisation. Consequently, a higher
ion production correlates with a more significant reduction
in freezing point, leading to the sequence: Acetic acid <
Trichloroacetic acid < Trifluoroacetic acid. (2 M) 9. p1 = p2 (½ M)
iC1RT = C2RT (½ M)
3 5 2
Nailing the Right Answer  (½ M)
322 M
U To ensure students get full credit, they should
2 × 322
explain that the depression in the freezing point M= ⇒ M = 42.9 g (½ M)
depends on the extent of ionization of the acids. 3× 5
Trifluoroacetic acid, being the strongest due to the
electron-withdrawing effect of fluorine, ionizes Short Answer Type Questions
the most, followed by trichloroacetic acid and then
acetic acid. The higher the ionization, the greater n
1. Molality(m) = (½ M)
the number of particles in the solution, leading to Wsolvent (in kg)
a larger freezing point depression. Students should For sucrose solution:
mention the relationship between ionization, 4
number of particles, and colligative properties. 342 4 1000
m= = × =0.122m
96 342 96
5. (i) I f the pressure greater than osmotic pressure is applied
1000
on the solution side, the direction of osmosis can be
(∆TF)1 = (273.15 – 271.15)K = 2K
reversed. At this stage, pure solvent will flow out of
the solution through a semipermeable membrane and (∆TF)1 = KF.m = KF × 0.122
this phenomenon is known as reverse osmosis. (1 M) 2 = KF × 0.122 ...(i)
(1 M)
(ii) In pure ethanol, H-bonding is present. When acetone is For glucose solution:
added, its molecules get in between the host molecules
5
which results in the breaking of some of the H-bonds.
180 5 1000
Due to this weakening of interactions,the solution will m= = × =0.292m
95 180 95
show positive deviation from Raoult’s law. (1 M)
1000
6. Henry’s law states that “the partial pressure of the gas in
vapour phase (p) is proportional to the mole fraction of the (∆TF)2 = KF × 0.292 ...(ii) (½ M)
gas (χ) in the solution” (½ M) Eqn (ii) / Eqn (i)
Mathematically, ( ∆TF )2 K F × 0.292
p = KH χ =
2 K F × 0.122
(½ M)
Here, KH is Henry’s law constant. (½ M)
0.292 × 2
Substitute the given values in above equation: ( ∆=
TF ) = 4.79
0.122
760 mm Hg ∴TF = 273.15 – 4.79 = 268.36K (½ M)
χ = p /KH = = 6.08 × 10−4 (1 M)
1.25 × 106 mm Hg Freezing point of glucose solution will be 268.36 K.
7. The method based on osmotic pressure is preferred over others 2. KH for O2 = 3.30 × 107 mmHg
for determining molar masses of biomolecules. (1 M) KH for N2 = 6.51 × 107 mmHg

53 Solutions P
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 Percentage of N2 = 79 i −1
Percentage of O2 = 20 α= … (ii)  (½ M)
n −1
Total pressure = 10 atm AlCl3 will undergo dissociation as follows:
1 atm = 760 mmHg AlCl3 → Al3+ + 3Cl–
Therefore, total pressure = 10 × 760 = 7600 mmHg From equation (ii) we get,
The partial pressure of oxygen, PO = (20/100) × 7600 i − 1 3.65 − 1 2.65
2 (½ M)
mmHg = 1520 mmHg (½ M) =
α = = = 0.883
n −1 4 −1 3
The partial pressure of Nitrogen, PN = (79 /100) × 7600 mmHg
2 Therefore, the percent dissociation = 88.3% (½ M)
= 6004 mmHg (½ M)
5. (a) Ethanol and acetone, or any other example, the volume
By using the Henry law,
will increase (The liquids show positive deviation from
Mole fraction of oxygen,
Raoult’s law) (1 M)
XO = PO / ΚΗ (½ M)
2 2
7 –5
(b) π1/π2 = i1/i2 = 4 (1 M)
1520 / (3.30 × 10 ) = 4.61 × 10  (½ M)
(c) 2 molal glucose because it will have lower depression
XN = PN /KH = 6004 / (6.51 × 107) = 9.22 × 10−5 (1 M)
2 2 in freezing point, higher melting point . (1 M)

Long Answer Type Questions

Key Takeaways 1. (a) (i) Phenol (C6H5OH) has the polar group −OH and
This question teaches students that to calculate the non-polar group –C6H5. Thus, phenol is partially soluble in
composition of gases in water, they must apply Henry’s water. (½ M)
law. Specifically, it involves converting total atmospheric (ii) Toluene (C6H5–CH3) has no polar groups. Thus, toluene
pressure into partial pressures of oxygen (PO2 = (20/100)
is insoluble in water. (½ M)
× 7600 mmHg) and nitrogen (PN2 = (79/100) × 7600 (iii) Formic acid (HCOOH) has the polar group −OH and
mmHg), then using the Henry’s law formula pA = KH × χA can form H-bond with water.
to find the molar fractions (χO = PO / KH(O ) and χN = Thus, formic acid is highly soluble in water. (½ M)
2 2 2 2
(iv) Ethylene glycol has a polar −OH group and can form
PO2 / KH(N )). This illustrates the relationship between gas
2 an H−bond. Thus, it is highly soluble in water. (½ M)
pressure, solubility, and temperature.
(v) Chloroform is insoluble in water. (½ M)
3. First calculate the observed molar mass: (vi) Pentanol (C5H11OH) has a polar −OH group, but it
∆Tf = iK f m also contains a very bulky nonpolar –C5H11 group. Thus,
(½ M)
pentanol is partially soluble in water. (½ M)
2.94 = i × 4.9 × (5 × 1000/122 × 35)
This implies, i = 0.512 (1 M) (b) The elevation (DTb) in the boiling point = 354.11 K –
353.23 K = 0.88 K (½ M)
i −1 0.512 − 1 (−0.488) (½ M)
=α = = = 0.976 1000 × w 2 × K b
1/ n − 1 1 ( −0.5) M2 = (½ M)
−1 ∆Tb × w1
2
Percent association of acid = 97.6% (1 M) 2.53 K kg mol−1 × 1.8 g × 1000 g kg −1
= M 2 = 58 g mol−1
4. We know that the formula for depression in freezing point 0.88 K × 90 g
is given as follows: (½ M)
DTf = i × Kf × m … (i) (½ M) Therefore, molar mass of the solute, M2 = 58 g mol–1
Where, i = Van’t Hoff factor (½ M)
m = molality 2. (a) Addition of glucose to water is an endothermic reaction.
DTf = depression in freezing point According to Le Chatelier’s principle, on increase in
We also know that ∆Tf = (T f
0
)
− Tf . Now substituting the temperature, solubility will increase. (1 M)

values for calculating DTf. (b) Q is ocean water, due to the presence of salts it freezes at
lower temperature (depression in freezing point) (1 M)
∆Tf = 0 − ( −0.068 )  °C= 0.068°C (½ M) (c) K3[Fe(CN)6] gives 4 ions in aqueous solution (½ M)
Now, from equation (i) i = 1 +(n – 1)a  (½ M)
∆Tf 0.068 0.068 i = 1+(4 – 1) × 0.852
=i = = = 3.65 (½ M)
K f × m 1.86 × 0.01 0.0186 i = 3.556 (½ M)
for dissociation, the relation between van’t hoff factor and DTb = iKb m = 3.556 × 0.52 × 1 = 1.85°C (1 M)
degree of dissociation may be written as: Tb = 101.85°C (½ M)
P
W CBSE Class – XII CHEMISTRY 54
 n1 = 100/18
x2 = 0.1 / 5.55 + 0.1 = 0.1 / 5.65 = 0.018 (½ M)
Mistakes 101 : What not to do! P° = 23.8 mm Hg
Students might misidentify which container has ocean Relative lowering of vapour pressure = (23.80 – P) /
water by not recognizing that ocean water freezes at a 23.80 = 0.018 (½ M)
lower temperature due to the presence of salts. Students 23.80 – P = 0.42 (½ M)
might calculate Van’t Hoff factor incorrectly by not using P = 23.80 – 0.42 = 23.37 mm Hg (1 M)
the given degree of dissociation (α = 0.852) or using an (b) Given, wB = 30 g and wA = 846 g
incorrect formula for boiling point elevation, leading to
incorrect results. Also, MB = 60 g/mol and MA = 18 g/mol
p°A − pS n2
As, = (½ M)
3. (a) (1 M) p°A n1 + n 2
23.8 − ps (30 / 60)
−
23.8 (846 / 18) + (30 / 60)
p = 23.55 mm Hg  (½ M)
S
(c) At dynamic equilibrium, the number of solute particles
going into the solution will be equal to the solute
particles separating out. As a result, the concentration
of solute in the solution will remain constant. (1 M)
5. (a) Osmotic pressure = CRT
If the concentration is doubled without a change in
temperature, the osmotic pressure will also be doubled.
(b) Osmotic pressure, Π = C × R × T  (½ M) Thus, osmotic pressure of the new solution will be
C = (Number of moles of urea + number of moles of 0.0052 atm. (1 M)
glucose + number of moles of sucrose)/Volume(L) (b) M = wRT/(πV), (1 M)
6 18 34.2 where w is the mass of the solute taken, V is the volume
+ +
= C 60= 180 342.3 0.6M of the solution taken. In one litre of solution, there are
(½ M)
0.5 12.5 g of solute
Osmotic pressure, π \ M = 12.5 g × 0.083 L bar mol–1 K–1 × 315 K/
= 0.6 molL–1 ×0.0821L-atm K–1 mol–1 × 300 K(½ M) ((0.00248 ×1.01 bar)(1 L))
M = 130,474 g mol–1  (1 M)
= 14.77 atm (½ M)
(c) (c) M1 = w1RT/(π1V1) - (1) for haemoglobin
M2 = w2RT/(π2V2) - (2) for protein (1 M)
Reverse For isotonic solutions, osmotic pressure is equal.
Aspect Osmosis
Osmosis
Dividing we get,
Definition Movement of Movement of
M2/M1 = (C2RT)/(C1RT), where C1=w1/V1 and C2
solvent molecules solvent molecules
= w2/V2
from a region from a region
of lower solute of higher solute 130000/M1 = 10/5
concentration to concentration to \ M1 = 65000 g/mol (1 M)
a region of higher a region of lower OR
solute concentration solute concentra- (a) The value of ‘i’ is expected to be the highest for solution
through a tion by applying C. From the given relations we can conclude that
semipermeable external pressure. πC > πA > πB (1 M)
membrane
Since, the osmotic pressure of solution C is the highest,
Pressure Occurs naturally Requires external therefore the value of ‘i’ will be highest in solution C.
Application without external pressure greater (1 M)
pressure. than the osmotic (b) Solution C is most likely to be potassium sulphate.
pressure.
(½ M)
(1 + 1 = 2) Solution A is most likely to be sodium chloride.(½ M)
4. (a) Relative lowering of vapour pressure = (P° – P) / P° = x2 Solution B is most likely to be glucose. (½ M)
(½ M) (c) Solution B is most likely to give a vapour pressure-mole
x2 = n2/(n1 + n2) fraction graph similar to that of an acetone-chloroform
n2 = 0.1 mixture. (½ M)

55 Solutions P
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 Acetone-chloroform solution has strong H-bonding in it. mass of solute (m)
H-bonding is seen only in the glucose solution but not in the n= =6×10-3
Molar mass of solute(M.M)
potassium sulphate and sodium chloride solutions. Hence,
it is expected to give a vapour pressure-mole fraction graph m = M.M. × 6 × 10–3
similar to that of an acetone-chloroform mixture. (1 M) m = 207 × 6 × 10–3
m = 1.24 g (½ M)
Case Based Questions
Case Based-II
Case Based-I (a) An example of negative deviations is the mixture of acetone
(a) Molarity is the most commonly used method of expressing and chloroform. (1 M)
concentration in laboratory experiments and industrial
(b) A non-ideal solution is a mixture that deviates from Raoult’s
applications because it is easy to measure and use in
law, where the vapour pressure of the components differs
calculations involving chemical reactions, stoichiometry,
from what is predicted by the ideal behaviour.  (1 M)
and solution preparation.
(c) (i) P ositive deviations occur when the interactions
OR
between unlike molecules (A-B) are weaker than
(a) Molality is preferred over molarity when temperature those between like molecules (A-A and B-B).
changes significantly because molality remains constant, (1 M)
while molarity changes with temperature due to change in (ii) Weaker A-B interactions lead to a higher tendency for
volume. (1 M) molecules to escape into the vapour phase, resulting
Mass of solute in an increased vapour pressure compared to Raoult’s
(b) Mass percentage
= × 100 law predictions. (1 M)
Total mass of solution OR
(½ M)
Total mass of Solution = mass of solute + mass of water (c) Positive deviations result in higher vapour pressures than
predicted, while negative deviations lead to lower vapour
= 2g + 250g = 252 g
pressures. The direction of deviation depends on the strength
2 of A-B interactions compared to A-A and B-B interactions.
=
Mass percentage × 100 = 0.79% (approx.) (½ M)
252 Weaker A-B interactions cause positive deviations, whereas
(c) (i) Let the mass of HCl = x g stronger A-B interactions cause negative deviations. (2 M)
mass of solute Case Based-III
=
Mass percentage × 100 (½ M)
Total mass of solution (a) (iii) Rate constant (k), it changes with temperature (1 M)
Total mass of solution = mass of solute + mass of H2O (b) The solubility will decrease with temperature at low
x temperatures, reach a minimum value and then increase at
25 = × 100
200 + x higher temperatures because the Henry’s constant typically
25(200 + x) = 100x increases with temperature at low temperatures, reaches
a maximum, the temperature at which maxima occurs for
5000 + 25x = 100x
n-octane-water pair is nearly 90°C and then decreases at
5000 = 75x higher temperatures. (1 M)
5000 OR
=x = 66.6 g
75 (b) It is not a good proposal. (½ M)
(½ M)
KH for CO2 is higher at 40°C,
n( solute )
(ii) conc. (ppm)
= × 106 (½ M) so the solubility of CO2 will be lower, thus fizz be less.
V( sol .) (½ M)
n (c) p = KH χ (1 M)
12
= × 106
500
 20°C, KH = 50 MPa, and at 60° C, KH = 100 M Pa
At
12 × 500 KH(60) / KH(20) = χ(20) / χ(60)
n= = 6 × 10−3
106 100 / 50 = 2 = χ(20) / χ(60) (1 M)

P
W CBSE Class – XII CHEMISTRY 56
Inverse Trigonometric
Functions
2
“Military and ballistics use inverse
trigonometry to calculate launch angles
for projectiles. In battles and shooting,
like when cannons or missiles are fired,
getting the right launch angle greatly
matters. Figuring out this angle is
where inverse trigonometry jumps in. It
helps experts decide how to aim these
projectiles accurately.”

SYLLABUS &
WEIGHTAGE

Years
List of Concept Names

2020 2022 2023 2024 2025

Domain, Range ,Principal Value and Properties of

Inverse Trigonometric Functions 1 Q (1 M) 1 Q (1 M) 1 Q (1 M)
– 1 Q (1 M)
(Definition, range, domain, principal, value branch, 1 Q (2 M) 1 Q (2 M) 1 Q (2 M)
Graphs of inverse trigonometric functions)
 To Access One
CONCEPT MAP Shot Revision Video
Scan This QR Code

y y
y y
π p
p
2 π
π 2
x' x x' 2
x π
–1 0 1
–1 0 1 –2 –1 2
x' x
–π 0 1 2
x'
0
x
–1 1
2
–π
y' y' 2
y'
y = sin–1 x y = cos–1 x
y' y = sec–1 x:
y
y y= cosec–1 x
π
π
2
2
Function Domain Range

Inverse Trigonometric
Domain and Range of
x' x  π π
–4 –3 –2 –1 0 1 2 3 4 x' x sin–1 x [–1, 1]  − 2 , 2 
–4 –3 –2 –1 0 1 2 3 4

Functions
−π cos–1 x [–1, 1] [0, p]
2
 π π
y' y' tan–1 x (–∞, ∞)  − , 
2 2
y = tan–1 x y = cot–1 x
cot–1 x (–∞, ∞) (0, p)

π
(i) sin x : R → [–1, 1]
sec–1 x R – (–1, 1) [ 0, π ] −  2 
Trigonometric Function

 
Graph of Inverse
(ii) cos x : R → [–1, 1]
Trigonometric Function  π π
(iii) tan x : R –  x : x = π 
(2n + 1) , n ∈ Z  → R
cosec–1 x R – (–1, 1)  − ,  − {0}
 2 
 2 2

(iv) cot x : R – {x : x = np, n∈Z} → R

INVERSE
(v) sec x : R –  x : x = π 
(2n + 1) , n ∈ Z  → R – (–1, 1) TRIGONOMETRIC Some Important Properties
 2  FUNCTIONS
(vi) cosec x : R – {x : x = np, n∈Z} → R – (–1, 1) ‰ Property IV:
(i) sin–1 (–x) = –sin–1 x, x ∈ [-1,1]
(ii) cos–1 (–x) = π – cos–1 x, x ∈ [-1, 1]
(iii) tan–1 (–x) = –tan–1 x, x ∈ R
‰ Property I: (iv) cosec–1 (coses q) = q, (iv) cosec–1 (–x) = –cosec–1 x, |x| ≥ 1
1 for all q ∈  π , π  , ≠ 0 (v) sec–1 (–x) = π – sec–1x, |x| ≥ 1
(i) sin–1 = cosec–1x, x ≥ 1 2 2
x   (vi) cot–1 (–x) = π – cot–1x, x ∈ R
or x ≤ –1 (v) sec–1 (sec q) = q, ‰ Property V:
π
(ii) cos–1
1
= sec–1x, x ≥ 1 for all q ∈ [ 0, π] , θ ≠ π
(i) sin–1 x + cos–1 x =
, for all x ∈ [–1,1]
x 2
(vi) cot–1 (cot q) = q, 2
or x ≤ –1 for all q ∈ (0, p) (ii) tan–1 x + cot–1 x = π , for all x ∈ R
2
1
(iii) tan–1 = cot–1x, x > 0
‰ Property III: (iii) sec–1 x + cosec–1 x = π , for all x ∈ ( – ∞, –1] ∪ [ 1, ∞ )
x (i) sin (sin–1 x) = x, 2
‰ Property II: for all x ∈ [– 1, 1] ‰ Property VI:
(i) sin–1 (sin q) = q, (ii) cos (cos–1 x) = x, x+ y
(i) tan–1x + tan–1 y = tan–1 , xy < 1
 π π for all x ∈ [– 1, 1] 1 − xy
for all q ∈  – , 
 2 2 (iii) tan (tan–1 x) = x, (ii) tan–1x – tan–1 y = tan–1 x − y xy > – 1
for all x ∈ R 1 + xy
(ii) cos–1 (cos q) = q,
(iv) cosec (cosec–1 x) = x, (iii) 2tan–1x = sin–1 2 x 2 , x ≤ 1
for all q ∈ [0, p ] 1+ x
for all x ∈ ( – ∞, –1] ∪ [1, ∞ )
2tan–1 x = cos–1 1 − x 2 , x ≥ 0
2
(iii) tan–1 (tan q) = q, (v) sec (sec–1 x) = x, (iv)
1+ x
for all q ∈  – π , π  for all x ∈ ( – ∞, –1] ∪ [1, ∞ ) (v) 2tan–1 = tan–1
2x
  , –1 < x < 1
 2 2 (vi) cot (cot–1 x) = x, for all x ∈ R 1− x
2

P
W CBSE Class – XII MATHEMATICS 56
 1 DOMAIN, RANGE, PRINCIPAL VALUE AND PROPERTIES OF
INVERSE TRIGONOMETRIC FUNCTIONS

Important Terms
‰ Inverse of a Function: If f: X → Y is one-one and onto (bijective) function, then there exists a unique function
f–1 : Y → X which assigns each element y ∈ Y to a unique element x ∈ X such that f(x) = y and is called inverse function of f.
i.e., f–1(y) = x ⇒ f(x) = y, x ∈ X and y ∈ Y
‰ Inverse Trigonometric Functions: Inverse trigonometric functions are defined as the inverse of their respective trigonometric
functions. For example, arcsin(x) gives the angle whose sine is x.
Inverse trigonometric functions are often denoted as arcsin(x), arccos(x), and arctan(x), but sometimes you might also see them
written as sin⁻¹(x), cos⁻¹(x), and tan⁻¹(x).

Important Concepts

‰ Principal Value Branches: Since trigonometric functions being periodic are in general not bijective (one-one onto) and thus
for existence of inverse of trigonometric function we restrict their domain and co-domain to make it bijective. This restriction of
domain and range gives principal value branch of inverse trigonometric function which are as follows:
Function Domain Range (Principal value branch)

y = sin–1 x [–1, 1]  –π π 
 2 , 2
 
y = cos–1 x [–1, 1] [0, p]

 –π π 
y = cosec–1 x R – (–1, 1)  2 , 2  − {0}
 

π
y = sec–1 x R – (–1, 1) [0, π] −  
2

 π π
y = tan–1 x R − , 
 2 2
y = cot–1 x R (0, p)

‰ Principal Values: The value of an inverse trigonometric function which lies in its principal value branch is called principal value
of the inverse trigonometric functions.
π π
(a) If sin–1 x = θ, then θ is its principal value when – ≤ θ ≤ .
2 2
(b) If cos–1 x = θ, then θ is its principal value when 0 ≤ θ ≤ π .
π π
(c) If tan–1 x = θ, then θ is its principal value when – < θ < .
2 2
Note:
(i) If no branch of an inverse trigonometric function is mentioned, we mean the principal value branch of that function.
1
(ii) sin −1 x ≠ or (sin x)–1 and same holds true for other trigonometric functions also.
sin x
(iii) If sin–1 x = y then x and y are the elements of domain and range of principal value branch of sin–1 respectively.
 π π
i.e., x ∈ [ −1,1] and y ∈  − , 
 2 2
Similar fact is also applicable for other inverse trigonometric functions.

57 Inverse Trigonometric Functions P

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 ‰ Principal Value Branch and Graph of Inverse Trigonometric Functions:
 −π π 
(i) Principal Value Branch and Graph of sin–1x: The range  ,  of sin–1x is called its principal- value-branch. This
 2 2
 −π π 
means, now every value of sin–1x will be taken in the interval  ,  .
 2 2
Graph of y = sin–1 x
y
π
2
 −π π 
x' x y sin −1 x : [ −1, 1] →  , 
=
–1 0 1  2 2
–π
2
y'
(ii) Principal Value Branch and Graph of cos–1x: The range of cos–1x, [0, π] is called principal value branch of cos–1x. This
means all the value of cos–1x will lie in the class interval [0, π].
Graph of y = cos–1 x
y
p
π

x' 2
x y = cos–1 x: [–1, 1] → [0, p]
–1 0 1

y'
 −π π 
(iii) Principal Value Branch and Graph of cosec–1x: The range of cosec–1x =  ,  − {0} is the principal value branch of the
cosec–1x.  2 2
Graph of y = cosec–1 x
y

π
2
–2 –1  −π π 
x' x y = cosec–1 x : R– (–1,1) →  ,  − {0}
0 1  2 2
2
–π
2
y'
π
(iv) Principal Value Branch and Graph of sec–1x: The range of sec–1x, i.e., [0, p] –   is the principal value branch of sec–1x.
2
Graph of y = sec–1 x:
y
p

π
2 π
x' x y = sec–1 x : R– (–1,1) → [0, p] –  
0 1 2
–1
y'

P
W CBSE Class – XII MATHEMATICS 58
  −π π 
(v) Principal Value Branch and Graph of tan–1x: The range of q = tan–1 x, i.e.,  ,  is the principal value branch of
 2 2
q = tan x.
–1

Graph of y = tan–1 x
y
π
2

 π π
x' x y = tan–1 x: R →  − , 
–4 –3 –2 –1 0 1 2 3 4  2 2

−π
2
y'
(vi) Principal Value Branch and Graph of cot–1x: The range of q = cot–1 x i.e., (0, p) is the principal value branch of
q = cot–1 x.
Graph of y = cot–1 x
y

π
2

x' x y = cot–1 x: R → (0, p)

–4 –3 –2 –1 0 1 2 3 4

y'
‰ Important substitution to simplify Trigonometric Expressions involving Inverse Trigonometric Functions:
Expression Substitution
a2 + x2 x = a tan θ or x = a cot θ
a2 – x2 x = a sin θ or x = a cos θ
x2 – a2 x = a sec θ or x = a cosec θ

a+x a−x
or x = a cos 2θ
a− x a+ x

Important Formulas
‰ Property I:
1 1
(i) sin–1 = cosec–1x, x ≥ 1 or x ≤ –1 (ii) cos–1 = sec–1x, x ≥ 1 or x ≤ –1
x x
1
(iii) tan–1 = cot–1x, x > 0
x
‰ Property II:
 π π
(i) sin–1 (sin q) = q, for all q ∈  – ,  (ii) cos–1 (cos q) = q, for all q ∈ [0, p ]
 2 2
π π π π
(iii) tan–1 (tan q) = q, for all q ∈  – ,  (iv) cosec–1 (coses q) = q, for all q ∈  ,  , ≠ 0
 2 2 2 2
π
(v) sec–1 (sec q) = q, for all q ∈ [ 0, π] , θ ≠ (vi) cot–1 (cot q) = q, for all q ∈ (0, p)
2

59 Inverse Trigonometric Functions P

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 ‰ Property III:
(i) sin (sin–1 x) = x, for all x ∈ [– 1, 1] (ii) cos (cos–1 x) = x, for all x ∈ [– 1, 1]
(iii) tan (tan–1 x) = x, for all x ∈ R (iv) cosec (cosec–1 x) = x, for all x ∈ ( – ∞, –1] ∪ [1, ∞ )
(v) sec (sec–1 x) = x, for all x ∈ ( – ∞, –1] ∪ [1, ∞ ) (vi) cot (cot–1 x) = x, for all x ∈ R
‰ Property IV:
(i) sin–1 (–x) = –sin–1 x, x ∈ [-1,1] (ii) cos–1 (–x) = π – cos–1 x, x ∈ [-1, 1]
(iii) tan–1 (–x) = –tan–1 x, x ∈ R (iv) cosec–1 (–x) = –cosec–1 x, |x| ≥ 1
(v) sec–1 (–x) = π – sec–1x, |x| ≥ 1 (vi) cot–1 (–x) = π – cot–1x, x ∈ R
‰ Property V:
π π,
(i) sin–1 x + cos–1 x = , for all x ∈ [–1,1] (ii) tan–1 x + cot–1 x = for all x ∈ R
2 2
π,
(iii) sec–1 x + cosec–1 x = for all x ∈ ( – ∞, –1] ∪ [ 1, ∞ )
2
‰ Property VI:
x+ y x− y
(i) tan–1x + tan–1 y = tan–1 , xy < 1 (ii) tan–1x – tan–1 y = tan–1 xy > – 1
1 − xy 1 + xy
2x 1 − x2
(iii) 2tan–1x = sin–1 , x ≤1 (iv) 2tan–1 x = cos–1 ,x ≥ 0
1 + x2 1 + x2
2x
(v) 2tan–1 = tan–1 , –1 < x < 1
1 − x2

Real Life Applications

‰ Optics And Camera Angles: In photography and optics, ‰ Astronomy and Stargazing: In astronomy, angles to
understanding angles of view and lens positioning involves celestial objects are determined using inverse trigonometry.
inverse trigonometry for capturing precise images and If an observer knows the distance to a star and the angle
adjusting focus. subtended by it, they can calculate the star's actual position.

‰ Determining Heights Using Shadows: In architecture,

inverse trigonometry aids in finding heights. Suppose an
architect assesses the height of a building.

Height

Angle of
elevation
Distance

P
W CBSE Class – XII MATHEMATICS 60
 Different Problem Types

Type I: Determining the Principal Value of the Given Inverse Trigonometric Function

 1 
Find the principal value of sin–1  .
 2

Solution:

 1  1
Step I: Let sin −1   = y then, sin y =
 2 2
 −π π  π 1
Step II: We know that the principal value branch of sin–1 is  ,  and sin   =
 2 2 4 2

1 π π  π π
Step III: Now, sin–1 = , since ∈ − , 
2 4 4  2 2

−1  1  π
Therefore, Principal value of sin   is .
 2 4

 3
Example: Find the principal value of cos −1  
 2 

 3 3 π
Solution: Let cos −1   = x Then, cosx
= = cos  
 2  2 6

π
We know that the principal value branch of cos–1x is [0, p] since, ∈ [0, p]
6

 3 π
Therefore, Principal value of cos–1   is
 2  6

Type II: Determining the Simplest form of the Inverse Trigonometric Expressions

 cos x  3π π
Simplify tan–1   where − < x<
 1 − sin x  2 2

Solution:

Step I: Apply trigonometric identities to simplify the expression:

  x x 
 cos 2   − sin 2    
 cos x   2   2  θ θ θ θ
 = tan   ∵ cos 2 θ + sin
= θ 1, cos
= θ cos 2 − sin 2 , =
sin θ 2sin cos 
−1 −1 2
tan 
 1 − sin x   2 x  2 x  x x   2 2 2 2
 cos  2  + sin  2  − 2sin 2 cos 2 
     

Step II: Again, simplify the expression using algebraic and trigonometric identities:

 x x x x    x  x
  cos + sin   cos − sin    cos  2  + sin  2  

= tan −1   2 2   2 2   −1
     ∵ a 2 − b 2 = ( a + b )( a − b ) , a 2 + b 2 − 2ab = ( a − b ) 
 = tan 
2
2
 x  x  
  x  x 

  cos  2  − sin  2  

  cos  2  − sin  2  
           

61 Inverse Trigonometric Functions P

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  x
Dividing by cos   in numerator and denominator we get,
2

  x x 
 cos  2  sin  2  
   +  
 x
   x   x
 cos   cos     1 + tan   
2 2 2  sin θ 
=
tan −1
  tan −1  ∵ tan θ = cos θ 
 cos  x   x     x  
  sin     1 − tan   
 2 − 2  2
 
 x   x
 cos   cos    
 
 2 2 

 x   π x 
 1 + tan 2   π   tan 4 + tan 2   tan A + tan B 
This =
expression can be written as: tan
= 
−1
 ∵ tan =1 −1
tan   = tan ( A + B ) 
∵ 1 − tan A tan
 1 − 1× tan x   4   1 − tan π tan x   B 
 2  4 2
  π x  π x
= tan–1  tan  +   = +
  4 2  4 2

Step III:
3π π 3π x π 3π π x π π π 2π  π x  2π π  x π π
Since, given − <x< ⇒− < < ⇒− + < + < + ⇒− < + < ⇒− < + <
2 2 4 2 4 4 4 2 4 4 4 4 4 2 4 2 2 4 2

π x π x  π π   π x  π x
Now, + is within the principal range of tan–1 i.e., + ∈  − ,  Therefore, tan–1  tan  +   = +
4 2 4 2  2 2   4 2  4 2

 cos x  3π π π x
Hence, simplest form of tan–1   where, − < x < is +
 1 − sin x  2 2 4 2

 sin x + cos x  π π
Example: Express sin-1   , where − < x < , in the simplest form.
 2  4 4

−1  sin x + cos x 
Solution: We have, sin  
 2 
 sin x cos x   π π
= sin −1  + =sin −1  sin x cos + cos x sin  sin ( A + B ) 
∵ sin A cos B + cos A sin B =
 2 2   4 4

  π   π
= sin −1  sin  x +  =  x + 
  4   4

π π π π π π π
∵− < x < ⇒ − + < x+ < +
4 4 4 4 4 4 4
 π π  π   π   −π π 
⇒ 0 <  x +  < ⇒  x +  ∈  0,  ⊆ 
 4  2  4   2   2 2 

π π  π π
Now, x + is within the principal range of sin–1 i.e., x + ∈  − , 
4 4  2 2

  π  π
∴ sin −1  sin  x +  =  x + 
  4   4

−1  sin x + cos x  π π π
Hence, simplest form of sin   , where − < x < is x +
 2  4 4 4

P
W CBSE Class – XII MATHEMATICS 62
 competency BASED SOlved Examples
 3π 
Multiple Choice Questions (1 M) 4. The value of sin–1  cos  is (Ev) (CBSE, 2020)
 5 
−1
1. The principal value of tan 3 − cot −1 ( − 3) is π 3π −π −3π
(a) (b) (c) (d)
(Un) (CBSE, 2018) 10 5 10 5
π π π
(a) –p (b) − (c) (d)   3π   −1   π π 
2 3 2 Sol. sin −1=
cos  5   sin cos  2 + 10  
      
π
Sol. We know, tan–1x + cot–1x = π 
2  π  
sin −1  − sin 
= ∵ cos  2 + x  =
− sin x 
Now, tan −1
3 − cot −1
(− 3)  10     

 ( )
∵ cot −1 − 3 = π − cot −1 3 
 =
 π
− sin −1  sin  ∵ sin −1 ( − x ) =− sin −1 x 
 10 
π π
= tan −1 3 − π + cot −1 3 = − π = −
2 2
( ) π   −π π  
=
− ∵ sin −1 ( sin x ) =
x, x ∈  , 
Topper’s Explanation (CBSE 2018) 10   2 2 

1
5. The value of the expression 2sec–1 2 + sin–1   is
 2
(Un) (NCERT Exemplar)
π 5π 7π
(a) (b) (c) (d) 1
6 6 6
1
Sol. We know, 2sec–1 (2) + sin–1  
2
Key Takeaways   π  −1   π 
= 2 sec–1  sec    + sin  sin   
π,    3   6 
We know, tan–1 x + cot–1 x = for all x ∈ R
2 π π 2π π 5π
=2 × + = + =
2. tan
−1
3 − sec −1 ( −2 ) is equal to 3 6 3 6 6

(An) (NCERT Exemplar) 6. sin (tan–1 x), where |x| < 1, is equal to (Ap)
π π 2π x 1 1 x
(a) p (b) − (c) (d) (a) (b) (c) (d)
3 3 3 1− x 2
1− x 2
1+ x 2
1 + x2
π π
Sol. tan–1 3 – sec–1 (–2) = tan–1 (tan ) – sec–1 (–sec ) 1
3 3 Sol. Let tan–1 x = θ ⇒ x = tan θ ⇒ cot θ =
x
π π π 2π π
= – sec–1 (sec (p – )) = – =– 1 1 1 x
3 3 3 3 3 ∴=
sin θ = = =
cosec θ 1 + cot 2 θ 1 1 + x2
3. If sin–1 x = y, then (An) 1+ 2
π π x
(a) 0 ≤ y ≤ p (b) − ≤ y≤
x
⇒ sin ( tan −1 x ) =
2 2
π π 1 + x2
(c) 0 < y < p (d) − < y <
2 2
Sol. Given sin –1 x = y, by f (x) sin −1 x − 1
7. The domain of the function defined =
 π π is (An) (NCERT Exemplar)
The range of the Principal value of sin–1x is  − ,  .
 2 2
(a) [1, 2] (b) [–1, 1]
π π
Therefore, − ≤ y ≤ (c) [0, 1] (d) None of these
2 2
63 Inverse Trigonometric Functions P
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 f ( x) sin −1 x − 1
Sol. ∵ =  x x  x
 cos 2 − sin 2  −1
 1 − tan 2 
−1
⇒ 0 ≤ x −1 ≤ 1 ∵ x − 1 ≥ 0 and − 1 ≤ x − 1=
≤ 1
=
tan   tan  
  cos x + sin x   1 + tan x 
⇒1≤x≤2  2 2  2
∴ x ∈ [1, 2]   π x  π x
=tan −1  tan  −   = −
8. If y = cot–1 x, x < 0, then (Re) (CBSE, 2022 Term-I)   4 2  4 2

(a)
π
< y≤π (b)
π
< y<π Answer Key
2 2 10. (a) 9. (d) 8. (b) 7. (a) 6. (d)
π π 5. (b) 4. (c) 3. (b) 2. (b)
(c) − < y<0 (d) − ≤ y<0
1. (b)
2 2
Sol. We have, y = cot–1 x, x < 0
⇒ x = cot y < 0 Assertion and Reason (1 M)
As we know that cotangent (i.e.cot y) is negative in IInd and
IVth quadrant, i.e cot y lies in IInd or IVth quadrant. Direction: In the following questions, a statement of Assertion
(A) is followed by a statement of Reason (R). Mark the correct
π 3π
⇒ < y < π or < y < 2π choice as.
2 2
(a) Both Assertion (A) and Reason (R) are true, and Reason (R)
  33π   is the correct explanation of Assertion (A).
9. The value of sin–1 cos  5   is
   (b) Both Assertion (A) and Reason (R) are true, but Reason (R)
is not the correct explanation of Assertion (A).
(An) (CBSE, 2022 Term-I)
(c) Assertion (A) is true, but Reason (R) is false.
3π −7 π (d) Assertion (A) is false, but Reason (R) is true.
(a) (b)
5 5
1. Assertion (A): sin–1 (sin 3) = 3.
π −π  −π π 
(c) (d) Reason (R): For principal value, when x∈  , ,
10 10  2 2
  33π   −1   3π   sin–1 (sin x) = x. [An]
Sol sin −1 cos
=    sin cos  6π +  
  5    5  Sol. We know sin (p – x) = sin (x)
  3π    −π π 
= sin −1 cos    Also, p – 3 belongs in  , 
  5   2 2
∴ sin (sin3) = 3
–1
  π 3π   −1   π 
= sin −1 cos  + =   sin sin  −   \ Assertion is true.
  2 10    10  
 −π π 
Reason: For principal value, when x∈  , 
π  π  π π  2 2
=− ∵ ∈ − ,  
10  10  2 2 ∴ sin–1 (sin x) = x

 1 + cos x + 1 − cos x  \ Reason is true.

10. Simplest form of tan–1   , Both Assertion (A) and Reason (R) are true, and Reason (R)
 1 + cos x − 1 − cos x 
is the correct explanation of Assertion (A).
3π
π
π< x<
2
is (Re)
( )
2. Assertion (A): Principal value of tan–1 − 3 is −
3
.

π x 3π x x x  π π
(a) − (b) − (c) − (d) π − Reason (R): tan–1x : R →  − ,  so for any x ∈ R,
4 2 2 2 2 2  2 2
 1 + cos x + 1 − cos x   π π
Sol. Given, tan–1 
 1 + cos x − 1 − cos x  tan–1 (x) represents an angle in  − ,  . (Ap)
   2 2
 x x  π π
 − 2 cos 2 + 2 sin 2  Sol. ∵ tan–1 x: R →  − ,  such that tan–1 (x) = θ
−1 3π  2 2
tan   , where π < x <
x x
 − 2 cos − 2 sin  2  π π
 2 2 where θ ∈  − ,  & x ∈ R
 2 2
P
W CBSE Class – XII MATHEMATICS 64
  π −π We know that the range of principal value branch of cot–1 is
( )
∵ tan–1 − 3 = θ ⇒ tan θ = − 3 = tan  −  ⇒ θ =
 3 3  2π  −1
(0, p) and cot   = . Hence, principal value of
Here, Both Assertion (A) and Reason (R) are true, and  3  3
Reason (R) is the correct explanation of Assertion (A).  −1  2π
cot–1   is . (1 M)
3. Assertion (A): Range of [sin–1 x + 2 cos–1 x] is [0, p].  3 3
Reason (R): Principal value branch of sin–1 x has range 9 π 9 −1  1  9 −1  2 2 
2. Prove that: − sin   = sin  
 π π 8 4  3 4
− 2 , 2  . (Ev) (CBSE, 2023)  3 
 
(Un) (CBSE, 2020)
Sol. Let f (x) = sin–1 x + 2 cos–1x = sin–1x + cos–1x + cos–1x
9π 9 −1 1 9  π 1
π  π Sol. L.H.S: − sin =  − sin −1 
f ( x) =
−1 −1
⇒ + cos −1 x ∵ sin x + cos x =
2 
8 4 3 4 2 3
2 
As 0 ≤ cos–1 x ≤ p π π
[Using sin–1 x + cos–1 x = ⇒ cos −1 x = − sin −1 x]
2 2
π π π π 3π
⇒ 0+ ≤ + cos −1 x ≤ π + ⇒ ≤ f ( x) ≤ 9 1
2 2 2 2 2 = cos −1   ...(i) (1 M)
4 3
 π 3π 
Range of f(x) =  ,  .
2 2 
Here, Assertion (A) is false but Reason (R) is true.
3
2 2

Key Takeaways
θ
π
We know, sin–1 x + cos–1 x = , for all x ∈ [–1,1] 1
2
1 1
R.H.S: Let cos–1 = θ ⇒ cos θ =
4. Assertion (A): All trigonometric functions have their 3 3
inverses over their respective domains. ∵ sin 2 θ + cos 2 θ =1
Reason (R): The inverse of tan–1 x exists for some x ∈ R. 1
2
8 2 2
(Re) (CBSE, 2023) ⇒ sin 2 θ +   = 1 ⇒ sin 2 θ = ⇒ sin θ =
3 9 3
 −π π 
Sol. A is true but R is false as tan–1 x exists for x ∈  ,  −1 2 2 1 ...(ii)
 2 2 =
⇒ θ sin
= cos −1  
3 3

Answer Key 9 −1  2 2  9 −1  1 
From equation (ii), sin   = cos   (1 M)
4  3
4. (c) 3. (d) 2. (a) 1. (a)  3  4
L.H.S. = R.H.S, Hence proved.

Subjective Questions
3. Find the value of tan
−1
3 − cot −1 − 3 . ( )
(Ap) (CBSE, 2018)
Very Short Answer Type Questions  (1 or 2 M) Sol. tan −1 3 − cot −1 − 3 ( )
 −1 
1. Find the Principal value of cot–1  .
= tan −1
(
3 − π − cot −1
3 )
 3
= tan −1 3 − π + cot −1 3 (1 M)
(Ap) (NCERT Intext) −1 −1
= tan 3 + cot 3−π
 −1 
−1
Sol. Let cot   = y . Then, π π
 3 = −π (∵ tan–1 x + cot–1x = )
2 2
−1 π  π  2π  π − 2π π
cot y = = − cot   = cot  π −  = cot   (1 M) = = − (1 M)
3 3  3  3  2 2

65 Inverse Trigonometric Functions P

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 −1  3π  −1  π −1   π π 
4. Find the principal value of sin  sin  . (An)= sin
=
 5  cos 3  sin sin  2 − 3   (1 M)
    
−1  3π 
Sol. sin  sin  −1  π π  π  π π 
 5  = sin
= sin 6  6 ∵ 6 ∈  − 2 , 2   (½ M)
    
 π π
We know that sin–1 (sin x) = x, x ∈  − ,   3
 2 2  1 
 + cos ( 0 )
−1 −1
8. Evaluate : 3 sin–1   + 2cos 
3π  π π   2  2 
Since ∉ − , which is the range of Principal value
5  2 2  (Ap) (CBSE, 2023)
branch of sin–1. (1 M)  3
 1 
 + cos ( 0 )
−1 −1
 3π   3π  2π Sol. We have, 3 sin–1   + 2 cos 
However, sin  =  sin  π − =  sin  2  2 
 5   5  5
 π  π π
2π  π π  =3sin −1  sin  + 2 cos −1  cos  + cos −1   (½ M)
and ∈ − ,  4  6 2
5  2 2 
π π π 3π π π
 3π   2π  2π =3 × + 2× + = + + (1 M)
∴ sin −1  sin  =sin −1  sin  = (1 M) 4 6 2 4 3 2
 5   5  5
9π + 4π + 6π 19π
−1  13π  = = (½ M)
5. Find the value of sin sin
7 
(Ev) 12 12

 3π 
  13π   9. Evaluate: sin–1  sin  + cos–1 (cos p) + tan–1 (1).
−1   π 
−1
=
Sol. sin sin    sin sin  2π −   (1 M)  4 
  7     7 
(An)
  π  π  3π 
sin −1 sin  −   =
= − (1 M) Sol. We have,sin–1  sin  + cos ( cos π ) + tan (1)
−1 −1
  7  7  4 
6. What is the principal value of   π   π
= sin −1 sin  π −   + π + tan −1  tan  (½ M)
 2π   2π    4    4
cos  cos  + sin −1  sin  ?
−1
(Un)
 3   3   π π π π π + 4π + π
= sin −1  sin  + π + = +π+ = (1 M)
 2π   2π   4 4 4 4 4
Sol. cos  cos  + sin −1  sin 
−1

 3   3  6π 3π
= = (½ M)
 2π    π    2π  π π   4 2
= cos −1  cos  + sin −1  sin  π −   ∵ ∉  − ,  
 3    3    3  2 2  π 
10. Solve for x: sin–1 4x + sin–1 3 x = − . (Un) (CBSE, 2020)
(1 M) 2
π
 2π   π  2π π Sol. sin–1 4x + sin–1 3x = –
cos −1  cos  + sin −1  sin  =
= + 2
 3   3 3 3
π
  π π  ⇒ sin −1 ( 4 x ) =− − sin −1 ( 3 x ) (½ M)
∵ sin ( sin=
x ) x if x ∈  − ,  
−1
2
  2 2 
and cos ( cos x ) = if x ∈ [ 0, π]  π 

−1
 ⇒ 4x = − sin  + sin −1 3 x 
2 
3π −1
(½ M)
= = π (1 M) = − cos(sin 3 x)
3
  ⇒ −4 x = 1 − 9 x 2 ⇒ 16 x 2 = 1 − 9 x 2 ⇒ 25 x 2 = 1
−1 −1
 3  
7. Evaluate: sin cos  sin    

   2    1
⇒ x2 = ⇒ x =±
1
(½ M)
25 5
(Ev) (NCERT Exemplar)
1
Sol. We have, −1 −1
As sin 4 x + sin 3 x < 0, x ≠
5
  3  −1   −1  π  
sin −1 cos(sin −1    = sin cos  sin  sin 3    (½ M)
1
So, x = − (½ M)
  2     5

P
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  1
11. Write the principal value of tan–11 + cos–1  −  (An) Short Answer Type Questions (2 or 3 M)
 2
1. Write the domain and range (Principal value branch) of
Sol. tan −1 1 + cos −1  − 
1
the following function: f(x) = tan–1 x (Re)
 2
Sol. The domain of tan–1x is all real numbers, represented as
 π   π  (–∞, ∞).
= tan  tan  + cos −1  cos  π −  
−1
(1 M)
 4    3 
π π π
The range of tan–1x is from − to but excluding and
 π  2π  2 2 2
= tan −1  tan  + cos −1  cos   (1 M) π
 4  3  − . (1 M)
2
 π π
π 2π  π π π 2π  The correct range of tan–1x is  − ,  
∵ 4 ∈  2 , 2  and 3 ∈ [ 0, π]
= + (1 M)
 2 2
4 3    
3π + 8π 11π  1 1
= = (1 M) 2. Prove that: 3 sin–1 x = sin–1 (3x–4x3), x ∈  − , 
12 12  2 2

 1  (An) (CBSE, 2018) (NCERT Intext)

12. Write cot–1   , x > 1 in the simplest form. Sol. R.H.S: sin–1 (3x – 4x3)
 x −1 
2

(Re) (NCERT Intext) Put x = sinq

= sin–1 (3sinq – 4sin3q) = sin–1 (sin3q) (1 M)
Sol. Let x = sec q, then x 2 − 1= sec 2 θ − 1= tan θ (1 M)
(∵ sin3x = 3sinx – 4sin3x)
1
Therefore, cot
−1
= cot–1 (cot θ) = θ = sec–1 x, which  1 1  π π  π π
x2 − 1 x sin θ ∈  − ,  ⇒ θ ∈  − ,  ⇒ 3θ ∈  − , 
=
 2 2   6 6   2 2
is the simplest form. (1 M) (1 M)
π 1  ⇒ R.H.S = 3q
13. Simplify: cos  + sin −1  Show your work.
2 3 Now, x = sinq ⇒ q = sin–1 x
(An) (CBSE CFPQ, 2024) ∴ 3q = 3sin–1 x
Sol. We have, ⇒ R.H.S = L.H.S
π 1   1  Hence, proved.
cos  + sin −1 =− sin  sin −1  (½ M)
(1 M)
 2 3   3 Topper’s Explanation (CBSE 2018)
 π  
∵ cos  2 + θ  = − sin θ 
   
We know, sin(sin–1q) = q, For all x ∈ [–1, 1]
 1  1
∴ − sin  sin −1 = − (½ M)
 3 3
 24   24 
14. What would be the value of cos–1   + tan–1   ?
 
25  7 
(An) (CBSE CFPQ, 2024) 3. Prove that sin 2 x=
−1
(
1 − x2 )
2cos −1 x ,
1
2
≤ x≤1
−1  24  24
Sol. Let, cos   = x ⇒ cos x = (½ M) (Ev) (CBSE, 2020)
 
25 25
 24  π
Now, cot x =
24
⇒x= cot −1   25 (1 M) Sol. Let x = cosθ. Then cos–1 x = θ, 0 ≤ q ≤  (½ M)
7  7  4
7
24
−1
(
We have sin 2 x =
1 − x2 ) (
sin −1 2 cos θ 1 − cos 2 θ )
π
We know, tan–1x
+ cot–1x
= , for x ∈ R (½ M)
2
= sin–1 (2cosq sinq) = sin–1 (sin 2q) = 2q (1 M)
 
24  
24
∴ cos −1   + tan −1  
 25   7   π π
 0 ≤ θ ≤ ⇒ 0 ≤ 2θ ≤ 
 24   24  π  4 2
= cot −1   + tan −1   = (½ M)
 7   7  2 = 2cos–1 x (1 M)

67 Inverse Trigonometric Functions P

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Topper’s Explanation (CBSE 2020)  −1  4  −1  5  
6. Find the value of sin  cos   + tan    . (Ap)
 5  12  
−1  4  4 3
Sol. Let cos   = α, ⇒ cos α = ,sin α = (½ M)
5 5 5
 5 5 5 12
and tan −1   = β ⇒ tan β = ,sin β = , cos β =
 12  12 13 13
−1  4  −1  12   33 
4. Prove that: cos   + cos   = cos −1    (Ev) (1 M)
 
5  
13  65  sin (a + b) = sin a cos b + cos a sin b (½ M)
4 12
Sol. Let cos–1 = x, cos −1 = y, x, y ∈ [ 0, π] 3 12 4 5 36 20 56
= × + × = + = (1 M)
5 13 5 13 5 13 65 65 65
4 12
–1 
⇒ cos x = , cos y = 63  −1  5  −1  3 
5 13 =
7. Prove that: sin  65  sin  13  + cos  5  (Ev)
     
2 2
4  12  −1  5  −1  3  5 3
∴ sin x =1 −   , sin y =1−   (1 M)
Sol. Let sin   = α, cos   = β ⇒ sin α = , cos β =
5  13   13  5 13 5
[∵ x, y ∈ [0, p] ⇒ sin x and sin y are + ve ] (½ M)
3 5 2 2
⇒ sin x = ,sin y = 5 3
5 13 ⇒ cos α= 1 −   ,sin β= 1−  
 13  5
Now, cos (x + y) = cos x. cos y –sin x. sin y
12 4 (1 M)
⇒ cos
= α =
,sin β
4 12 3 5 33
= × − × ⇒ cos ( x + y ) = (1 M) 13 5
5 13 5 13 65 Now, sin (a + b)= sin a cos b + cos a sin b
 33  5 3 12 4 15 48 63  63 
⇒ x+ y = cos −1   = . + .= + = β sin −1  
⇒ α +=
 65  13 5 13 5 65 65 65  65 
Putting the value of x and y, we get (1 M)
4
−1 12  33  Putting the value of a and b, we get
cos + cos −1 = cos −1  = RHS
5 13  65  5 3  63 
⇒ sin −1 + cos −1 =sin −1   (½ M)
Hence, L.H.S = R.H.S proved. (1 M) 13 5  65 

Key Takeaways Key Takeaways

We know, cos (A + B) = cos A cos B – sin A sin B
We know, sin (A ± B) = sin A cos B ± cos A sin B

 −1 3 
5. Solve: cos (tan–1x) = sin  cot


4
(Un) (CBSE, 2017) Long Answer Type Questions  (4 or 5 M)
 −1 3   −1 4 2
Sol. Given cos (tan–1 x) = sin  cot  1. Find the value of sin  cos + tan −1 
 4  5 3
(Re) (CBSE, 2019)
π 3
⇒ cos ( tan −1 x ) =
cos  − cot −1  (1 M)
 2 4 
5
 π   
∵=sin θ cos  − θ   4  2  3
π −1 3  2  Sol. sin cos −1   + tan −1   
−1
⇒ tan x = − cot    5  3 
2 4  −1 −1 π
and tan x + cot x =
4
  2 
2  = sin  tan −1   + tan −1   
3
 (1 M)
  4  3 
(1 M)
π π 3 3   3 2 
⇒ − cot −1 x = − cot −1 ⇒ cot −1 x = cot −1  −1  4 + 3  
2 2 4 4  −1  17  
= sin =  tan    sin  tan    (1 M)
⇒ x=
3  3 2
 1 − ×     6 
(1 M)  
4   4 3  

P
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 π 1 a π 1 a  2b
3. Prove that: tan  + cos −1  + tan  − cos −1  =
  17  
325  4 2 b   4 2 b a
= sin sin −1  17 (1 M)
 (Re)
  325  
π 1 −1 a  π 1 −1 a 
6 Sol. L.H.S : tan  + cos  + tan  − cos 
17 17  4 2 b  4 2 b
= = (1 M)
325 5 13
π  π  1 a
= tan  + x  + tan  − x  , where=
x cos −1 (½ M)
1 + x2 4  4  b
(
2. Prove the following: cos  tan −1 {sin cot −1 x } = ) 2 + x2
2

π π
tan + tan x tan − tan x
(Ap) 4 4 1 + tan x 1 − tan x
= + = +
π π
Sol. L.H.S: cos [tan–1 {sin (cot–1x)}] 1 − tan ⋅ tan x 1 − tan ⋅ tan x 1 − tan x 1 + tan x
4 4
Let cot–1 x = q ⇒ x = cot θ
(1 M)
  1 
= cos  tan=
−1
{sin θ} cos  tan −1  (1 + tan x ) + (1 − tanx )
2 2
 (1 M)
  cos ecθ  = (½ M)
1 − tan 2 x
 −1 1   −1 1 
=
cos  tan  cos  tan  (1 M) 1 + tan 2 x + 2 tan x + 1 + tan 2 x − 2 tan x
1 + cot θ  1 + x2  = (1 M)
 
2
1 − tan 2 x
−1 1 2 (1 + tan 2 x )
Let tan = α =
1 + x2 1 − tan 2 x
1 1 2 2
⇒ = tan α ⇒ = tan 2 α = =
1+ x 2 1 + x2 cos 2 x 1 a
cos 2  cos −1 
2 b
1 sin 2 α 1 sin 2 α
⇒ = ⇒ = +1 +1  (1 M) 2
1+ x 2
cos α
2
1+ x 2
cos 2 α =
 −1 a  (1 M)
cos  cos 
2+ x 12
1+ x 2
 b
⇒ = ⇒ cos
= α (1 M)
1+ x 2
cos 2 α 2 + x2 [∵ cos(cos–1x) = x if x ∈ [–1, 1]

 1 + x2   −1 1 + x 2  a
=⇒ α cos −1  =  ⇒ cos α cos  cos  Here, ∈[–1, 1]]
 2 + x2   2 + x 2  b
  
2 2b
1 + x2 = = = R.H.S
= = R.H.S a a
2 + x2 b
Hence, L.H.S = R.H.S proved. (1 M) Hence, L.H.S = R.H.S.,proved. (1 M)

Mistakes 101 : What not to do! Nailing the Right Answer

When working on trigonometric inverse problems, students Students should know the basic trigonometric identities
should be careful with the substitution of variables. and trigonometric ratios for these types of problems.

69 Inverse Trigonometric Functions P

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 miscellaneous Exercise
10. The value of cot (sin–1x) is
Multiple Choice Questions (1 M)
1 + x2 x 1 1 − x2
−1  1 (a) (b) (c) (d)
1. The principal value of sin  −  is x 1− x 2
x x
 2
π π π π
(a) (b) (c) − (d) − Assertion and Reason (1 M)
3 6 3 6
2. The domain of function cos–1 (2x–1) is Direction: In the following questions, a statement of Assertion
(A) is followed by a statement of Reason (R). Mark the correct
(a) [0,1] (b) [–1,1] (c) (–1,1) (d) [0, p]
choice as.
−1  1  −1  1  (a) Both Assertion (A) and Reason (R) are true, and Reason (R)
3. The Principal value of cos   + sin  −  is
 
2  2 is the correct explanation of Assertion (A).
(CBSE, 2022 Term-I)
(b) Both Assertion (A) and Reason (R) are true, but Reason (R)
π π π
(a) (b) p (c) (d) is not the correct explanation of Assertion (A).
12 3 6
(c) Assertion (A) is true, but Reason (R) is false.
−1  9π 
4. The principal value of tan  tan  is
 8  (d) Assertion (A) is false, but Reason (R) is true.
π 3π π 3π
(a) (b) (c) − (d) − 1. Assertion (A): The value of the expression (cos–1 x)2 is equal
8 8 8 8
to sec2x.
5. What is the domain of the function cos–1 (2x – 3)?
(CBSE, 2022 Term-I) Reason (R): The least numerical value, either positive or
negative of angle θ is called the Principal value of the inverse
(a) [–1,1] (b) [1,2) (c) (–1,1) (d) [1,2]
trigonometric function.
−1
 1+ x − 1− x 
6. The simplest form of tan   is −1  8  −1  3   77 
2. Assertion (A): sin   + sin   = sin −1  
 1+ x + 1− x   17  5  85 
(CBSE, 2022 Term-I)
 x+ y 
π x π x Reason (R): tan–1 x + tan–1 y = tan–1  
(a) − (b) +  1 − xy 
4 2 4 2
π 1 π 1 3. Assertion (A): The range of the function
(c) − cos −1 x (d) + cos −1 x
4 2 4 2 3π  π 5π 
f ( x ) 2sin −1 +
= , where x ∈ [ −1,1] , is  , 
7. The principal value of  tan −1 3 − cot −1
 ( 3 ) is 2 2 2 
Reason (R): The range of the principal value branch of
π
(a) p (b) sin–1 (x) is [0, p]. (CBSE, 2023)
6
4. Assertion (A): Maximum value of (cos–1x)2 is p2.
(c) 0 (d) 2 3
Reason (R): Range of the principal value branch of cos–1 x
−1  13π   −π π 
8. The value of sin  cos  is. (CBSE, 2022 Term-I) is  ,  . (CBSE, 2023)
 5   2 2
3π π 3π π
(a) − (b) − (c) (d)
5 10 5 10 Subjective Questions
9. If sin–1 x > cos–1, then x should lie in the interval
(CBSE, 2022 Term-I)
Very Short Answer Type Questions  (1 or 2 M)
 1   1 
(a)  −1, −  (b)  0, − 
 2  2 1. Evaluate: tan −1 3 − sec −1 ( −2 ) .

 1   1  −1   17 π  
(c)  ,1 (d)  ,0 2. Find the value of sin sin  −  (CBSE, 2020)
 2   2    8 
P
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 −1  1  −1  1  −1   π  −1  3a x − x  −a a
2 3
3. Evaluate: tan  −  + cot   + tan  sin  −   . 3. Write tan  3 , a > 0; <x< in the
2 
 3   3    2   a − 3ax  3 3
4. Find the principal value of cosec–1 (2). (NCERT Intext) simplest form. (NCERT, Intext)
−1  1
5. Find the principal value of cos  −  . π
4. Solve for x : sin (1 − x ) − 2sin ( x ) =
−1 −1
 2 (CBSE, 2020)
2
−1  1 1 
6. Draw the graph of f(x) = sin x, x ∈  − ,  . Also,
 2 2 Case Based Questions
write range of f(x). (CBSE, 2023)
7. Draw the graph of cos–1 x, where x ∈ [–1,0]. Also, write its
Case Based-I
Two men on either side of a temple 30 meters high observe its top
range. (CBSE, 2023) at the angles of elevation a and b respectively. (as shown in the
figure above).
8. Write the domain and range (Principal branch) of the
following functions: f(x) = tan–1x (CBSE, 2023)
−1  3π  −1  3π 
9. Evaluate: sin  sin  + cos  cos  + tan (1)
−1

 4   4 
(CBSE, 2023)
−1   7π  
10. Evaluate: cos cos  −   (CBSE, 2023)
  3 
−1  1  −1  1 
11. Write the Principal value of cos   + 2sin   .
2 2
 2 x +1 
12. Prove that: sin −1  x 
= 2 tan −1 ( 2 x ) , where x ≤ 0
1 + 4 
(CBSE CFPQ, 2024)
13. If cos–1 a + cos–1 b + cos–1 g = 3p, then find the value of The distance between the two men is 40 3 meters and the
distance between the first person A and the temple is 30 3
a(b + g) – b(g + a) + g (a + b) (CBSE APQ, 2023)
meters. (CBSE, QB)
 1 + sin x + 1 − sin x  π Based on the above information, answer the following questions.
14. Reduce cot −1   where < x < π in
 1 + sin x − 1 − sin x  2 (i) Find ∠ CAB = a in terms of sin–1
to simplest form. (CBSE APQ, 2023) (ii) Find ∠ CAB = a in terms of cos–1
(iii) Find ∠ BCA = b in terms of tan–1
Short Answer Type Questions (2 or 3 M) OR
Find the domain and range of cos–1x.
1. Find the principal value of cosec–1(–1).
Case Based-II
  1 
2. Find the principal value of tan 2 cos  2sin −1   .
−1
In a school, Manish was asked to construct a triangle ABC in
  2 
1 1
3. Find the principal value of sec–1(–2). which two angles B and C are given by tan–1   and tan–1  
2 3
−1 1 1 respectively.
4. Find the principal value of cos − 2sin −1 .
2 2 Based on the above information, answer the following questions .
1  1 (i) Find the value of sin B.
5. Find the value of sin −1   − cos −1  −  .
3  3
(ii) Find the value of cos C.
Long Answer Type Questions  (4 or 5 M) (iii) Find the value of B + C.
OR
1  −1 2 x 1 − y2 
1. tan sin + cos −1  ,| x |< 1, y > 0 and xy < 1 Find the value of cos (B + C).
2 1+ x 2
1+ y2 
(NCERT Intext) Case Based-III
2. Find the value of the given expression: We know that all trigonometric functions are periodic functions.
So, they are neither one-one nor onto and hence, are not bijections.
 3 3 But we can make bijection functions by restricting their domains
tan  sin −1 + cot −1 
(NCERT Intext)
 5 2 and co-domains.

71 Inverse Trigonometric Functions P

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Based on the above information, answer the following questions . (iii) The range of the function cosec–1 x, is:
 π π
(i) The function f: D → [–1,1] defined by f (x) = cos x is one- (a)  − ,  (b) R – (–1, 1)
 2 2
one and onto if D is:
 π π
 π π (c)  − ,  − {0} (d) R
(a) [0, p] (b)  − ,   2 2
 2 2
(iv) For x ∈ [– 1, 1], sin–1 (–x) is:
 π  π  (a) sin–1 (x) (b) – sin–1x
(c) 0,  (d)  − , 0 
 2  2  1
(c) − −1 − {0} (d) cos–1 (x)
(ii) The domain of the function tan–1 x, is: sin ( x )
 π π (v) cos (cos–1x) = x for:
(a)  − ,  (b) (0, p)
 2 2 (a) x ∈ [– 1,1] (b) x ∈ R
(c) [– 1, 1] (d) R  π π
(c) x ∈ R – (– 1,1) (d) x ∈  − , 
 2 2

Answer keys
Multiple Choice Questions
1. (d) 2. (a) 3. (a) 4. (a) 5. (d) 6. (c) 7. (b) 8. (b) 9. (c) 10. (d)

Assertion and Reason

1. (d) 2. (b) 3. (c) 4. (c)

Case Based Questions

Case Based-III
(i) (a) (ii) (d) (iii) (c) (iv) (b) (v) (a)

Hints & Explanations

Multiple Choice Questions
 1
Nailing the Right Answer
1. (d) Let sin −1  −  =
y.
 2 Students should remember the domain and range of all
inverse trigonometric functions.
1 π  π
Then sin y =− =− sin   =sin  −  .
2 6  6
−1  1   −1  cos −1  cos π  − sin −1  1 
We know that the range of the Principal of sin–1x. 3. (a) cos   += sin −1      
2  2  3  2
 π π  π 1
is  − ,  and sin −  =
−
π π π π π
 2 2  6 2 
= − sin −1  sin  = − =
3  4  3 4 12
−1  1 π
Therefore, the Principal value of sin  −  is − .
 2  6  9π  −1   π 
4. (a) tan −1=
 tan  tan  tan  π +  
2. (a) Let, f(x) = cos–1 (2x–1)  8    8 
–1 ≤ 2x – 1 ≤ 1 ⇒ 1 + 1 ≤ 2x ≤ 1 + 1  π  π  π  −π π  
= tan −1  tan
=  ∵ ∈ , 
0 ≤ 2x ≤ 2 ⇒ 0 ≤ x ≤ 1  8  8  8  2 2  
∴ Domain of the given function is [0, 1].

P
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 5. (d) Let, f(x) = cos–1 (2x – 3) As we know that sin–1x and cos–1x exist for x∈ [–1,1].
–1 ≤ 2x – 3 ≤ 1
 1   1  1 
⇒ 2 ≤ 2x ≤ 4 ⇒ 1 ≤ x ≤ 2. ∴ x∈ ,1 ⇒  ,1 ⊂  ,1
 2   2  2 
∴ x ∈ [1, 2] or domain of x is [1, 2].
10. (d) Let sin–1 x = q, then sin q = x.
1 1
Mistakes 101 : What not to do! ⇒ cosec θ=
x
⇒ cosec 2 θ= 2
x
Students often make mistakes in understanding the domain
of inverse trigonometric functions correctly. 1 1 − x2
⇒ 1 + cot 2 =
θ ⇒ cot =
θ
x2 x
 1+ x − 1− x 
−1
6. (c) We have, tan   1 − x2
⇒ cot ( sin −1 x ) =
 1+ x + 1− x  x
1
Put=x cos 2θ,so that=
θ cos −1 x
2 Assertion and Reason
 1 + cos 2θ − 1 − cos 2θ 
tan −1   1. (d) Assertion:
 1 + cos 2θ + 1 − cos 2θ  2
2  −1 1 
∵ =
cos −1 x  sec x  ≠ sec x
2
 2 cos 2 θ − 2sin 2 θ   
= tan −1  
 2 cos 2 θ + 2sin 2 θ 
  Reason: We know that the smallest numerical value either
positive or negative, of q is called the principal value of the
−1  cos θ − sin θ  −1  1 − tan θ 
= tan
=   tan  
 cos θ + sin θ   1 + tan θ  function.

π π 1 2. (b) As we know,
= tan −1 (1) − tan −1 ( tan θ )= − θ= − cos −1 x
4 4 2
(
sin–1x + sin–1 y = sin–1 x 1 − y 2 + y 1 − x 2 )
 2
Nailing the Right Answer
2
8 −1 3 8  3 3  8 
∴ sin −1 + sin
= sin −1  1−   + 1−   
17 5 17  5 5  17  
Students should carefully substitute the variable with its 
corresponding trigonometric ratio.
 8 4 3 15   77 
= sin −1  × + = ×  sin −1  
 17 5 5 17   85 
7. (b) tan −1 3 − cot −1 ( 3) Hence, Assertion is correct and Reason is correct
 π  π  π π 2π π π but Reason is not the correct explanation of Assertion.
= tan −1  tan  − cot −1  cot  = − = − =
 3  6 3 6 6 6 6 3π
f ( x ) 2sin −1 x +
3. (c) We have,=
8. (b) We have, 2
 13π  −1   2π   π π 2π 2π
sin −1 =
 cos  sin cos  3π −   As − ≤ sin −1 x ≤ ⇒ − ≤ 2sin −1 x ≤
 5    5  2 2 2 2

 2π    π 2π   2π 3π 3π 2π 3π
⇒ sin −1  − cos  = sin −1  − sin  −   ⇒− + ≤ 2sin −1 x + ≤ +
 5    2 5  2 2 2 2 2
π 5π  π 5π 
⇒
 π
sin −1  − sin = −1   π   −π ⇒ ≤ f ( x) ≤ ⇒ Range of f ( x ) is  , 
 sin sin  − 10=
 2 2 2 2 
 10      10
9. (c) We have, Here, Assertion (A) is true but Reason (R) is false

sin–1 x > cos–1 x 4. (c) Let cos–1 x = q ⇒ x = cos q

⇒ q ∈ [0, p] ⇒ 0 ≤ q ≤ p ⇒ 0 ≤ q2 ≤ p2
π π π
⇒ sin −1 x > − sin −1 x ⇒ 2sin −1 x > ⇒ sin −1 x >
2 2 4 ⇒ Maximum value of q2 is p2
π 1 ⇒ Maximum value of (cos–1x)2 is p2.
⇒ x > sin ⇒x>
4 2 ⇒ Here, Assertion (A) is true but Reason (R) is false.

73 Inverse Trigonometric Functions P

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 Subjective Questions Y

Very Short Answer Type Questions

π
−1
1. Given, tan −1
3 – sec–1 (–2) = tan 3 – (p – sec–1 2) 4

π sin −1 x

= tan  tan  − π + sec −1 2 
−1 (½ M) −1
 3
2
π  1  −2π  π X' X
= − π + cos −1 =  + cos −1  cos  (½ M) 0
3 2 3  3 π
2π π π 4
=− + =− (1 M)
3 3 3 −π
  −17 π   −1   17 π   4
2. sin −1 sin    = − sin sin   (½ M) (1 M)
  8    8  Y'

  π   π π  1 1   π π
= − sin −1 sin  2π +   = − sin −1 sin  = − (½ M) f ( x ) sin −1 x, x ∈  −
Range of = ,  is  − ,  (1 M)
  8    8  8  2 2   4 4
7. Given function,
−1  1  −1  1  −1   π 
3. Given, tan  −  + cot   + tan  sin  −   f(x) = cos–1 x, x ∈ [–1,0]
 3  3   2 
When x = –1 ⇒ f(–1) = cos–1 (–1) = cos–1 (cos p) = p
 π  π
=− tan −1  tan  + cot −1  cot  + tan −1 ( −1) (½ M)  π π
 6  3 When x = 0 ⇒ f(0) = cos–1 (0) = cos–1  cos  =
 2 2
π π π π π π
=− + + tan–1 (–1) = − + − = (1½ M) π 
6 3 6 3 4 12 Its range is  , π  .
2 
4. Let y = cosec–1 (2) ⇒ cosec y = 2
π π π 
We know that, cosec = 2 , So cosec y = cosec (½ M) ∴ Range of cos–1x is  , π  . (1 M)
6 6 2 
∴ Graph of f(x) = cos–1 x shown alongside.
 π π
Since range of Principal value of cosec–1 is  − ,  . (½ M)
 2 2 Y
π  π π
∵ ∈ − ,  p
6  2 2
π cos–1x π
Therefore, Principal value of cosec–1 (2) is . (1 M)
6 2
−1  −1  1 π
− ⇒ cos y =
5. y = cos   ⇒ cos y = − cos
 2  2 3 X
–1 O
 π  2π  (1 M)
=
cos y cos  π − = cos    (1 M)
 3  3 
−1  π π
8. ∵ tan x : R →  − ,  , R is one-one and onto.
−1  −1  2π  2 2
Therefore, Principal value of cos   is . (1 M)
 2  3  π π
So its inverse exists and is given by tan–1 x: R →  − , 
 1 1   2 2
f ( x ) sin −1 x, x ∈  −
6. Given function, = , 
 2 2 (1 M)
 π π
 −1   1    π  π Domain = R, Range =  − ,  (1 M)
⇒ f =sin −1  − =sin −1  sin  −   =
−  2 2
 2  2   4  4
 1  −1  1  −1   π  π
and f  =  sin  =  sin  sin  − =

 2  2   4  4 Nailing the Right Answer
 1 1  Students should remember the domain and range of inverse
f ( x ) sin −1 x, x ∈  −
Graph of = ,  trigonometric functions to avoid mistakes.
 2 2

P
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 −1  3π  −1  3π   x x  x x 
9. We have, sin  sin  + cos  cos  + tan (1)   sin 2 + cos 2  +  sin 2 − cos 2  
−1

 4   4  = cot −1     

 x x   x
  sin + cos  −  sin − cos   x 
  π   3π    π
= sin −1 sin  π −   + cos −1  cos   + tan −1  tan    2 2  2 2  
  4    4    4
 x
(1 M) −1
 2sin 2  −1  x
= cot
=   cot  tan 
x
 2 cos   2 
(½ M)
 π  3π   π
=sin  sin  + cos −1  cos  + tan −1  tan 
−1
(½ M)  2
 4  4   4
  π x  π x
π 3π π 5π =cot −1  cot  −   = −
= + + = (½ M)   2 2  2 2
4 4 4 4 (½ M)

−1   −7 π   −1   π  Short Answer Type Questions

10. We have, cos cos=
   cos cos  −2π −   (1 M)
  3    3  1. We know that the range of principal value of cosec–1x is
 −π π 
 π π π   2 , 2  – {0}.
= cos −1  cos=  ∈ [ 0, π]  (1 M)  
 3 3 3 
Let cosec–1 (–1) = q ⇒ cosec q = –1 (1 M)
−1  1  −1  1  −1  π π  π  π  π
11. cos =   cos = =
  cos  cos  ∵ ∈ [ 0, π] ⇒ cosec q = – cosec ⇒ cosecq = cosec −  (1 M)
2 2  3 3  3  2  2
π
(½ M) ⇒ θ=− (1 M)
2
−1  1  −1  π π  π  π π 
=
Also, sin   sin =  sin  ∵ 6 ∈  − 2 , 2  
 
2  6 6   
Mistakes 101 : What not to do!
(½ M)
Not knowing the domain and range of inverse trigonometric
1 1 π  π  π π 2π functions can lead to students making mistakes.
∴ cos −1   + 2sin −1   = + 2   = + = (1 M)
2 2 3 6 3 3 3
1 π
 2 x +1 
am ⋅ an , ( am ) =
∵ a m + n = a mn  2. We know that, sin–1   =
n
12. L.H.S = sin −1  x   
2
  6
1 + 4 
  1 
  ∴ tan 1 2 cos  2 sin 1   (½ M)
2.2 x   −1  2θ     2 
= sin −1  ∵ sin  1 +=2 
2 tan −1 θ  (1 M)
1 + ( 2 x ) 
2
  θ  
  1    
= tan 2 cos  2    (1 M)
= 2 tan–1 (2x) = R.H.S   6 

\ L.H.S = R.H.S Hence, proved. (1 M) 1   1  1 π

= tan  2 cos  = tan  2   = tan–1(1) =  (1½ M)
13. cos–1 a + cos–1 b + cos–1 g = 3π  3  2 4

⇒ cos–1a = p,cos–1 β = π and cos–1g = π

\ α = β = γ = –1 (1 M) Mistakes 101 : What not to do!
α (β + γ) – β (g + a) + γ (α + β) Students should know the principal value concept of
= (–1)(–1–1) – (–1)(–1–1) + (–1)(–1–1) trigonometric functions.
= 2 – 2 + 2 = 2 (1 M)
 π  π 
3. Here, sec sec ( −=
2 ) sec −1  − sec 
−1 −1

−1  1 + sin x + 1 − sin x 
 ∵ sec= 2
14. cot    3  3 
 1 + sin x − 1 − sin x  (1 M)
 x x
2
x x 
2   π   π
   = sec −1  sec=
π −  sec −1  sec  
 sin + cos  +  sin − cos  
(1 M)
−1   2 2  2 2    3   3
=

cot  (1 M)
 x x
2
x x 
2 2π  2π  π 
∵ 3 ∈ [ 0, π] −  2 
  =
  sin + cos  −  sin − cos   3   
(1 M)
  2 2  2 2 

75 Inverse Trigonometric Functions P

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 1 1 π  −1 3 3  3 2
4. Let cos–1   = x. Then, cos x = = cos   . 2. tan  sin cot −1  tan  tan −1 + tan −1  
+= (1 M)
2 2 3  5 2   4 3
1 π   3 2 
\ cos–1   = (1 M)
2 3  −1  4 + 3  
= tan  tan  
1   3 2   (1½ M)
1 π  1− × 
Let sin–1 =   = y. Then, sin y = = sin   .   4 3 
 
2 2 6
1 π   9 + 8 
\ sin–1   = (1 M) = tan  tan −1   (1½ M)
2 6   12 − 6  
 1  1  π 2π π π  −1 17  17
\ cos –1   − 2sin −1   = − = − = 0 (1 M)
= tan
= 
 2  2 3 6 3 3  tan  (1 M)
 6 6
1  1
5. Given, sin −1   − cos −1  −  3. After dividing numerator and denominator by a3 we have
3  3
  x   x 3 
We know that cos–1(–q) = p – cos–1q  3  −   
a a x  x
1   1  tan −1       put = tan θ and =
θ tan −1  
= sin −1   −  π − cos −1    (1 M)  x
   a a
3   3   1 − 3  
 a 
1 1
= sin −1   − π + cos −1   (1 M) (1½ M)
3 3
 3 tan θ − tan 3 θ 
−1  1 π π
−1  1  = tan −1  
= sin   + cos   − π = − π = − (1 M)  1 − 3 tan θ 
2 (1 M)
 
3  
3 2 2
= tan–1 (tan 3q) (1 M)
= 3q (½ M)
Mistakes 101 : What not to do!
 x
Students make mistakes by using the wrong principal value = 2 tan −1   (1 M)
a
of trigonometric functions.
π
4. Given that, sin (1 − x ) − 2sin ( x ) =
−1 −1

2
Long Answer Type Questions
π
1. Let x = tan q ⇒ q = tan–1 x ⇒ sin −1 (1 − x ) = + 2sin −1 ( x )
2
2x  2 tan θ 
∴ sin −1 sin −1 
=  π 
1 + x2  1 + tan θ 
2 ⇒ 1−
= x sin  + 2sin −1 ( x )  (1 M)
2 
= sin–1 (sin 2q) = 2q = 2 tan–1x (1 M)
 π  
Let y = tan φ ⇒ φ = tan–1 y x cos ( 2sin −1 ( x ) ) ∵ sin  + θ=
⇒1 −=  cos θ  (1 M)
 2  
1 − y2 −1  1 − tan φ 
2
∴ cos −1
1+ y2
=
cos  
 1 + tan φ 
2 (1 M) = {
⇒ 1 − x cos cos −1 (1 − 2 x 2 ) }
= cos–1 (cos 2 φ) = 2φ = 2 tan–1 y (½ M) ∵2sin
 = −1
x cos −1 (1 − 2 x 2 ) 

1 2x −1 1 − y 
2
⇒ 1 – x = 1 – 2x2 ⇒ x = 2x2 ⇒ 2x2 – x = 0 (1 M)
∴ tan  2 tan −1 + cos  (½ M)
2 1 + x2 1+ y2  1
⇒ x ( 2 x − 1) = 0 ⇒ x = 0, x = (1 M)
1 2
= tan  2 tan −1 x + 2 tan −1 y  (1 M)
2 for x = 0, we have
= tan [tan–1 x + tan–1y] π π π π π
sin −1 (1 − 0 ) − 2sin −1 0 = −0⇒ −0 = −0⇒ =
2 2 2 2 2
 −1  x + y   x + y
= tan
=  tan   (1 M)
  1 − xy   1 − xy Hence, L.H.S = R.H.S proved. (1 M)

P
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 Case Based Questions cos C
(ii)= =
3 3
1 + 32
2
10
Case Based-I 1 1
=
(iii) ∵ tan B = , tan C
(i) We have, 2 3
tan B + tan C
∵ tan ( B + C ) =
B
1 − tan B tan C

1 1 5 5
+
= 2 3 = 6= 6= 1
1 1 6 −1 5
1− ×
30 m 2 3 6 6
π
+ C tan −=
⇒ B= 1
(1)
b 4
a
A D C OR
30 3m 10 3m
∵ From (i) and (ii) we have
Now in ∆ABD (right angled)
1 3
BD 30 1 sin B
= = , cos C
tan=
α = = 5 10
AD 30 3 3
1 2
⇒ tan=
α
1
= tan 30º ⇒ =
α 30º cos B = 1 − sin 2 B = 1− =
3 5 5

1 1 sin C = 1 − cos 2 C = 1 −

9
=
1
⇒ sin
= α sin
= 30º ⇒ α =sin −1  
2 2 10 10

3 ∴ cos (B + C) = cos B cos C –sin B sin C

(ii) We have from (i) a = 30º ⇒ cos
= α cos
= 30º
2 2 3 1 1 6 1
= × − × = −
 3 5 10 5 10 50 50
⇒ α =cos −1  
 2 
5 25 1 1
(iii) In right ∆BCD, we have = = = =
50 50 2 2
BD
tan =
β
DC
⇒ tan =
β
30
10 3
= 3 ⇒=
β tan −1
( 3) Case Based-III
OR (i) (a) As for f(x) =cos x
Let cos–1 x = y ⇒ x = cos y At x → 0, cos x is 1
–1 ≤ cos y ≤ 1 ⇒ –1 ≤ x ≤ 1 ⇒ Domain = [–1,1] At x → p, cos x is 1
0 ≤ y ≤ p ⇒ Range = [0, p] So, in [0, p] the function is one-one and onto.

Case Based-II (ii) (d) The domain of tan–1x is R.

(i) We have, tan −1   =

1 1  π π
2
B ⇒ tan B =  − 2 , 2  − {0}
(iii) (c) The range of the function cosec–1 x =
2  
(iv) (b) As sin–1 (– x) = –sin–1 x
1 1 1 1
tan −1   =
C ⇒ tan C =∴ sin=
B = (v) (a) As for cosine function x ranges from –1 to 1.
3 3 1 + 22
2
5

77 Inverse Trigonometric Functions P

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 Sexual Reproduction in
Flowering Plants
1
“In flowering plants, we see a prime
example of insect pollination, such as a
bee actively harvesting pollen from purple
aster. The bee’s body is designed for
pollination, showcasing its hairy texture
which traps pollen grains, facilitating
the transfer of pollen between flowers, a
critical process for plant reproduction.”

SYLLABUS &
WEIGHTAGE

Years
List of Concept Names 2022
2020 2023 2024 2025
(Term I)
Pre-Fertilisation: Structures and Events 2Q
(Flower structure; development of male and female (1 M Each)
– 7 Q* 1 Q (5 M) 1 Q (3 M)
gametophytes; pollination - types, agencies and 1 Q (4 M)
examples; out breeding devices; pollen-pistil interaction) 1 Q (5 M)
Double Fertilisation
Post-Fertilisation: Structures and Events
(Double fertilisation; post fertilisation events - 3Q
development of endosperm and embryo, development 1 Q (3 M) 4 Q* 1 Q (3 M) (1 M Each) –
of seed and formation of fruit; special modes- apomixis, 1 Q (5 M)
parthenocarpy, polyembryony; Significance of seed
dispersal and fruit formation)
For the year 2021, the exam was not conducted

* All questions were of MCQ type and carried equal marks.

 P
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Pre-fertilisation: Structures and Events Flower-A Fascinating Organ of Angiosperms

Gamete transfer through pollination

Sexual Reproduction Female Reproductive
Stigma
Gametogenesis / Gamete Formation Transfer of pollen grains to stigma of a Anther Part (Pistil)
Occurs in specialised structures: anthers pistil for fertilisation.
in Flowering Plants • Gynoecium in a flower
Filament Style

Pistil

Stamen
(male) and ovules (female). represents the female
Petal reproductive part.
Megasporogenesis and Monosporic Embryo sac Sepal Ovary • It can be a single pistil
Development Ovule (monocarpellary)
Megaspore mother cell (2n) • Ovule usually has a single embryo sac formed or multiple pistils
from a megaspore. (multicarpellary).

CBSE Class – XII BIOLOGY

Meiosis
Chalazal pole Male Reproductive Part (Stamen) • Pistils may be fused
Megaspore tetrad i.e., 4 Megaspores (n)
• Comprises a filament and a bilobed anther. (syncarpous) or
Outer
Inner • Anther is dithecous, containing four separate (apocarpous).
A single chalazal megaspore is functional; rest degenerate integument
integument
microsporangia that become pollen sacs.
Embryo sac
Functional Megaspore (n) Nucellus
Hilum Autogamy
1st nuclear division
Micropyle • Self-pollination within the same
Two nucleated stage Funicle
CONCEPT MAP

Pollinating agents Cross pollination Self pollination flower

Megasporangium (ovule) showing embryo sac • Ensures fertilisation but reduces
2nd nuclear division
genetic diversity.
Four nucleated stage • Requires perfect synchrony

walls are laid yet

3rd nuclear division Biotic Agent Abiotic Agents: Wind, Water Xenogamy Geitonogamy in pollen release and stigma

free nuclear and no cell

• Cross-pollination • Pollination

These divisions are strictly

Insects (Bees, butterflies, flies, ants, wasps, Wind pollinated f lower receptivity.
Eight nucleated stage • Flowers are not colorful and do between different between different
etc.), birds (sunbirds humming birds), bats, some • Some plants like Viola (Common
Cell walls are now laid down, forming primates (lemurs), arboreal rodents, and reptiles not produce nectar. flowers on different flowers on the Pansy), Oxalis, and Commelina
8 nucleated, 7-celled embryo sac (gecko lizard and garden lizard), etc. • Pollen grains are light and non- plants. same plant. produce chasmogamous (with
sticky. • Geitonogamy exposed anthers and stigma,
• Have exposed stamens and large is functionally allowing cross-pollination) and
Floral Adaptation feathery stigmas to facilitate cross-pollination cleistogamous (flowers do not open
• Nectar Production: Flowers produce nectar pollen dispersal and capture, e.g., but genetically and ensure autogamy) flowers.
to attract pollinators. grasses. self-pollination.

24
• Color and Scent: Bright colors and fragrances Water pollinated f lower
lure pollinators.

megaspore (monosporic development)

• Flowers are not very colorful and

Development of embryo sac from a single

• Mimicry: Some flowers mimic female insects do not produce nectar.
to facilitate pollination e.g., Ophrys
• Pollen grains are long, ribbon
• Offers Oviposition (egg-laying) site: In
some species floral rewards are in providing like released inside water and are
Filiform safe places to lay eggs; Some flowers, like the carried passively; some of them
apparatus reach the stigma and achieve
6-feet Amorphophallus and Yucca.
pollination.
Structure of Microsporangium • E.g., Vallisneria, Hydrilla and
• Each microsporangium has four wall layers (epidermis, endothecium and middle layers Zostera.
protect the microsporangium and aid in anther dehiscence and the tapetum nourishes developing
pollen grains) and a central sporogenous tissue that gives rise to pollen grains.
Pollen Grain (25-50 mm in diameter)
Epidermis
Component Main Function Location
Endothecium Exine (made of Provides protection due to its resistance to extreme conditions Outer layer of the pollen grain wall.
sporopollenin) and aids in the preservation of pollen grains as fossils.
Middle layers
Connective Intine (made of cellulose and Essential for the maturation of the pollen grain and pollen tube Inner layer of the pollen grain wall, beneath the
Microspore
Epidermis pectins) germination. exine.
mother cells
Endothecium Germ pores Serve as sites for pollen tube emergence, facilitating the Specific regions on the exine where sporopollenin
Sporogenous process of germination. is absent.
Middle layers tissue Tapetum Vegetative cell Provides nutrients and energy for the growth of the pollen Larger cell within the pollen grain, contains a large
tube. nucleus and abundant food reserves.
(a) Tapetum
Generative cell Divides mitotically to produce two male gametes for Smaller cell that floats in the cytoplasm of the
To Access One

(b)
fertilisation. vegetative cell, typically spindle-shaped.
Scan This QR Code

(a) T.S. of Young Anther   (b) Enlarged view of Microsporangium

Shot Revision Video

• 
Some plants release 2-celled pollen (one generative and one vegetative cell), while others disperse 3-celled pollen (one vegetative and two male gametes).
Microsporogenesis and Microgametogenesis:
Meiosis
Pollen mother cell (2n) Microspore tetrad i.e., 4 microspores (n) Microsporogenesis pollen yield: Pollen Viability and Storage:
Anther matures P=N×4 • 
Pollen viability varies, influenced by temperature and humidity.
and dehydrates ( each microspore mother cell produces 4 microspores/pollens) • 
Rice, wheat pollen: lose viability within 30 minutes of their release.
Pollen 
where, • 
Pollen in some members of Rosaceae, Leguminoseae, Solanaceae: viable for months.
Male Vegetative cell (n) mitosis I 4 Pollen grains (n)
gamete (n) Pollen
P = Number of pollen grains • 
Pollen can be stored for years in liquid nitrogen (-196°C), can be used as pollen
(Unequal or Male gametophyte N = Number of microspore mother cell banks in crop breeding programmes.
Male mitosis II Generative cell (n) division)
gamete (n)
 Endosperm Development Artificial Hybridisation: Steps For
Outbreeding Devices (To Avoid Inbreeding Depression)
Controlled Pollination in Bisexual Flowers
• Occurs before embryo development due to its crucial role in nutrition of embryo. • Encourage cross-pollination to overcome inbreeding depression.
• Primary endosperm cell (PEC) divides and forms a triploid endosperm tissue. • Flowering plants have developed many devices to discourage self pollination.
Emasculation:
• Contains reserve food materials for the developing embryo. Hereditary Device:
Remove anthers with forceps before pollen release.
• In the most common type of endosperm development: • Self-incompatibility: Genetic mechanism blocking self-pollen fertilisation by hindering
{ Primary endosperm nucleus (PEN) undergoes multiple nuclear divisions. Bagging: germination or pollen tube growth.
{ Results in formation of many free nuclei. • Cover emasculated flowers with butter paper bags to Physiological Device:
{ This phase is known as free-nuclear endosperm. avoid unwanted pollen contamination.
• Asynchronous pollen release and stigma receptivity.
{ Following free-nuclear stage, cell walls begin to form.
• Once stigma is receptive, dust selected pollen. • Spatial separation of anther and stigma i.e., the anther and stigma are placed at different positions
{ Endosperm shifts from a free-nuclear to a cellular structure.
• Rebag flowers for fruit development. so, that the pollen cannot come in contact with the stigma of the same flower.
Note: If the female parent produces unisexual flowers,
{ The number of nuclei before this transition varies widely. • Production of unisexual flowers in monoecious (castor, maize) and dioecious (papaya) plants.
there is no need for emasculation.
• Endosperm can be persistent (e.g., castor, coconut) or consumed by the embryo
(e.g., pea, groundnut, beans).

Sexual Reproduction in
Post-fertilisation Events Flowering Plants
Fruit Embryo development Pollen-Pistil Interaction
Ovary develops into a fruit, with the • Develops at the micropylar end of the embryo sac.
ovary wall becoming the pericarp. • Typically, zygotes divide only after some endosperm is formed.
• Early embryonic stages are similar in monocots and dicots. Double Fertilisation and
True Fruit • Components of Embryonal Axis: Epicotyl + Plumule + Fate of Male Gametes
Fruit develops solely from the Hypocotyl + Radicle • Chemical components of pollen and
ovary of the flower, e.g., mango. pistil determine compatibility. Pollen grains
• Compatible pollen triggers pollen tube
False Fruit formation on the stigma.
Fruit develops from both the Monocot Embryo • Certain plants release two-celled pollen Pollen tube
ovary and surrounding floral parts, • Monocotyledonous embryo have one cotyledon. Involves two fertilisation events in the
embryo sac, syngamy and triple fusion. (vegetative and generative cell).
like the thalamus, e.g., apple, • The generative cell divides into two male
Antipodal
Scutellum

25
strawberry, cashew, etc. Syngamy gametes during pollen tube growth. Polar nuclei
(A single cotyledon in grasses, situated laterally
of the embryonal axis) • Fusion of one male gamete with the • Three-celled pollen already contains
Parthenocarpic Fruits Egg cell
egg cell forming a diploid zygote (2n). two male gametes.
Fruit develops without fertilisation, Coleoptile (A hollow foliar structure that encloses Synergid
occurs in species like banana, a shoot apex and a few leaf primordia)
• Zygote develops into the embryo. • The pollen tube enters the ovule through
Shoot apex
the micropyle, then enters one of the
leading to seedless fruits. Triple fusion
synergids through the filiform apparatus.
• Fusion of the other male gamete with L.S. of a flower showing
Epiblast two polar nuclei in central cell forming growth of pollen tube
Seed (Represents rudiments of a triploid primary endosperm
second cotyledon) nucleus (PEN) (3n).
• Final product of sexual reproduction,
formed inside fruits. Radicle • After triple fusion, the central cell Polyembryony
• Ovules inside the ovary develop into becomes the primary endosperm cell
Root cap (PEC).
seeds.
• Consist of seed coat(s), cotyledon(s), and Coleorhiza (An undifferentiated sheath that • PEC develops into the endosperm, a
covers radicle and root cap) nutritive tissue. In plants like Citrus & Mango, nucellar cell division yields multiple
embryonal axis.
embryos in a single ovule, notably observable in squeezed orange seeds.
Seed dormancy: As seeds mature, they reduce water content (10-15% moisture by
Dicot Embryo
mass), and embryo metabolic activity slows down and may enter a state of dormancy.
• Dicotyledonous embryo has an embryonal axis with two
cotyledons. Apomixis
Non-albuminous Seed: No residual endosperm as it is completely consumed during
embryo development, e.g ., pea, groundnut, beans.

• Form of asexual reproduction in some flowering plants, mimicking sexual

Albuminous Seed: Retain a part of the endosperm as it is not completely used up
reproduction but occurring without fertilisation.
during embryo development, e.g., wheat, maize, barley, castor coconut. • Diploid egg cell in some species develops directly into embryo without reduction
division and fertilisation.
Cotyledons: Thick and swollen, store food reserves. • Mostly observed in Asteraceae and grasses.
• Valuable in agriculture, maintains hybrid traits efficiently.
Perisperm: Some seeds like black pepper and beet, remnants of the nucellus persist
as the perisperm.

Sexual Reproduction in Flowering Plants

Seed coat: Integuments of ovules harden to form tough seed coats, and the micropyle
allows oxygen and water entry during germination.

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1 PRE-FERTILISATION: STRUCTURES AND EVENTS

NCERT Definitions (Commonly asked in 1 mark)

U Stamen: The male reproductive part of a flower, consisting of a long and slender stalk called filament, and a bilobed anther that
produces pollen.
U Microsporangium: A structure within the anther, located at the corners, where pollen grains are produced and which further
matures to form pollen sacs.
U Dithecous: An anther with two thecae, or compartments, each containing two microsporangia.
U Tapetum: Innermost wall layer of microsporangium that nourishes pollen grains.
U Sporogenous tissue: A group of compactly arranged homogenous cells located in the centre of each microsporangium
U Microspore tetrad: Arrangement of microspores in a cluster of four cells.
U Pollen grain: The male gametophyte in angiosperms that carries male germ cells to the female reproductive structures.
U Intine: The inner wall of a pollen grain, characterised by a thin and continuous layer composed of cellulose and pectin.
U Exine: Hard outer layer of a pollen grain made up of sporopollenin.
U Pistil: The female reproductive part of a flower, consisting of stigma(serves as a landing platform for pollen grains), style
(an elongated stalk-like structure that supports the stigma) and the ovary (basal bulged part that contains the ovules).
U Megasporangium (Ovule): Structure within the ovary of the flower where the female gametophytes are developed.
U Funicle: The stalk-like structure that attaches the ovule to the placenta.
U Hilum: The region where the body of the ovule fuses with the funicle.
U Integuments: Protective envelopes that surround the ovule.
U Embryo sac: The female gametophyte within the ovule, where fertilisation occurs.
U Micropyle: A small opening in the integuments of the ovule through which the pollen tube enters.
U Nucellus: Mass of cells having abundant food reserves, enclosed by integuments.
U Chalaza: Basal part of the ovule that lies opposite to the micropyle.
U Anatropous: The type of ovule where the body of ovule is inverted at 180° and micropyle lies near the funicle while chalazal
end is present on the opposite side.
U Locule: Ovarian cavity where placenta is located.
U Monosporic development: Method of embryo sac formation from a single megaspore.
U Stigma receptivity: The ability of the pistil to accept or reject pollen grains based on chemical dialogue. This dialogue is
mediated by chemical components of the pollen interacting with those of the pistil.
U Inbreeding depression: Refers to the reduced survival and fertility of offspring of related individuals.
U Artificial hybridisation: Method in crop improvement programmes where different species or genera are crossed to combine
desirable traits and produce commercially ‘superior’ varieties.
U Emasculation: Removal of anthers from the bisexual flower bud before the anther dehisces using a pair of forceps.
U Bagging: It is the procedure where emasculated flowers are covered with a butter paper bag to prevent contamination from
unwanted pollen.

Important Facts

The most resistant organic materials found in the outer layer of the pollen grain capable of withstanding
01 high temperatures, strong acids, alkalis, and enzymes. ~ Sporopollenin

02 The number and length of stamens vary across different species. ~ Stamens variability

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 The wall layer of microsporangium possesses dense cytoplasm and generally have more than one nucleus.
03 ~ Tapetum

04 An ovule generally has a single embryo sac formed from a megaspore.

In over 60% of angiosperms, pollen grains are shed at a 2-celled stage while in others, the generative cell divides
05 mitotically to form two male gametes before shedding, resulting in a 3-celled stage.

06 One meiotic event produces four pollen grains, and each pollen grain can produce two male gametes.

07 The formation of an ‘8-nucleate, 7-celled’ embryo sac involves one meiotic and three mitotic divisions.

The cellular thickenings at the micropylar tip play an important role in guiding the pollen tubes into the synergid.
08 ~ Filiform apparatus

The plant that provides floral rewards as safe places for insects to lay eggs and has a flower about 6 feet in height.
09 ~ Amorphophallus

In Vallisneria, female flowers reach the surface of water via long stalk, while male flowers or pollen released on to the
10 surface of water, carried by water currents to achieve pollination while in sea grasses (Zostera), submerged female flowers
receive elongated, ribbon-like pollen grains directly in water.

In some cereals such as rice and wheat, pollen grains lose viability within 30 minutes of their release, and in some
11
members of Rosaceae, Leguminoseae and Solanaceae, they maintain viability for months.

Classification

Ploidy of parts of ovule

Funicle Hilum Chalaza Integuments Micropyle Nucellus Embryo sac

(2n) (2n) (2n) (2n) (2n) (2n) (n)

Egg apparatus Secondary nucleus Antipodal cells

Egg cell (n)
•  (2n) : Product of fusion
(n)
Synergids (n)
•  of 2 haploid (n) polar
nuclei

27 Sexual Reproduction in Flowering Plants P

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 Types of Pollination depending on the source of pollen
Pollination Pollen source Genetic outcome in Common in plants Requirement of
type offsprings pollinating agents
Autogamy Same flower Same genetic material Viola, Oxalis, and Commelina May or may not be
(self-pollination) required
Geitonogamy Different flower on the Same genetic material Common in plants where flowers are Required
same plant (cross-pollination) closely spaced
Xenogamy Different flower on a Genetically diverse (cross- Preferred in diverse ecosystems with Required
different plant pollination) varied plant species

Types of pollination based on agents of pollination

Aspect Wind pollination Water pollination Animal pollination
Pollen Light, non-sticky pollen Ribbon-like, have mucilaginous covering Sticky or spiny pollen to
characteristics Pollen may be released on surface of adhere to animals
water or underwater.
Flower No scent or nectar; exposed stamens. Subtle features, non-showy Bright colourful, scented,
characteristics Large, feathery stigmas to catch pollen. nectar-rich
Often single ovule in each ovary.
Examples Grasses, corn, etc Freshwater (Vallisneria, Hydrilla), Orchids, roses, etc
Marine (Zostera)

Devices to prevent self-pollination

Asynchrony between pollen Different positions of Self-incompatibility Production of

release and stigma receptivity anther and stigma unisexual flowers

Difference Between
Microsporogenesis vs. Megasporogenesis
Feature Microsporogenesis Megasporogenesis
Formation of microspores from pollen mother Formation of megaspores from megaspore mother cells
Definition
cells (PMC) (MMC)
Location Anthers within the stamen Ovules within the ovary
Microspores that develop into male Megaspores that develop into female gametophyte
Product
gametophyte (pollen grains) (embryo sac)
Number of functional Four megaspores: One (functional) and three
Four functional microspores
gametes produced (degenerate)
Cleistogamous vs. Chasmogamous flowers
Aspect Cleistogamous Flowers Chasmogamous Flowers
Definition Flowers do not open at all Flowers has exposed anthers and stigma
Pollination Always autogamous, as they self-pollinate. Can be autogamous, geitonogamous and xenogamous
Seed Set Assured seed-set even without pollinators Often dependent on pollinator availability or other factors
Vegetative cell vs. Generative cell
Vegetative cell Generative cell
Size Large Smaller and floats in the cytoplasm of the vegetative cell
Presence of reserve food
Contains abundant food reserve Does not have a food reserve
material
Shape of nucleus Large, Irregular Spindle shaped with dense cytoplasm

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 Monoecious vs. dioecious
Monoecious Dioecious
Both male and female flowers are present on the same plant. Male and female flowers are present on separate plants.
Prevents autogamy but not geitonogamy Prevents both autogamy and geitonogamy
E.g., Castor, maize E.g., Papaya

Real Life Application Based Questions

1. Have you ever heard that some people have allergies to pollen grains? Are these allergies seasonal? What are the common
symptoms?
Ans. Yes, pollen grains from many species can cause severe seasonal allergies leading to chronic respiratory disorders such as asthma
and bronchitis. Symptoms typically include sneezing, coughing, and breathing difficulties.
2. How does pollen storage help conserve plant biodiversity?
Ans. Storing pollen aids in protecting plant species, especially those at risk of extinction. It allows for future plant reproduction through
artificial pollination, supporting efforts to maintain or increase plant diversity in nature and botanical gardens.
3. Have you ever seen bees frequently visiting certain flowers in the garden, like roses or sunflowers? Why do you think bees
are particularly attracted to these flowers, and how does this benefit both the bees and the flowers?
Ans. Yes, bees are attracted to these flowers primarily for nectar, which serves as a food source. In return, bees aid in pollination by
transferring pollen from one flower to another, thus playing a crucial role in the reproductive cycle of plants.
4. Why might a flower evolve to have a specific shape, size, or color? Consider the orchid Ophrys, which mimics female bees
to attract male bees. How does this specific adaptation benefit the orchid?
Ans. The orchid Ophrys benefits from mimicking female bees as it ensures that male bees, attempting to mate with the flower,
inadvertently pollinate it. This specificity in floral morphology ensures efficient and targeted pollination by exploiting the mating
behaviors of bees.
5. When we observe certain plants, such as Yucca, closely, we can see the eggs and larvae of some insects on them. What leads
these specific insects to choose these plants for laying their eggs, and how does it benefit the insects?
Ans. Insects like moths lay eggs on specific plants such as Yucca because these plants provide a secure habitat for their offspring.
Both species, moths and the plant ,cannot complete their life cycles without each other. The moth deposits its eggs in the locule
of the ovary and the flower, in turn, gets pollinated by the moth. The larvae of the moth come out of the eggs as the seeds start
developing.
6. Farmers often seek to increase crop yield through selective breeding. How can understanding the mechanism of pollen-pistil
recognition help farmers ensure that only desired pollen successfully fertilises crops?
Ans. Understanding pollen-pistil recognition allows farmers to selectively breed crops by ensuring compatibility between specific
pollen and pistils. This targeted approach can increase crop yields, improve genetic traits, and enhance resistance to diseases by
promoting successful cross-pollination with desired traits.
7. In ecosystems where both monoecious and dioecious plants are present, how might the presence of unisexual flowers affect
the plant community’s structure and reproductive success?*
Ans. The presence of unisexual flowers in monoecious and dioecious plants can influence the plant community by reducing the
likelihood of self-pollination and enhancing cross-pollination between different plants. This can lead to a more genetically diverse
and resilient plant community, which is crucial for ecological stability and adaptation.

Myth Buster

U Myth: All flowers are colorful and have a pleasant scent.

Fact: While many flowers are colorful and scented to attract pollinators, some are not. Some flowers, like those pollinated by
wind or water, may be small, lack scent, and appear dull because they do not need to attract animal pollinators.
U Myth: All pollination leads to fertilisation.
Fact: Not all pollination events result in fertilisation. For fertilisation to occur, the pollen that lands on the stigma must be
compatible. Incompatible pollen, whether from a different species or from the same plant in cases of self-incompatibility, does
not lead to successful fertilisation.

* Concepts beyond the board exam syllabus, offering deeper insight and critical thinking on NCERT topics.

29 Sexual Reproduction in Flowering Plants P

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 U Myth: All brightly colored and scented flowers are pollinated by insects.
Fact: While insects are the most common pollinators, many other animals, such as birds, bats, primates (lemurs), arboreal
(tree-dwelling) rodents, or even some reptiles (gecko and garden lizard), can also pollinate flowers.
U Myth: Pollen’s only role is to fertilise the ovule.
Fact: Besides fertilisation, pollen grains are rich in nutrients and are used as dietary supplements in the form of tablets and
syrups, believed to enhance athletic and race horse performance.
U Myth: Pollination by water is common among all aquatic plants.
Fact: Pollination by water is quite rare and limited to about 30 genera of mostly monocotyledons. Many aquatic plants, such as
water lilies and water hyacinth, have flowers that emerge above water and are pollinated by insects or wind, not by water.
U Myth: Cleistogamous flowers (flowers that do not open) are less evolved than open, pollinated flowers.
Fact: Cleistogamous flowers represent a highly evolved strategy to ensure seed production, especially under adverse conditions
or when pollinators are scarce. These flowers self-pollinate internally, guaranteeing reproduction without the need for pollinator
services, which can be an advantage in stable but resource-limited environments.

Mnemonics

U Wall layers of microsporangium: “Every Elephant Might meet Tigers”

Every Elephant Might meet Tigers
↓ ↓ ↓ ↓
Epidermis Endothecium Middle layers Tapetum
U Examples of plants that produce two types of flowers – Chasmogamous and Cleistogamous: “Power, Control, Ownership”
Power Control Ownership
↓ ↓ ↓
Common pansy (Viola) Commelina Oxalis
U To remember examples of monoecious and dioecious plants: “Make Connections Manifest Dreams Pursue”
Make Connections Manifest Dreams Pursue
↓ ↓ ↓ ↓ ↓
Monoecious Castor Maize Dioecious Papaya
U To remember arrangement of cells and nuclei in the embryo sac:
“At Chalazal’s Base,
Three Antipodals Race,
in the Centre,
Two Polar Nuclei Embrace,
near the Micropyle,
Synergids and an Egg Face,
Making the Embryo sac,
a Fascinating Place.”
U Examples of a single ovule ovary: “Man Won iPad”
Man Won iPad
↓ ↓  ↓
Mango Wheat  Paddy
U Examples of a multiple ovule ovary: “Papa Waters Orchids” Embryo sac
Papa Waters Orchids
↓ ↓ ↓
Papaya Watermelon Orchids

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 competency BASED SOlved Examples
(c) Cleistogamous flowers exhibit both autogamy and
Multiple Choice Questions geitonogamy.
(1 M)
(d) Chasmogamous flowers never exhibit autogamy.
1. Pollen grains retain viability for months in plants
8. The figure given below shows megasporogenesis
belonging to different families given below:
and development of typical female gametophyte in
(Re) (CBSE, 2022 Term-I)
angiosperms. In which of the following options all
(i) Solanaceae (ii) Leguminoseae
divisions (D 1 to D 5) and structure (S) are correctly
(iii) Gramineae (iv) Rosaceae
identified? (Un)
(v) Liliaceae
The correct option is:
(a) (i), (ii) and (v) (b) (i), (ii) and (iv) D1 D2 D3
(c) (ii), (iv) and (v) (d) (i), (iii) and (v)
2. Study the given diagram and choose the correct option Megaspore Megaspore Megaspore Functional
against ‘A’ and ‘B’ (Re) (CBSE APQ, 2023) mother cell dyad tetrad megaspore
(a) A-Egg apparatus; B-Polar body A (MMC)
(b) A-Antipodals; B-Egg apparatus
D4 D5
(c) A-Synergids; B- Egg apparatus
(d) A-Central cell; B-Antipodals B
3. The outermost and innermost wall layers of S
microsporangium in an anther are respectively: (a) D1 - Meiosis I (b) D1 - Meiosis I
(Re) (NCERT Exemplar) D2 - Meiosis II D2 - Meiosis II
(a) Endothecium and tapetum D3 - Mitosis D3 - Mitosis
(b) Epidermis and endodermis D4 - Mitosis D4 - Mitosis
(c) Epidermis and middle layer D5 - Mitosis D5 - Mitosis
(d) Epidermis and tapetum S - Microgametophyte S - Embryo
4. Which of the following structures contains PMC? (Re)
(c) D1 - Meiosis I (d) D1 - Mitosis
D2 - Meiosis II D2 - Meiosis
B D3 - Mitosis D3 - Mitosis
D4 - Mitosis D4 - Mitosis
A D5 - Mitosis D5 - Mitosis
C
S - Embryo sac S - Embryo sac
(a) Only A (b) Only B 9. During the pollen grain formation, the generative cell
(c) Both A and C (d) Only C divides to give rise to the two male gametes. What is the
5. Starting from the innermost part, the correct sequence ploidy of the generative cell? (Re)
of parts in an ovule are,  (Un) (NCERT Exemplar) (a) n (b) 2n
(a) egg, nucellus, embryo sac, integument
(c) 3n (d) 4n
(b) egg, embryo sac, nucellus, integument
(c) embryo sac, nucellus, integument, egg 10. Choose the correct option according to P & Q. (Un)
(d) egg, integument, embryo sac, nucellus
6. “Cells of the tapetum of a microsporangium are usually
multinucleate”.
Which of the following can be a reason for the tapetal cells
to become multinucleate?  (An) (CBSE CFPQ, 2023)
(a) They fuse with the polar cells of the megasporangium.
(b) They do not undergo karyokinesis.
(c) They do not undergo cytokinesis. P Q
(d) They do not undergo mitosis. (a) P- Multicarpellary, syncarpous pistil of Papaver.
7. Choose the correct statement from the following: (Un) (b) Q- Multicarpellary, syncarpous gynoecium of Michelia.
(a) Cleistogamous flowers always exhibit autogamy. (c) P- Multicarpellary, apocarpous pistil of Papaver.
(b) Chasmogamous flowers always exhibit geitonogamy. (d) Q- Multicarpellary, apocarpous androecium of Michelia.

31 Sexual Reproduction in Flowering Plants P

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 11. Consider three plants with the following modes of 9. Assertion (A): Corn cobs have a multiple ovule in each
pollination:  (Un) (CBSE CFPQ, 2024) ovary and numerous flowers packed into an inflorescence.
Plant P: autogamy Reason (R): Corn plants tassels wave in the wind to trap
Plant Q: xenogamy pollen grains. (Re)
Plant R: geitonogamy 10. Assertion (A): The flower of Amorphophallus provides a
safe place for insects to lay eggs, which assists in pollination.
Which of the above case/s is/are most likely to NOT show
genetic variation in the offspring? Reason (R): Providing a safe place for egg-laying by insects
increases the chances of these insects visiting the flower,
(a) only P (b) only Q thereby enhancing pollination. (Un)
(c) only P and R (d) only Q and R

Assertion and Reason Subjective Questions

(1 M)
Direction: The following questions consist of two statements Very Short Answer Type Questions  (2 M)
– Assertion (A) and Reason (R). Answer these questions by
selecting the appropriate option given below: 1. You are conducting artificial hybridization on papaya
and maize. Which one of them would require the step of
(a) Both A and R are true, and R is the correct explanation emasculation and why? However for both you will use
of A. the process of bagging. Justify giving one reason.
(b) Both A and R are true, but R is not the correct explanation (Un) (CBSE, 2019)
of A.
Ans. Maize requires emasculation because it is a monoecious
(c) A is true, but R is false. plant having unisexual flowers; removal of stamens
(d) A is false, but R is true. prevents self-pollination and ensures cross-pollination in
1. Assertion (A): Pollen tube germinates through the germ hybridization whereas papaya is unisexual so there is no
pores on the pollen grains. need for emasculation.  (1 M)
Reasoning (R): Pollen-pistil compatibility chemicals help However, for both papaya and maize, the process of bagging
to dissolve sporopollenin for the pollen tube to germinate. is needed to prevent cross-pollination from other plants.
(Un) (CBSE CFPQ, 2023) Bagging helps in ensuring controlled pollination, which is
2. Assertion (A): The microsporangia are located at the corners necessary for artificial hybridization. (1 M)
of the anther. 2. The exine layer of pollen grains contains sporopollenin
Reason (R): The length of the stamen depends on the number which is a highly resistant chemical. Sporopollenin allows
of microsporangia in the anther. (Un) pollen grains to be well-preserved as fossils.
3. Assertion (A): Tapetum is the innermost layer of anther (a) Can fossilised pollens fertilise an ovum of the same
which is single layered. species in the present day? Justify.
Reason (R): Tapetum is polyploid and multinucleated. (b) How do scientists preserve pollen grains for later
(Un) use?  (Ap) (CBSE CFPQ, 2023)
4. Assertion (A): Chalaza is found opposite the micropyle in Ans. (a) No
an ovule. Reason: Fossilised pollen grains, despite being well-
Reason (R): Integuments encircle the nucellus and shield preserved due to the sporopollenin, are no longer viable for
it. (Un) such a long time as those taken for fossilisation. (1 M)
5. Assertion (A): The ploidy of the megaspore mother cell (b) Scientists preserve pollen grains by storing them at low
(MMC) is haploid (n). temperatures in liquid nitrogen at -196°C. (½ M)
Reason (R): The MMC undergoes meiotic division during Such stored pollen can be used as pollen banks,similar to
megasporogenesis. (Re) seed banks, in crop breeding programmes. (½ M)
6. Assertion (A): Pollen grains are well preserved as fossils. 3. “Continued self-pollination results in inbreeding
Reason (R): The exine of pollen grains is made up of depression”.
sporopollenin, which can withstand high temperatures, (a) Mention ONE impact of inbreeding depression on
strong acids and alkali. (Re) the upcoming generations in farmland.
7. Assertion (A): In maize, the presence of unisexual flowers (b) State ONE way in which cross-pollination helps in
on the same plant helps in preventing self-pollination. avoiding inbreeding depression.
Reason (R): Maize plants are dioecious. (Un) (Ap) (CBSE CFPQ, 2023)
8. Assertion (A): The introduction of a mutation in the tapetum Ans. (a) Inbreeding depression can result in loss of fertility and
cells of anthers can affect pollen viability. vigour in the existing population. (1 M)
Reason (R): The tapetum layer provides essential nutrients (b) Cross pollination brings about variation of characters
for the developing pollen grains. (Un) that help in increased vigour of the population. (1 M)
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 Short Answer Type Questions (b) Employing Bagging for Cross-Pollination:
(3 M)
z After emasculation, the flowers of Variety A should
1. How is pollination carried out in water plants? be covered with a bag to prevent any pollen from
(Un) (NCERT Exemplar) contaminating the stigma. (½ M)
Ans. Pollination in water plants can occur in many ways z During the flowering period, remove the bag and
depending on the plant species. manually introduce pollen from Variety B to ensure
In Vallisneria and Hydrilla (Freshwater): Male flowers cross-pollination. After pollination, re-bag the flowers
release pollen onto the water surface, which is carried by to protect against any other pollen source. (½ M)
currents to female flowers that have reached the surface via
long stalks, achieving pollination. (1 M)
In Zostera (Marine seagrasses): Female flowers remain Key Takeaways
submerged in water, and the long, ribbon-like pollen grains
are released directly into the water and carried passively to From the answer, students should recognize the importance
reach the stigma and achieve pollination. (1 M) of distinguishing between bisexual and unisexual flowers
in crop hybridization, implementing emasculation and
In water hyacinth and water lily: Flowers emerge above the
bagging where necessary to ensure effective cross-
level of water and are pollinated by insects or wind as in
most of the land plants. (1 M) pollination.

2. One of the major approaches of crop improvement 3. Draw a schematic transverse section of a mature anther
programmes is Artificial Hybridisation. Explain the of an angiosperm. Label its epidermis, middle layers,
steps involved in making sure that only the desired pollen tapetum, endothecium, sporogenous tissue and the
grain pollinates the stigma of a bisexual flower by a plant connective. (Cr) (CBSE, 2020)
breeder. (Ap) (CBSE, 2023) Ans.
OR
A botanist is conducting a research project to create
a new hybrid plant species with desirable traits. The
botanist is working with two different plant varieties,
Variety A has bisexual flowers and Variety B has
unisexual flowers having stamens . The goal is to achieve
successful artificial hybridization while preventing self-
pollination and ensuring cross-pollination. (An)
(a) How can the botanist prevent self-pollination in
variety A and variety B?
(b) How can the botanist ensure that only cross-
pollination occurs between Variety A and Variety B?
Ans. Artificial hybridisation, a key crop improvement strategy, (3 M)
involves controlled pollination, safeguarding the stigma Topper’s Explanation
from unwanted pollen through two steps:
Emasculation: If the female parent bears bisexual flowers,
removal of anthers from the flower bud before the anther
dehiscence using a pair of forceps is necessary. This step is
referred to as emasculation. (1 M)
Bagging: Emasculated flowers have to be covered with a
bag of suitable size, generally made up of butter paper, to
prevent contamination of its stigma with unwanted pollen.
This process is called bagging. (1 M)
When the stigma of bagged flower attains receptivity, mature
pollen grains collected from anthers of the male parent are
dusted on the stigma, and the flowers are rebagged, and the
fruits allowed to develop. (1 M)
OR
(a)
z In Variety A (bisexual flowers):The botanist should Nailing the Right Answer
perform emasculation, removing the male reproductive Please ensure that the diagram of the transverse section of
parts (anthers) to prevent self-pollination. (1 M) a mature anther is neat and clear. Avoid confusion among
z In Variety B (unisexual flowers): No emasculation is the different layers of the microsporangium.
needed.  (1 M)

33 Sexual Reproduction in Flowering Plants P

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 4. When and where do tapetum and synergid cells develop z Xenogamy: Pollen is transferred to a flower on a
in flowering plants? Mention their functions. different plant, ensuring cross pollination and genetic
(Un) (CBSE, 2019) diversity. (1 M)
OR
Long Answer Type Questions  (5 M)
Where are the following structures present in a male
gametophyte of an angiosperm? Mention the function 1. With a neat, labelled diagram, describe the parts of a
of each one of them. (Un) (CBSE, 2019) mature angiosperm embryo sac. Mention the role of
(a) Germ pore synergids. (Cr, Re) (NCERT Exemplar)
(b) Sporopollenin
(c) Generative cell
Ans. Tapetum development: Ans.
z During microsporogenesis within the microsporangium.
(½ M)
z Develops inside the anthers of a stamen, specifically
within the microsporangia as one of the four wall layers
surrounding the sporogenous tissue. (½ M)
Function of tapetum:
z Nourishes the developing pollen grains. (½ M)
Synergid cells development:
z During megasporogenesis within the megasporangium
(ovule). (½ M)
z Located within the embryo sac of the ovule at the (2½ M)
micropylar end, forming part of the egg apparatus.
(½ M) Parts of the embryo sac:
Function of synergid: Have a filiform apparatus that plays an (i) Egg Apparatus: Located at the micropylar end, includes
important role in guiding the pollen tubes into the synergid. one egg cell and two synergids.
(½ M) (ii) Central Cell: Contains two polar nuclei, positioned in
OR the middle of the embryo sac.
(a) Germ pore (iii) Antipodal Cells: Found at the chalazal end, typically
z Location: Present on the pollen grain exine. (½ M) comprising three cells. (1½ M)
z Function: Serves as an aperture for the pollen tube to Role of synergids:
emerge during pollen germination. (½ M)
The synergids have special cellular thickenings at the
(b) Sporopollenin
micropylar tip called filiform apparatus, which play an
z Location: Found in the outer wall (exine) of pollen important role in guiding the pollen tubes into the synergid.
grains. (½ M) (1 M)
z Function: Provides protection against environmental
2. Draw the longitudinal section of the flower showing the
stresses; it is one of the most resistant organic materials
known. (½ M) growth of the pollen tube. (Cr)
(c) Generative cell Ans.
z Location: Located inside the pollen grain, initially as
part of the vegetative cell’s cytoplasm. (½ M)
z Function: Divides to form two male gametes necessary
for fertilisation. (½ M)
5. Explain three different modes of pollination that can
occur in a chasmogamous flower. (Un) (CBSE, 2020)
Ans. Modes of Pollination in chasmogamous flowers:
z Autogamy: Pollen is transferred to the stigma of the
same flower, promoting self-pollination. (1 M)
z Geitonogamy: Pollen is transferred to a different
flower but on the same plant, effectively acting as self-
pollination genetically. (1 M) (5 M)

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 3. Analyse the different parts of an anatropous ovule given z Finally, cell walls are formed, organizing the six out
below. Answer the following questions it follows: (An) of the eight nuclei into cells, with the remaining two
A (polar nuclei) situated in the central cell below the egg
E apparatus. (½ M)
D
(b)

B
F

(a) Identify and describe the parts labelled A to F.

(2 M)
(b) Explain the significance of B in the context of megaspore
development. 5. Some structures of anther and ovule are given below.
Ans. (a) A: Hilum - It is the region where the body of the ovule Pollen mother cell, microspore, generative cell, nucellus,
fuses with a funicle. (½ M) synergids, antipodal cells. (Re, Un)
B: Nucellus - It is the tissue that surrounds the embryo sac (a) Mention the ploidy of given structures.
and provides nourishment to the developing gametophyte. (b) Explain the vegetative cell, its roles. What would
(½ M) happen if we destroy the vegetative cell by a laser?
C: Chalaza - Basal part of the ovule lying opposite to the (c) What percentage of angiosperms shed 2-celled pollen,
micropylar end. (½ M) and what occurs in others before release?
D: Micropyle - A small opening in the integuments of the
Ans. (a) Pollen mother cell - 2n
ovule that allows for the entry of the pollen tube during the
Microspore - n
process of fertilisation. (½ M)
Generative cell -n
E: Funicle - The stalk that attaches an ovule to the placenta
Nucellus - 2n
in the ovary. (½ M)
Synergids - n
F: Embryo sac - The female gametophyte that contains the Antipodal cells - n (1/2 × 6 = 3 M)
egg cell and is the site of fertilisation. (½ M) (b) When the pollen grain is mature it contains two cells,
(b) The nucellus, labelled as B plays a critical role in the vegetative cell and generative cell. The vegetative cell
the development of the megaspore within the ovule is bigger and has abundant food reserves. If it is destroyed,
as it provides nutritional support and protection to the formation of pollen tube and transport of male gametes
developing megaspore. It is the tissue in the ovule where towards embryo sac will be disrupted. (1 M)
megaspore mother cell (MMC) undergoes meiosis to (c) Approximately 60% of angiosperms release 2-celled
yield four megaspores, essential for female gametophyte pollen grains. In the remaining species, the generative cell
formation. (2 M) undergoes mitosis, resulting in the 3-celled stage before
4. (a) Describe the process of megasporogenesis till the pollen grains are shed, reflecting the diverse reproductive
development of embryo sac, in an angiosperm. strategies for successful pollination. (1 M)
(b) Draw the diagram of pollen grain showing the
2- celled stage.  (Un, Cr)
Ans. (a) Megasporogenesis in an angiosperm involves the Key Takeaways
following steps: From this answer, students learn the ploidy levels of
z An ovule differentiates a single megaspore mother cell various reproductive structures in angiosperms.
(MMC) in the micropylar region of the nucellus. (½ M)
z MMC undergoes meiotic division, resulting in four
megaspores. (½ M) Hints & Explanations
z Of the four megaspores produced, one is functional
Multiple Choice Questions
while the other three degenerate. (½ M)
1. (b) Pollen grains maintain viability for months in some
z The functional megaspore develops into the female
members of the Leguminosae, Rosaceae and Solanaceae
gametophyte or embryo sac through monosporic families.
development. (½ M)
2. (b) A-Antipodals; B-Egg apparatus
z Mitotic divisions of the functional megaspore’s nucleus
form a 2-nucleate, then a 4-nucleate, and eventually 3. (d) From the outermost to the innermost, the layers of a
an 8-nucleate embryo sac without forming cell walls microsporangium are as follows: epidermis, endothecium,
middle layers and tapetum.
immediately (free-nuclear divisions). (½ M)

35 Sexual Reproduction in Flowering Plants P

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 4. (a) Pollen mother cells, also known as microspore mother a single ovary. In contrast, Michelia (Q) features an
cells, are diploid cells derived from the sporogenous tissue apocarpous gynoecium, with separate, unfused carpels.
(A) in the anther. They undergo meiotic divisions to form four
11. (c) Autogamy (self-pollination) and geitonogamy
haploid microspores, which then develop into pollen grains.
(pollination within the same plant) do not promote genetic
5. (b) Starting from the innermost part of the ovule, the correct variation, whereas xenogamy (cross-pollination) does.
sequence is the egg, which is located inside the embryo
sac. The embryo sac is surrounded by the nucellus, and the Assertion and Reason
outermost layers are the integuments.
1. (c) Pollen-pistil compatibility chemicals play a role in the
germination of the pollen tube, but they do not dissolve the
sporopollenin in the pollen exine. Instead, the pollen tube
Key Takeaways growth is facilitated by the compatible interaction between
the pollen and the pistil.
After reading the answer, students learnt the precise
anatomical structure of an ovule, starting from the 2. (c) Microsporangia are indeed situated at the corners of the
innermost to outermost part. anther in flowering plants. The length of the stamen, which
holds the anther, is primarily determined by genetic factors
6. (c) The tapetal cells become multinucleated due to the and not by the number of microsporangia within the anther.
failure of cytokinesis, which is the division of the cell’s 3. (b) Microsporangium generally is surrounded by four wall
cytoplasm. layers; the innermost layer is tapetum. It is single layered
7. (a) Chasmogamous flowers, with open blossoms, undergo and it provides nourishment to the developing pollen grains.
either self-pollination (autogamy) or cross-pollination It is typically a multinucleated or polypoid.
(geitonogamy or xenogamy). Cleistogamous flowers 4. (b) The chalaza is located opposite the micropyle in an
remain closed, promoting self-pollination due to proximity ovule. It is a distinct region at the basal end of the ovule.
of anthers and stigma. Integuments encircle the nucellus except at the tip where a
small opening is present called micropyle.
5. (d) The ploidy of the megaspore mother cell (MMC) is
Mistakes 101: What not to do! diploid (2n), and it undergoes meiotic divisions to produce
A common mistake is misinterpreting the terms 4 haploid megaspores.
“cleistogamous” and “chasmogamous.” Students often
6. (a) Pollen grains are effectively preserved as fossils
confuse their breeding behaviors, leading to errors in
due to their exine, the outer wall, which consists of
selecting the correct option that describes their pollination
sporopollenin—a highly durable organic substance. This
characteristics accurately.
material resists degradation from high temperatures, strong
8. (c) The diagram illustrates megasporogenesis in acids, and alkalis, and no known enzyme can break it down.
angiosperms. Starting with a megaspore mother cell (MMC) 7. (c) Maize has unisexual flowers on the same plant
which undergoes meiosis I and meiosis II, producing a (monoecious), which prevents self-pollination i.e.,
megaspore dyad and then a tetrad. One megaspore develops autogamy but not geitonogamy.
into a functional megaspore through mitotic divisions, it 8. (a) Introduction of mutations in the gene of tapetum cells
develops into a female gametophyte or embryo sac. can affect pollen viability since the tapetum layer is crucial
for providing nutrients necessary for pollen development,
directly impacting pollen grain health and functionality.
Mistakes 101: What not to do!
9. (d) Corn cobs do not have multiple ovules in each ovary;
Students might confuse the process of meiosis with
they typically have a single ovule per ovary.
mitosis, leading to incorrect identification of the divisions.
10. (a) Amorphophallus flowers and insects engage in a
9. (a) The generative cell in the pollen grain is haploid, mutually beneficial relationship. The flower provides a safe
meaning it has a single set of chromosomes (n). This cell place for insects to lay their eggs, while the insects assist
divides mitotically to form two haploid male gametes. in pollination by transferring pollen between flowers. This
10. (a) Papaver (P) exhibits a multicarpellary, syncarpous arrangement ensures the survival and reproduction of both
pistil, where multiple carpels are fused together to form species.

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2 DOUBLE FERTILISATION &
POST-FERTILISATION: STRUCTURES AND EVENTS

NCERT Definitions (Commonly asked in 1 mark)

U Double fertilisation: A unique process in angiosperms where two fertilisation events syngamy and triple fusion occur
simultaneously.
U Primary Endosperm Nucleus (PEN): A triploid nucleus resulting from the fusion of a male gamete with two polar nuclei(triple
fusion), initiating endosperm formation.
U Primary Endosperm Cell (PEC): The central cell after triple fusion becomes the primary endosperm cell.
U Endosperm: A tissue that provides nutritional support to the developing embryo, developing from the primary endosperm cell
after triple fusion.
U Free-nuclear endosperm: An initial stage of endosperm development where the Coconut water
primary endosperm nucleus divides without forming cell walls, resulting in many (free nuclear
free nuclei. endosperm)
U Proembryo: The earliest stage of embryo development.
U Epicotyl: The portion of embryonal axis above the level of cotyledons,which
terminates with the plumule or stem tip. White
Kernel
U Hypocotyl: The cylindrical part below the level of cotyledons that terminates at its (cellular
lower end in the radicle or root tip. endosperm)
U Coleorhiza: An undifferentiated sheath covering the root tip in monocot embryos.
U Coleoptile: A hollow foliar structure covering the shoot apex and a few leaf primordia in monocot embryos.
U Perisperm: Remnants of nucellus present in some seeds like black pepper and beet.
U Seed dormancy: A state of reduced metabolic activity allowing seeds to survive unfavorable conditions until germination.

Important Facts

A small pore in the seed coat facilitates entry of oxygen and water into the seed during germination.
01 ~ Micropyle

Most zygotes divide only after a certain amount of endosperm is formed so that developing embryo gets
02 nourishment from the nutritive endosperm.

Hybrid seeds have to be produced every year as the genes expressing the hybrid characters in the plants
03 of the progeny will segregate and do not maintain hybrid characters.

The oldest viable seed, excavated from the Arctic Tundra, germinated and flowered after an estimated
04 record of 10,000 years of dormancy. ~ Lupinus arcticus

A recent record of 2000 years old viable seed discovered during the archeological excavation at King
05 Herod’s palace near the Dead Sea. ~ Phoenix dactylifera

06 Zygote give rise to the proembryo and subsequently to globular, heart-shaped and mature embryo.

37 Sexual Reproduction in Flowering Plants P

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 Classification

Types of Fruits

Aspects True fruits False fruits Parthenocarpic fruits

Definition Develop solely from the ovary Develop from parts of the flower •  Develop without fertilisation and
other than the ovary, often involving are typically seedless
the thalamus •  Parthenocarpy can be induced by
growth hormones

Examples Mango, watermelon etc. Apple, Strawberry, Cashew Banana

Types of seeds
(Based on the presence or absence of endosperm at the time of seed maturity)

Aspect Non/Ex-albuminous Seeds Albuminous Seeds

Definition Seeds without residual endosperm, as it is consumed Seeds retain a part of the endosperm, as it is not
during development. completely used up.

Examples Pea, Groundnut Wheat, Maize, Barley, Castor

Difference Between

Endosperm vs. Embryo

Endosperm Embryo

Forms before the embryo and provides nutrition to the developing Develops after the endosperm formation has begun.
embryo.

Usually triploid, formed from the fusion of a male gamete with Diploid, formed from the fusion of a male gamete with an egg
two polar nuclei. cell.

May or may not persist in the mature seed. Always remains as a crucial part of the seed.

Generally a nutritive tissue. Give rise to various parts of plant tissues.

Syngamy vs. Triple fusion

Syngamy Triple fusion

Fusion of a male gamete (sperm) with a female gamete (egg) Fusion of a male gamete with the two polar nuclei in the central
cell of the embryo sac

Results in the formation of a diploid zygote Results in the formation of a triploid primary endosperm nucleus
(PEN)

Zygote develops into the embryo Primary endosperm cell develops into the endosperm

Takes place in the egg apparatus of the embryo sac Takes place in the central cell of the embryo sac

A universal process in sexual reproduction across various Unique to angiosperms (flowering plants) and part of the double
organisms fertilisation process

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 Integument vs. Testa*
Integument Testa
Covers the nucellus of the ovule. Outer covering of seed
Thin and one or two layered Thick and one layered
Contains living cells Contains dead cells
Sclereids are absent Cells are rich in sclereids
It is a prefertilised structure It is post fertilised structure

Monocot embryo vs. Dicot embryo

Monocot embryo Dicot embryo

Have one cotyledon(scutellum in grasses) Have two cotyledons

Consists of a plumule and radicle which are often covered by Have a more distinct embryonal axis with a clear demarcation
protective sheaths called the coleoptile and coleorhiza respectively. of the plumule (stem tip) above the cotyledons and the radicle
(root tip) below.
Plumule appears lateral due to excessive growth of the single Plumule is terminal and lies in between the two elongated
cotyledon cotyledons

Apomixis vs. Polyembryony

Apomixis Polyembryony
• Formation of seeds without fertilisation Formation of seed occurs, more than one embryo in each seed.
•  
It is a type of asexual reproduction that mimics sexual
reproduction

Example: Some species of Asteraceae and grasses Example: Citrus and Mango

Real Life Application Based Questions

1. Have you ever noticed new varieties of vegetables and fruits in the market? What are the economic benefits and challenges
that farmers face when using hybrid seeds to produce these varieties?
Ans. Hybrid seeds enhance crop yields and quality but require annual repurchase due to trait segregation in subsequent generations,
increasing costs for farmers. Research in apomixis could allow reuse of seeds without losing hybrid vigor, reducing expenses.
2. Do you know what the juicy and tasty part of the mango that you enjoy in summer is called? Is the mango a monocot or a
dicot? Is the mango considered a true fruit or a false fruit?
Ans. The juicy and tasty part of the mango is called the mesocarp. Mango is a dicot. It is considered a true fruit because it develops
from the ovary of a flower.
3. Have you enjoyed the refreshing taste of coconut water and the delicious white part? Can you identify which parts of the
coconut these correspond to in terms of the endosperm?
Ans. Yes, coconut water and the white part are both enjoyable and nutritious. In a coconut, the water represents the free-nuclear endosperm,
while the white edible part is the cellular endosperm, each offering hydration and nutrients.
4. When you look at a sliced kiwi fruit, you can see it’s filled with many small seeds. What does this tell you about the number
of ovules in a kiwi fruit’s ovary?
Ans. The number of seeds in a kiwi fruit directly corresponds to the number of ovules present in its ovary. Each seed develops from a
single fertilised ovule. Therefore, the numerous small seeds you see in a sliced kiwi fruit indicate that its ovary contained a large
number of ovules.

* Concepts beyond the board exam syllabus, offering deeper insight and critical thinking on NCERT topics.

39 Sexual Reproduction in Flowering Plants P

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Myth Buster

U Myth: Sexual reproduction is the only way plants can produce seed.
Fact: Apomixis (asexual reproduction) allows plants to produce seeds without fertilisation.
U Myth: All parts of a flower contribute to fruit formation.
Fact: In most flowering plants, only the ovary develops into the fruit. However, in some cases (e.g., apples and strawberries),
other parts of the flower such as the thalamus may contribute. Such fruits are known as false fruits.
U Myth: Fruits are always the results of fertilisation.
Facts: Some fruits, such as bananas, develop without fertilisation, a phenomenon known as parthenocarpy. These fruits are
seedless, as they bypass the usual fertilisation process that leads to seed formation.
U Myth: Seeds produced through apomixis are not true seeds.
Fact: Apomixis leads to the production of seeds without fertilisation, but these are still true seeds genetically identical to the
parent plant. This method can be highly beneficial for maintaining desirable traits in plant breeding.
U Myth: The process of double fertilisation is universal in all plants.
Fact: Double fertilisation is unique to angiosperms (flowering plants) and does not occur in gymnosperms or non-flowering
plants.
U Myth: Seeds immediately germinate after they are shed from the parent plant.
Fact: Many seeds enter a period of dormancy where their metabolic activity slows, allowing them to withstand periods that are
not conducive to germination. This dormancy can vary greatly in duration.

Mnemonics

U Example of Albuminous seeds: “Mc Cheese n Wheat Burger”

Mc Cheese n Wheat Burger

Maize Castor Wheat Barley

U Example of Polyembryony: P.C.M.

P Polyembryony

C Citrus

M Mango

U The sequence of parts in a dicot (dicotyledonous) embryo from anterior to posterior:

“Penguins Explore cold, Harsh Regions Regularly”
Penguins Explore Cold Harsh Regions Regularly

Plumule Epicotyl Cotyledons Hypocotyl Radicle Root cap

U The sequence of parts in a monocot (monocotyledonous) embryo from anterior to posterior:

“Students Can Study Extra Resources, Recalling Chapters Rapidly”
Students Can Study Extra Resources, Recalling Chapters Rapidly

Scutellum Coleoptile Shoot apex Epiblast Radicle Root cap Coleorhiza

P
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 competency BASED SOlved Examples
A B C D
Multiple Choice Questions (1 M) (a) Seed coat Scutellum Epicotyl Hypocotyl
1. Non-albuminous seeds are present in (Re) (b) Seed coat Scutellum Hypocotyl Epicotyl
(a) Maize (b) Wheat (c) Seed coat Cotyledon Endosperm Hypocotyl
(c) Rice (d) Groundnut (d) Seed coat Endosperm Cotyledon Hypocotyl
2. Remnants of nucellus are persistent during seed 9. “X” is an example of a very old viable seed that was
development in: (Re) (CBSE APQ, 2023) excavated from the Arctic Tundra. The seed of “X” is
germinated and flowered after 10,000 years of dormancy.
(a) pea (b) groundnut
Identify “X”. (Re)
(c) wheat (d) black pepper
(a) Nymphaea (b) Lupinus arcticus
3. In an embryo sac, the cells that degenerate after
fertilisation are: (Un) (NCERT Exemplar) (c) Phoenix dactylifera (d) Strobilanthes kunthiana
(a) Synergids and primary endosperm cell 10. Choose the correct labellings for the parts X, Y and Z in
the given figure of the stages in embryo development in
(b) Synergids and antipodals
a dicot: (Re) (CBSE, 2022 Term I)
(c) Antipodals and primary endosperm cell
(d) Egg and antipodals.
4. In a typical dicotyledonous embryo, the portion of Y
embryonal axis above the level of cotyledons is:
Z
(Re) (CBSE, 2022 Term-I)
(a) Plumule (b) Coleoptile
(c) Epicotyl (d) Hypocotyl X
5. Which of the following are haploid, diploid and triploid
structures in a fertilised embryo sac? (Un)
(a) X is suspensor, Y is radicle and Z is cotyledon
(a) Synergid, zygote and primary endosperm nucleus
(b) X is radicle, Y is cotyledon and Z is suspensor
(b) Synergid, polar nuclei and zygote
(c) X is cotyledon, Y is suspensor and Z is radicle
(c) Antipodal, synergid and primary endosperm nucleus
(d) X is zygote, Y is radicle and Z is cotyledon
(d) Synergid, polar nuclei and zygote
6. During apomictic seed formation, there is no reduction
division and the gametes (both egg cell and the pollen/ Assertion and Reason  (1 M)
sperm cells) are diploid.
Direction: The following questions consist of two statements
What is the ploidy of the endosperm formed through – Assertion (A) and Reason (R). Answer these questions by
apomixis? (An) (CBSE CFPQ, 2023) selecting the appropriate option given below:
(a) 2n (b) 3n (c) 4n (d) 6n (a) Both A and R are true, and R is the correct explanation
7. Which of the following is TRUE for a flower giving rise of A.
to a false fruit in apple? (Un) (CBSE CFPQ, 2023) (b) Both A and R are true, but R is not the correct explanation
(a) The ovary is infertile. of A.
(b) The ovary does not undergo fertilisation. (c) A is true, but R is false.
(c) The thalamus undergoes fertilisation. (d) A is false, but R is true.
(d) The thalamus forms a part of the fruit. 1. Assertion (A): Apomixis is a form of asexual reproduction
8. Identify the parts labelled as A, B, C and D in the given that mimics sexual reproduction.
figure and select the correct option. (Re) Reason (R): Apomixis involves the production of seeds
without the fusion of gametes. (Re)
A
2. Assertion (A): Primary endosperm nucleus is triploid in
B angiosperms.
Reason (R): It is the product of fusion of male gamete with
an egg cell. (Un)
C
3. Assertion (A): The free-nuclear endosperm stage involves
cell wall formation immediately after nuclear divisions.
D Reason (R): In free-nuclear endosperm, numerous nuclei
are present without any cellular boundaries initially. (Un)

41 Sexual Reproduction in Flowering Plants P

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 4. Assertion (A): Endosperm in a flowering plant is formed Ans. The labeled part ‘A’ in the given diagram represents plumule.
before the formation of the embryo. (1 M)
Reason (R): The endosperm provides food to the developing The structure in which it develops as the plant matures:
embryo. (Un) (CBSE CFPQ, 2023) Stem (1 M)
5. Assertion (A): Perisperm is the protective covering of seed 3. In a case of polyembryony, if an embryo develops from
which helps in its dispersal. the synergid and another from the nucellus which is
Reason (R): A ripened ovary wall forms a pericarp which haploid and which is diploid? (Ap)
functions as a fruit wall. (Un) (CBSE APQ, 2023) Ans. Polyembryony involves embryos arising from diploid
6. Assertion (A): Endosperm is completely consumed during cells like the nucellus or integuments, resulting in diploid
the development of embryo in ex-albuminous seeds. embryos. If synergids are involved in polyembryony, the
Reason (R): Castor, pea and beans are all examples of embryos would be haploid due to the haploid genetic
ex-albuminous seeds. (Re) (CBSE, 2023) makeup of synergids. (2 M)
7. Assertion (A): The pollen tube releases two male gametes
into the synergid.
Mistakes 101: What not to do!
Reason (R): The pollen tube needs to transfer its contents
to the ovule for fertilisation. (Un) Confusion between the haploid and diploid nature of the
synergid and nucellus embryos. Lack of understanding of
8. Assertion (A): The coleorhiza is a hollow foliar structure the concept of polyembryony in flowering plants.
in monocotyledons.
Reason (R): It encloses and protects the young shoot apex 4. Which is the triploid tissue in a fertilised ovule? How is the
and a few leaf primordia during the initial stages of growth. triploid condition achieved? (Un) (NCERT Exemplar)
(Re) Ans. Triploid tissue in fertilised ovules is endosperm. (½ M)
The triploid condition is achieved through triple fusion.
(½ M)
Subjective Questions
In triple fusion, a male gamete fuses with two polar nuclei
Very Short Answer Type Questions  to form the primary endosperm nucleus. (1 M)
(2 M) 5. (a) You are given castor and bean seeds. Which one of
1. Observe the following figure and answer the questions the two would you select to observe the endosperm?
that follow. (An) (b) The development of endosperm precedes that of
embryo in plants. Justify. (Ap, Un) (CBSE, 2019)
Ans. (a) Between castor and bean seeds, the castor seed should
be selected as the endosperm in these seeds persists in the
mature seeds and can be easily observed. (1 M)
(b) In plants, the endosperm development begins after
double fertilisation and before the embryo development. It
(a) What can you suggest about the location of the ovary provides the essential nutrition to the developing embryo
in the flower from which the above fruit is formed? ensuring its proper growth and development. (1 M)
(b) What do you call this fruit and why? 6. (a) From which end of the ovule, and how does the pollen
Ans. (a) By observing the figure, we can infer that the apple tube gain its entry into the embryo sac of a Hibiscus flower?
core (ovary) is completely enclosed by the thalamus i.e. the (b) State the fate of the male nuclei present in the pollen
location of the ovary is in the thalamus. (1 M) tube. (Ap) (CBSE, 2023)
(b) Apples are considered false fruits. In botanical terms Ans. (a) The pollen tube enters the embryo sac of a Hibiscus
a fruit is the mature ovary of a flowering plant, typically flower through the micropylar end, guided by the filiform
containing seeds. In fruits such as apples, the thalamus also apparatus of the synergid. (1 M)
contributes to fruit formation. (1 M) (b) One of the male nuclei fuses with the egg cell to form
2. The embryonal axis of a dicotyledonous plant is crucial the zygote, which develops into the embryo. The other male
for the correct development of the seedling. What does the nucleus fuses with the two polar nuclei to form the primary
part labeled ‘A’ develop into as the plant matures?  (An) endosperm nucleus in the central cell which then develops
into endosperm. (1 M)
A

Key Takeaways
After studying the solution, students should understand the
specific entry path of the pollen tube in the Hibiscus flower
and the fate of male nuclei in fertilisation, distinguishing
between the formation of zygote and endosperm.

P
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 Short Answer Type Questions (b) Non-albuminous seeds, such as peas and groundnuts,
(3 M) have no residual endosperm because it is completely
1. Sudha cracked open a coconut and found the following consumed during the development of the embryo. (1 M)
content as shown in the image below:
In contrast, albuminous seeds, like wheat, maize, castor and
(An)(CBSE CFPQ, 2023) barley, retain a part of the endosperm as it is not completely
used up during embryo development. (1 M)
Q 4. (a) What is perisperm? In which seeds is it found?
(b) How do false fruits differ from true fruits in terms
P of their development?
(c) What are parthenocarpic fruits, and how can they
be artificially induced? (Un)
(a) Identify the parts of the seed labelled P and Q. Ans. (a) The perisperm is not a universal feature in all seeds
(b) What is most likely to have happened to the coconut but is specifically the residual, persistent nucellus found in
water? certain seeds, such as black pepper and beet. (1 M)
(c) What is the ploidy of the coconut water that we drink
(b) False fruits are distinguished from true fruits based on
from the tender coconut? Justify.
their development; false fruits involve contributions from
Ans. (a) P: Endosperm (½ M) the thalamus also, as seen in apples, strawberries, and
Q: Embryo (½ M) cashews. True fruits, on the other hand, develop exclusively
(b) The coconut water would have been consumed by the from the ovary. (1 M)
developing embryo. (1 M) (c) Parthenocarpic fruits are seedless fruits that
(c) 3n (½ M) develop without the process of fertilisation. This unique
It is a free-nuclear endosperm. (½ M) development can be artificially induced through the
2. (a) Which part in the following figure represents application of growth hormones, resulting in fruits like
cotyledon. What is the other name of cotyledons in bananas that do not contain seeds. (1 M)
the grass family? 5. Differentiate between the following:(Un) (CBSE, 2023)
(b) It is situated to which side of the embryonal axis? (a) Perisperm and Pericarp
(c) Identify the part which has few leaf primordia (b) Syncarpous pistil and Apocarpous pistil
enclosed. (Un)
(c) Plumule and Radicle
A
Ans. (a)
B Perisperm Pericarp
C Remnants of nucellus The wall of the fruit formed
from the ovary wall after
D
fertilisation
Found in some seeds like Found in guava, orange,
E black pepper and beet mango, etc
F
G (½ × 2 = 1 M)
Ans. (a) Part labeled as ‘A’ in the following figure represents (b)
cotyledon.The other name of cotyledons in the grass family Syncarpous pistil Apocarpous pistil
is scutellum. (1 M)
Carpels are fused Carpels are free and not fused
(b) Scutellum is situated towards one side (lateral) of the
together
embryonal axis. (1 M)
(c) Coleoptile (B) (1 M) Example: Papaver Example: Michelia
3. (a) What is the significance of seeds becoming dry (½ × 2 = 1 M)
(10-15% moisture by mass) upon maturation? (c)
(b) How do non-albuminous and albuminous seeds differ
Plumule Radicle
in terms of endosperm presence? (Un)
Ans. (a) Significance of seeds becoming dry upon maturation: The embryonic shoot The embryonic root below
above the cotyledons the cotyledons
z Reduces the general metabolic activity of the embryo.
(½ M) Develops into the shoot Develops into the root after
z Allowing it to survive for extended periods until after germination germination
favorable conditions for germination arise. (½ M) (½ × 2 = 1M)

43 Sexual Reproduction in Flowering Plants P

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 6. Banana and many citrus fruits are formed without 3. The embryo sac represents the female gametophyte in a
fertilisation. flowering plant. (Un) (CBSE CFPQ, 2023)
(a) Name the process of fruit/seed formation in both. (a) What are the constituents of the egg apparatus in the
(b) Mention ONE similarity and ONE difference between embryo sac?
the two processes. (Un) (CBSE CFPQ, 2024) (b) What is the ploidy of the cells of the egg apparatus?
Ans. (a) Banana: Parthenocarpy (c) The formation of the embryo sac involves mitotic
divisions that are “free nuclear” till the 8-celled
Citrus fruits: Apomixis (1 M)
stage. What does the term “free nuclear” mean?
(b) Similarity: Both are asexual modes of reproduction. (d) The filiform apparatus at the micropylar end forms
(1 M) an important part of the embryo sac. What is the
Difference: importance of the filiform apparatus?
Parthenocarpy: In bananas, it involves fruit development Ans. (a)
without seed formation. (½ M) z Two synergids (1 M)
Apomixis: In citrus, it involves seed formation without z One egg cell (1 M)
fertilization, producing seeds genetically identical to the (b) Haploid (1 M)
parent. (½ M)
(c) Free nuclear: Nuclear divisions that are not followed by
Long Answer Type Questions  cell wall formation/cytoplasmic division. (1 M)
(5 M)
(d) The filiform apparatus guides the pollen tube into the
1. Give a reason for each of the following: synergids. (1 M)
(a) The exine of pollen grains is very hard. 4. (a) Explain the different ways by which apomictic seeds
(b) The endosperm of flowering plants is triploid. can develop. Give an example of each.
(c) A pollen grain landing on a stigma does not ensure (b) Mention one advantage of apomictic seeds to farmers.
fertilisation. (c) Draw a labeled mature stage of a dicotyledonous
(d) Sexual reproduction brings in variation. embryo. (Un, Cr)
(e) Seeds of hybrid varieties need to be produced afresh Ans. (a) (i) Diploid egg cell is formed without reduction division
every year. (Un) (CBSE CFPQ, 2023) and develops into embryo without fertilisation, e.g.
Ans. (a) Exine, composed of sporopollenin, protects pollen Asteraceae / grasses. (½ + ½ M)
grains from harsh environmental conditions, preserving the (ii) In citrus / mango, some of the diploid nucellar cells
generative cell that gives rise to male gametes for fertilisation surrounding the embryo sac start dividing, protrude
during transit from the anther to the stigma. (1 M) into embryo sac & develop into a embryo
(b) One of the male gametes fuse with two polar nuclei (½ + ½ M)
forming (n + n + n) nucleus of the endosperm. (1 M)
(b) No segregation of character in hybrid seeds,
(c) There is a pollen-pistil compatibility factor that allows economically beneficial / desired varieties are cultivated.
fertilisation. (1 M) (1 M)
(d) Fusion of two gametes coming from two parents (c)
ensures mixing of characters. (1 M)
(e) The characters in the progeny separate out and do not
maintain hybrid characters. (1 M)
2. Starting with the zygote, draw the diagrams of the
different stages of embryo development in a dicot.
(Cr) (NCERT Exemplar)
Ans.
(2 M)

Hints & Explanations

Multiple Choice Questions
1. (d) A non-albuminous seed is a type of seed that does not
contain residual endosperm as it is completely consumed
during embryo development, e.g., pea, groundnut, etc.
2. (d) In some seeds like beet and black pepper, remnants of the
nucellus persist during seed development. This persistent
(5 M)
part of the nucellus is referred to as the perisperm.

P
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3. (b) The cells that degenerate after fertilisation in an embryo is the hypocotyl, which is the part of the stem of a seedling
sac are synergids and antipodals. that is below the cotyledons and directly above the root.
4. (c) In a typical dicotyledonous embryo, the part of the 9. (b) Lupinus arcticus is a very old viable seed that was
embryonal axis above the level of cotyledons is the epicotyl. excavated from the Arctic Tundra.
5. (a) The synergid is haploid (n), as it is part of the 10. (c) X is cotyledon, Y is suspensor and Z is radicle.
gametophyte structure.The zygote is diploid (2n), resulting
from the fusion of haploid male and female gametes.The Assertion and Reason
primary endosperm nucleus is triploid (3n), formed after
1. (a) Apomixis is the mode of production of seeds without
the fusion of two polar nuclei (each haploid) with one male
fertilisation in some families of grasses. Therefore, it is an
gamete(haploid).
asexual reproduction that mimics sexual reproduction.
6. (d) According to the question, since no reduction division
2. (c) The primary endosperm nucleus in angiosperms is
occurs during the formation of the gametes, the fusion
triploid, resulting from the fusion of a single male gamete
of the diploid male gamete from the pollen with the two
diploid polar nuclei in the embryo sac results in a hexaploid with two polar nuclei, not from fusion with an egg cell.
(6n) endosperm. 3. (d) The free-nuclear stage is characterized by numerous
free nuclei before cell wall formation occurs.
4. (a) Endosperm formation in flowering plants occurs
before the embryo forms because it provides the necessary
Nailing the Right Answer nutritional support for the developing embryo.
While answering, students should carefully consider the 5. (d) Perisperm is not the protective covering of seed nor
ploidy levels of all components involved in the formation does it help in dispersal; it is actually the remnant of the
of the endosperm. As it is a hypothetical question, do not nucellus that persists in the seed after fertilisation.
confuse the apomixis phenomenon occurring without
6. (c) Castor is an albuminous seed.
fertilisation.
7. (a) The primary function of the pollen tube is to deliver the
7. (d) In an apple, the thalamus, which is a part of the flower, male gametes to the ovule, where one gamete fuses with
enlarges and forms a significant portion of the fruit, known the egg cell to form the zygote (embryo), and the other
as a false fruit. fuses with the polar nuclei to form the primary endosperm
8. (d) A is the seed coat, which protects the seed. B is the cell(endosperm).
endosperm, which provides nutrition to the developing 8. (d) The coleorhiza is not a hollow foliar structure; it is a
embryo. C is the cotyledon, which stores food reserves.D sheath that encloses the radicle and root cap.

45 Sexual Reproduction in Flowering Plants P

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 miscellaneous Exercise
(c) Hybrid seeds need not be produced every year
Multiple Choice Questions (1 M) (d) Their germplasm is easily stored in labs for further
research
1. The wall layer of microsporangium which nourishes the 7. From among the situations given below, choose the one that
pollen grain is: (Re) (CBSE, 2023) prevents both autogamy and geitonogamy: (An)
(a) epidermis (b) endothecium (a) Monoecious plant bearing unisexual flowers
(c) middle layers (d) tapetum (b) Dioecious plant bearing only male or female flowers
2. Kiwi is a dioecious species. Which of the following methods (c) Monoecious plant with bisexual flowers
can be definitely RULED OUT as a possible mode of
(d) Dioecious plant with bisexual flowers
pollination in its case? (Un) (CBSE, APQ 2023)
8. In a flower, if the megaspore mother cell forms megaspores
P. cleistogamous autogamy without undergoing meiosis and if one of the megaspores
Q. chasmogamous autogamy develops into an embryo sac, its nuclei would be:
R. geitonogamy (Un) (NCERT Exemplar)
S. xenogamy
(a) Haploid
(a) only P and R (b) only P and Q
(b) Diploid
(c) only Q and S (d) only P, Q and R
(c) A few haploid and a few diploid
3. A dicotyledonous plant bears flowers but never produces
fruits and seeds. The most probable cause for the above (d) With varying ploidy.
situation is: (An) (NCERT Exemplar) 9. In the dioecious aquatic plant shown, identify the
characteristics of the male flowers that reach the female
(a) Plant is dioecious and bears only pistillate flowers
flowers for pollination. (Un)
(b) Plant is dioecious and bears both pistillate and staminate
Pistillate
flowers flowers
Water level
(c) Plant is monoecious
(d) Plant is dioecious and bears only staminate flowers
4. Which of the following outbreeding devices are used
by majority of flowering plants to prevent inbreeding
depression? (Un) (CBSE, 2022 Term-I)
(i)Pollen release and stigma receptivity are not synchronised.
(ii)Different positions of anther and stigma. Staminate
(iii)Production of different types of pollen grains. flowers
(iv) Formation of unisexual flowers along with bisexual
flowers.
Size of Colour of Characteristic feature of
(v) Preventing self-pollen from fertilising the ovules by
flower flower pollen grain
inhibiting pollen germination.
(a) (i), (ii) and (v) (b) (ii), (iii) and (v) (a) Small Brightly Lightweight and non-
coloured sticky
(c) (i), (iii) and (v) (d) (iii), (iv) and (v)
(b) Large Colourless Large and sticky
5. Arrange the following diagrams showing microspore
development according to developmental stages. (Un) (c) Small White Small, covered with
mucilage
(d) Large Colourless Non-sticky
10. Select the option that shows the correctly identified
‘ U’ , ‘X’ , ‘Y’ and ‘Z’ in a developing dicot embryo.
(Re) (CBSE, 2023)
A B C
(a) A → C → B (b) A → B → C X
(c) C → B → A (d) C → A → B
Z
6. Scientists are trying to transfer apomictic genes into hybrid
varieties of several of our food and vegetable crops so that: Y
(Un)
(a) They get resistance against common pathogens
(b) The yield of these plants can be increased many folds U
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 (a) X - Plumule (2n), Y - Suspensor (n), Z - Cotyledon (2n),
U - Radicle (2n). Subjective Questions
(b) X - Plumule (2n), Y - Suspensor (2n), Z - Radicle (2n),
U - Cotyledon (2n). Very Short Answer Type Questions  (2 M)
(c) X - Suspensor (2n), Y - Cotyledon (2n), Z - Radicle (2n), 1. Observe the image given below and answer the following
U - Plumule (2n). questions. (An)
(d) X-Cotyledon (2n), Y-Radicle (n), Z-Plumule (n),
U-Suspensor (n).
11. Choose the option that gives the correct number of pollen
grains that will be formed after 325 microspore mother
cells undergo microsporogenesis. (Ev) (CBSE, 2023)
(a) 325 (b) 650
(c) 1300 (d) 975 A

Assertion and Reason (1 M)

Direction: The following questions consist of two statements B
– Assertion (A) and Reason (R). Answer these questions by
selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation
of A.
(a) State the function of B.
(b) Both A and R are true, but R is not the correct explanation
(b) Based on the image, identify ‘X’ from the given flow
of A.
chart and name the tissue they are derived from.
(c) A is true, but R is false.
Vegetative cell
(d) A is false, but R is true. Pollen Pollen
X tetrad grain
1. Assertion (A): The embryonal axis is part of the embryo
Generative cell
that develops into the mature plant.
2. Pollen grains in wheat are shed at 3-celled stage while in
Reason (R): The primary function of the embryonal axis is
peas they are shed at 2-celled stage.” Explain. Where are
to store nutrients during seed development. (Un)
germ pores present in a pollen grain? (Un)
2. Assertion (A): The presence of a scutellum can be seen in
3. Why are pollen grains produced in enormous quantities in
monocotyledonous seeds.
the given plant? (An)
Reason (R): The portion of the embryonal axis above the
level of attachment of scutellum is the epicotyl. (Re)
3. Assertion (A): The early stages of embryo development
process is identical in both monocotyledons and dicotyledons.
Reason (R): The structure of the mature embryo differs
significantly between monocots and dicots. (Re)
4. Assertion (A): All angiosperm seeds contain a persistent
endosperm.
Reason (R): Some angiosperm seeds, such as those of peas
and beans, consume the endosperm before maturation.(Re)
5. Assertion (A): All hormonal changes in plants lead directly
to flower formation. 4. State two advantages of an apomictic seed to a farmer.
(Un) (CBSE, 2020)
Reason (R): Hormonal changes along with structural
changes in plants can lead to the differentiation and further 5. (a) Name the structure in which the following is transformed
development of the floral primordium. (Un) after fertilisation
6. Assertion (A): Pollen grains of some species cause severe (i) Ovary
allergies and respiratory disorders in humans. (ii) Ovary wall
Reason (R): Pollen grains of Parthenium can be helpful in (b) Give one example of each: fleshy and one dry fruit.
treating allergies and respiratory disorders. (Re) (Re)

47 Sexual Reproduction in Flowering Plants P

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 6. What is micropyle and what is its significance in seeds? Long Answer Type Questions 
(Un)
(5 M)
1. (a) As a senior biology student you have been asked
7. Discuss the advantage of seeds being relatively dry and
to demonstrate to the students of secondary level
entering a state of dormancy. (Un)
in your school, the procedure(s) that shall ensure
8. During germination of a monocot seed, such as wheat, which cross-pollination in a hermaphrodite flower. List the
part in the given figure grows into the primary root, and what different steps that you would suggest and provide
encloses it? Also identify the part labeled as ‘F’. (An) reasons for each one of them.
A
(b) Draw a diagram of a section of a megasporangium of
an angiosperm and label funicle, micropyle, embryo sac
B
and nucellus. (Un, Cr)
C
2. A flower of tomato plant following the process of sexual
D
reproduction produces 200 viable seeds. (Ev)
Answer the following questions giving reasons:
E
F (a) What would have been the minimum number of ovules
G present in a pre-pollinated pistil?
9. The longitudinal section of a maize grain is shown below. If a (b) How many microspore mother cells would minimally
mutation leads to the absence of ‘A’ , predict the implications be required to produce the requisite number of pollen
for seed viability. (An) grains?
(c) How many pollen grains must have minimally pollinated
A the carpel?
(d) How many male gametes would have used to produce
these 200 viable seeds?
(e) How many megaspore mother cells were required in
this process?
3. Differentiate between microsporogenesis and mega-
sporogenesis.  (Un)
4. (a) Explain any four devices that flowering plants have
Short Answer Type Questions developed to encourage cross-pollination.
(3 M)
(b) Why do plants discourage self-pollination? State any
1. Differentiate between wind pollinated and insect pollinated
one reason. (Un)
flowers. (Un) (CBSE, 2020)
5. (a) Explain the monosporic development of embryo sac in
2. Describe the sequence of steps involved in artificial the ovule of an angiosperm.
hybridisation to ensure the use of desired pollen grains for
pollination without contamination. (Un ) (b) Draw a diagram of the mature the embryo sac of an
angiospermic ovule and label any four parts in it.
3. Draw LS. of an embryo of grass and label its parts. (Un) (CBSE, 2023)
(Cr) (CBSE, 2019)
6. Study the diagram given below showing the modes of
4. Pollen banks are playing a very important role in promoting pollination. Answer the questions that follow.
plant breeding programmes the world over. How are pollens
preserved in the pollen banks? Explain. How are such banks 2 3
1
benefitting our farmer ? Write any two ways.
(Un) (CBSE, 2019)
5. Explain the process of pollination in Vallisneria. How is it
different in water-lily, which is also an aquatic plant?
(Un) (CBSE, 2017)
6. (a) Draw a L.S. of pistil showing pollen tube entering into (a) The given diagram shows three methods of pollen
the embryo sac. Label the following: transfer in plants. What are the technical terms used for
(Cr, Re) (CBSE, 2019) pollen transfer methods ‘1’ , ‘2’ and ‘3’ ?
(i) Nucellus (ii) Antipodals (b) How do the following plants achieve pollination
(iii) Synergids (iv) Micropyle successfully?
(b) Write the functions of the following: (i) Water lily
(i) Synergids (ii) Micropyle (ii) Vallisneria
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 (c) Flowering plants have developed many devices to avoid (i) Define the processes shown by the figure. (2)
inbreeding depression. Explain one hereditary and one
physiological device which helps plants to achieve this (ii) How many meiotic and mitotic divisions are required to
target. (An, Un) (CBSE, 2023) complete the above processes? (1)

(iii) How many male gametes are eventually produced by a single

Case Based Questions
meiosis? (1)
Case Based-I OR
In a research lab, scientists are analyzing the seed development
in Arabidopsis, a member of the Brassicaceae family. (iii) Why can C be well-preserved as a fossil? (1)
They came across the figure as given below. Answer the following
questions based on the given figure. (An) Case Based-III
A
Study the following figure and answer the following questions.
(Ap)
C

B Plasma
membrane
Egg
nucleus B
(i) Derive a simple formula to calculate the number of
chromosomes of B. (1)
A
(ii) Suppose the genotype of a female plant is bb and male is BB.
Both these plants are associated with the formation of seeds D
given in the figure. Based on this information, calculate the
genotype of the A. (1) E

(iii) If A is completely consumed by B, the resultant seed will be (i) 

The structure labeled ‘A’ assists in the guidance of which
called? Give two examples of such seeds. (2) structure? (1)
OR
(ii) 
If the structure labeled as ‘B’ is non-functional how might
(iii) Explain the functions of part A and B. (2)
this impact fertilisation? (1)
Case Based-II
The given figure shows microsporogenesis and microgameto- (iii) 
Explain how the structures labeled ‘C’ and ‘D’ are crucial
genesis. (Un) for the process of double fertilisation. (2)
A B C
OR

(iii) D
 escribe the function of the structure labeled ‘E’ during
PMC fertilization. (2)

Answer keys
Multiple Choice Questions
1. (d) 2. (d) 3. (d) 4. (a) 5. (d) 6. (c) 7. (b) 8. (b) 9. (c) 10. (c)
11. (c)

Assertion and Reason

1. (c) 2. (b) 3. (b) 4. (d) 5. (d) 6. (c)

49 Sexual Reproduction in Flowering Plants P

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 Hints & Explanations

mother cells undergo this process, the total number of

Multiple Choice Questions pollen grains formed will be 325 × 4 = 1300.

1. (d) Tapetum, the innermost wall layer of microsporangium

nourishes the developing pollen grains.
Assertion and Reason
2. (d) In dioecious species like kiwi, the male and female 1. (c) The nutrient storage is primarily the role of the
reproductive organs are in separate plants., so it can’t self- cotyledons and endosperm during seed development.
pollinate. This means cleistogamous (P) and chasmogamous
autogamy (Q), and geitonogamy (R) can’t happen because 2. (b) The assertion is correct as monocotyledonous seeds,
they all need both sexes on one plant. like maize, feature a scutellum, a specialized cotyledon that
absorbs nutrients. The reason is also correct, describing the
3. (d) The most likely reason a dicotyledonous plant bears
flowers but never produces fruit or seeds is that it is embryonal axis’s upper part above the scutellum attachment
dioecious and only bears staminate, or male flowers, and as the epicotyl, related to seedling development. However
lack pistillate or female flowers necessary for fertilisation. the reason does not explain the reason correctly.
Therefore, cannot produce fruits or seeds. 3. (b) The early embryo development stages are similar in
4. (a) The correct outbreeding devices that prevent inbreeding monocots and dicots, involving proembryo formation
depression by promoting cross-pollination in flowering and body plan establishment. However, mature monocot
plants are: embryos have one cotyledon, while dicots have two.
(i) Pollen release and stigma receptivity are not synchronised. 4. (d) Not all angiosperm seeds contain a persistent endosperm;
(ii) Different positions of anther and stigma. in many cases, such as peas and beans, the endosperm is
(iv) Preventing self-pollen from fertilising the ovules by consumed during embryo development.
inhibiting pollen germination. 5. (d) Not all hormonal changes in plants lead directly to
5. (d) C → A → B flower formation. Hormonal changes initiate the process,
The images depict microspore maturation into a pollen but structural changes and environmental factors also play
grain. Image C shows the microspore, Image A marks crucial roles in the development of floral primordia.
initial maturation with spindle formation and vacuole 6. (c) Pollen grains of certain species like Parthenium can
enlargement, and Image B represents a mature pollen grain cause allergies and respiratory disorders.
with vegetative and generative cell formation.
6. (c) Scientists are trying to transfer apomictic genes into Subjective Questions
hybrid varieties of several of our food and vegetable crops
so that hybrid seeds need not be produced every year.
Very Short Answer Type Questions
7. (b) Dioecious plants with only male or female flowers
cannot undergo autogamy because they lack both male 1. (a) B is the innermost wall layer of microsporangium
and female reproductive organs on the same flower, and known as the tapetum. It nourishes the developing pollen
geitonogamy is prevented as there are no neighbouring grains. (1 M)
flowers of the opposite sex on the same plant. (b) X-A (Microspore mother cells) are derived from
8. (b) If a plant’s megaspore doesn’t split its chromosomes sporogenous tissue in the anther. (1 M)
in half through meiosis and one megaspore grows into an 2. In flowering plants, pollen grains typically contain two
embryo sac, then all the nuclei inside that sac will have two cells: a vegetative cell and a generative cell.
sets of chromosomes, making them diploid.
Wheat: The pollen grains contain two male gametes and
9. (c) In Vallisneria, the size of the male flower is small, the one vegetative cell(3-celled stage) at the time of shedding.
color of the flower is white, the characteristic feature of (½ M)
pollen grain is small and covered with mucilage.
Pea: Pollen grains, at shedding, have one vegetative and
10. (c) X - Suspensor (2n), Y - Cotyledon (2n), Z - Radicle (2n), one generative (2-celled stage)cell that divides to form two
U - Plumule (2n). male gametes after the pollen grain has been deposited on
11. (c) In microsporogenesis, each microspore mother cell the compatible stigma of a flower. (½ M)
undergoes meiosis to produce four microspores, which Germ pores are the small opening in the exine(outer layer)
develop into pollen grains. Therefore, if 325 microspore of pollen grains. (1 M)
P
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 Flowers are not very Flowers are bright, colourful
colourful and do not and produce nectar.
Mistakes 101: What not to do! produce nectar.
Students may confuse the stages at which wheat and pea Stigma is large and feathery Stigma is sticky and precisely
pollen grains are shed, mixing up the 3-celled and 2-celled
to effectively trap airborne positioned to collect pollen
stages.
pollen grains. from visiting insects.
3. The given plant is of maize which is pollinated by wind.
(1 × 3 = 3M)
The pollen grains produced in enormous quantities in
wind-pollinated plants as Topper’s Explanation
z It is a wind pollinated plant, the pollen grains coming
in contact with the stigma is a chance factor. (1 M)
z It increases the chances that some pollen will land on a
receptive stigma, leading to fertilisation. (1 M)
4. Two advantages of apomictic seeds to a farmer are:
(a) They produce genetically identical offspring, ensuring
consistent traits. (1 M)
(b) Farmers can avoid the cost and effort of hybrid seed
production. (1 M)
5. (i) Ovaries develop into fruit (½ M)
Ovary wall develop into pericarp (½ M)
(ii) Example of fleshy fruit: Orange (½ M)
Example of dry fruit: Groundnut (½ M)
6. The micropyle is a small pore in the seed coat. (1 M)
The micropyle plays a critical role by allowing water and
oxygen to enter the seed, which is essential for activating
the embryo’s metabolic processes and initiating seed
germination. (1 M)
7. Dehydration and dormancy of mature seeds are important
factors for storing seeds,
z ensuring it can be used as food throughout the year
(1 M)
z preventing premature germination under unfavorable
conditions, ensuring that seeds can withstand extreme
environments and periods of scarcity. (1 M)
8. In the monocot embryo, the radicle (labeled as ‘E’) is the
part that develops into the primary root and it is initially
enclosed and protected by the coleorhiza (labeled as ‘G’). 2. Steps involved in artificial hybridisation, ensuring the
(1½ M) use of desired pollen grains for pollination without
‘F’ in the given figure is the root cap. (½ M) contamination:
9. Implications for seed viability: z Emasculation: Remove anthers from the female
The endosperm(A) is the primary source of nutrients parent’s bisexual flowers using forceps before anthers
for the developing embryo; without these nutrients, the open, to avoid self-pollination.
embryo would be unable to develop, and the seed would z Bagging: Cover the emasculated flowers with a
not germinate. (2 M) butter paper bag to protect against contamination by
Short Answer Type Questions unwanted pollen.
z Pollination: When the stigma becomes receptive,
1.
introduce pollen collected from the desired male parent
Wind pollinated Flowers Insect pollinated flowers onto the stigma.
Pollen grains are small, Pollen grains are larger, z Rebagging: After pollination, cover the flowers
light, and non-sticky to heavier, and sticky to adhere again with the bag to prevent any external pollen
facilitate transport by wind. to the bodies of insects. contamination. (3 M)

51 Sexual Reproduction in Flowering Plants P

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 3. 6. (a)

(½ × 4 = 2 M)
(3 M)
4. Pollen Preservation: Pollen banks preserve pollen by (b) (i) Have special cellular thickenings at the micropylar
freezing it in liquid nitrogen at –196°C. This method keeps tip called filiform apparatus which plays an important role in
the pollen viable for extended periods, facilitating its use in guiding the pollen tube into the synergid. (½ M)
cross-breeding and genetic diversity programs. (1 M)
(ii)
Benefits to Farmers:
z Facilitates entry of oxygen and water into the seed
z Enhanced crop varieties: Pollen banks enable the cross-
breeding of plants to develop crops with improved during germination. (½ M)
traits like disease resistance and higher yields. (1 M)
Long Answer Type Questions
z Biodiversity and adaptability: Access to diverse pollen
helps farmers cultivate plants better suited to changing 1. (a) The procedure to ensure cross-pollination in
environmental conditions, increasing agricultural
sustainability. (1 M) hermaphrodite flower
z Emasculation: If the female parent bears bisexual
5. Pollination in Vallisneria:
flower, removal of anther from the flower before the
z Female flowers reach the water surface through long anther dehisces can ensure cross-pollination. (1 M)
stalks. (½ M) z Bagging: Emasculated flowers have to be covered with
z Male flowers or pollen grains are released onto the a bag of suitable size generally made up of butter paper
surface of water. (½ M) to prevent contamination of its stigma with unwanted
z Pollen grains passively transported by water currents to pollen. (1 M)
female flowers. (½ M) z When stigma of bagged flower attains receptivity,
Difference with Water lily: mature pollen grains collected from anthers of male
z Water lily flowers emerge above the level of water. parent and dusted to stigma and flowers are rebagged
(½ M) and fruits are allowed to develop. (1 M)
z Pollinated by insects or wind, similar to terrestrial (b)
plants. (½ M)
zWater lily’s pollination strategy involves biotic
agents (insects/wind), not abiotic water currents like
Vallisneria. (½ M)
Topper’s Explanation

(½ × 4 = 2 M)
2. (a) The minimum number of ovules would also have been
200, as 200 viable seeds are formed. After fertilisation, the
ovule turns into seeds. (1 M)
(b) Each microspore mother cell gives rise to 4 microspores,
each of which develops into a pollen grain. Thus, to obtain
the requisite number of pollen grains the minimum number
of microspore mother cell required would be 200 × ¼ = 50.
(1 M)
(c) The minimum number of pollen grains that might be
involved in pollination of carpel are 200, as 200 viable
seeds are formed. This is because each pollen grain contains
P
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 2 male gametes, out of which one fuses with egg forming Production of unisexual flowers: Plants may produce either
zygote that gives rise to seeds. (1 M) male or female flowers, which encourage cross-pollination
(d) The number of male gametes involved would be 200. as self-fertilisation is not possible within a single flower.
Each male gamete fuses with one egg nuclei to form a (1 M)
zygote that gives rise to seed. (1 M)
(b) Self-pollination is discouraged because it can lead to
(e) 200 megaspore mother cells were involved. They inbreeding depression, where genetic diversity is reduced,
undergo meiotic division to form 4 haploid megaspores.
leading to offspring that may be less vigorous and less
Out of them, only 1 becomes a functional megaspore and
adaptable to environmental changes. (1 M)
the rest 3 degenerate. (1 M)

Key Takeaways Key Takeaways

After reading the answer, students should understand the From the answer, students should understand that plants
quantitative relationships in plant reproduction, including have evolved complex mechanisms to prevent self-
the ratios of microspore mother cells to pollen grains pollination, thereby increasing genetic diversity.
and the significance of male and female gametes in the
development of seeds. 5. (a) In a majority of flowering plants, one of the
megaspores is functional while the other three degenerate.
3.
Only the functional megaspore develops into the female
Megasporogenesis Microsporogenesis gametophyte (embryo sac). This method of embryo sac
formation from a single megaspore is termed monosporic
Process of formation of Process of formation of microspores
development. (1 M)
megaspores from the from the pollen mother cell (PMC)
(Meiosis)
megaspore mother cell Megaspore mother cell Megaspore tetrad
(MMC) (2n) (n)

Occurs in the ovule within Occurs in the anther of a flower Degeneration of 3 megaspores
3 mitosis without cell
the ovary of a flower (one functional megaspore left)
wall formation/
(n)
Results in formation of Results in formation of megaspores Free nuclear division
microspores that develop that develop into female
into male gametophytes gametophytes (embryo sac) (Cell wall formation)
(pollen grains) 8 nuclear stage Embryo sac
(7 – celled, 8 –
Four functional microspores One functional and three non- nucleate stage)
are produced functional megaspores are produced (2 M)

Pollen grain contains Embryo sac contain 3 antipodals, (b)

vegetative cell and secondary nucleus (containing
generative cell two polar nuclei), egg apparatus
(containing two synergids and one
egg cell)

(1 × 5 = 5 M)
4. (a) Pollen release and stigma receptivity: Pollen may be
released either before the stigma is receptive or after the
stigma has become receptive, preventing the pollen from
contacting the stigma of the same flower. (1 M)
Different positions of anther and stigma: The anther and
stigma are positioned in such a way that pollen cannot come (2 M)
into contact with the stigma of the same flower. (1 M)
6. (a) The technical terms used for pollen transfer methods
Self-incompatibility: This genetic mechanism prevents are:
pollen from the same flower or other flowers of the same
1. Autogamy. (½ M)
plant from fertilizing the ovules by inhibiting pollen
germination or the growth of the pollen tube in the pistil. 2. Geitonogamy. (½ M)
(1 M) 3. Xenogamy. (½ M)

53 Sexual Reproduction in Flowering Plants P

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 (b) Pollination mechanisms: Case Based-II
(i) 
Water lily: Achieves pollination through insects (i) Microsporogenesis refers to the process of microspore
(or wind), such as beetles, which are attracted to the formation within the microsporangium (or pollen sac) by
flowers that float above the water. (1 M) meiotic or reduction division.  (1 M)
(ii) Vallisneria: Utilizes water for pollination. Male flowers Microgametogenesis is the plant reproduction process
detach and float to the surface, where they come into where a pollen grain develops into a three-celled
contact with female flowers, ensuring successful microgametophyte stage. (1 M)
pollination. (1 M) (ii) The mature male gametophyte of angiosperms is formed by
(c) Hereditary device (½ M) one meiotic and two mitotic divisions. (1 M)
Self-incompatibility: Pollen from the same flower or (iii) A single meiosis in total produces 8 male gametes because
from different flowers of the same plant is recognized and one meiosis produces 4 pollen grains and each pollen grain
rejected, preventing self-fertilisation. (½ M) produces 2 male gametes. (1 M)
Physiological device OR
Pollen release and stigma receptivity: Pollen may be (iii) Pollen grains (C) are well-preserved as fossils because of
released either before the stigma is receptive or after the the presence of sporopollenin. (1 M)
stigma has become receptive, preventing the pollen from
contacting the stigma of the same flower. (½ M)
Mistakes 101: What not to do!
Case Based Questions One often mixes up the terms “microsporogenesis” and
“microgametogenesis” or uses them interchangeably.

Case Based-I Case Based-III

(i) Number of chromosomes of embryo (B): Total number of (i) The structure labeled ‘A’ (filiform apparatus) assists in the
chromosomes in male plant + Total number of chromosomes guidance of pollen tube (containing male gametes) into the
in female plant/2. (1 M) synergid. (1 M)
(ii) Endosperm (A) is usually a triploid tissue formed as a result (ii) If the synergid (labeled ‘B’) is non-functional, it could
of triple fusion involving two polar nuclei and a sperm cell. impair the guidance mechanism for male gametes,
As it is 3n in structure, the genotype could be bbB. (1 M) potentially disrupting or preventing successful fertilisation.
(1 M)
(iii) When the endosperm(A) is completely consumed by the
developing embryo(B), the seed is called non-endospermic (iii) In the double fertilisation process, one male gamete (labeled
or non- albuminous. Examples include Pea and groundnut, ‘D’) fuses with the egg nucleus, resulting in the formation
where nutrients are stored in the cotyledons. (2 M) of diploid zygote. The other male gamete fuses with the two
polar nuclei (labeled ‘C’), forming a triploid endosperm
OR
nucleus. Both these fertilisation events are essential for the
(iii) The embryo’s primary function is to develop into a mature successful reproduction in flowering plants. (2 M)
plant under favorable conditions. It contains the earliest OR
forms of the plant’s roots, leaves, and stem. (1 M)
(iii) The stucture labelled as ‘E’ is the pollen tube, which
Endosperm contains reserve food materials and are used for delivers the male gametes to the egg cells and central cell
the nutrition of the developing embryo. (1 M) for double fertilisation. (2 M)

P
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Article Writing
4
Preview
An article is a written work that provides a detailed analysis, perspective, or argument on
a particular topic or subject. It is meant for a wide audience and is typically published
in newspapers, magazines, journals, or online platforms. Articles aim to inform, entertain,
persuade, or raise awareness.

Steps to attempt Article Writing

 U
 nderstand the Format: Familiarize yourself with the typical structure of an article:
Title, Introduction, Main Body, and Conclusion. Ensure your article adheres to this
format.
 Engaging Title: Begin with a catchy and relevant title. It should give readers a hint
about the topic and make them want to read more.
 Introduction: Start with a strong introduction that provides a brief overview of what
the article is about. It should hook the reader’s attention and convey the essence of
the topic.
 Use Subheadings: In the main body, use subheadings to break down the content. This
makes the article easy to navigate and helps to organize your ideas.
 L
 ogical Flow: Ensure that the paragraphs in the main body have a logical flow. Each
paragraph should have a clear focus, and transitions between paragraphs should be
smooth.
 S
upport with Facts: While the article allows you to express personal opinions, it’s
crucial to back up your statements with relevant facts, statistics, or examples for
credibility.
 C
 oncise and Clear: Avoid verbosity. Be clear and concise in your expressions. Overly
complex sentences can confuse readers.
 U
 se of Language: Ensure your language is formal and free of slang. However, you can
use rhetorical questions, idiomatic expressions, or quotations to make your point more
effectively.
 Conclusion: Summarize the main points discussed in the article in your conclusion. You
can also end with a call to action or a thought-provoking statement that leaves an
impact on the reader.
 Proofread: Always set aside a few minutes at the end to proofread your article. Check
for spelling, grammar, punctuation errors, and ensure that the content flows smoothly.
 SOLVED EXAMPLE
1. You are Ananya Sharma of class XII-A. Write an article for your school bulletin on the significance of adopting ecofriendly
habits in daily life, emphasizing the urgent need to address environmental issues and the advantages it brings to our future.
Use the given cues along with your thoughts to pen this article.

Importance of green living

Benefits -
For self: health improvements, cost savings
For the planet: reduced pollution, conservation of resources
How can students contribute?

Ans. Going Green: A Choice We Must Make

By Ananya Sharma, XII-A
In today’s world, adopting eco-friendly habits isn’t just a choice but a necessity. With alarming rates of pollution and natural
resource depletion, green living stands as our beacon of hope. Embracing sustainable practices not only ensures a healthier
lifestyle but also leads to substantial cost savings. Simple habits like recycling, conserving energy, or using public transport can
make a world of difference.
For students, the possibilities are endless. From setting up recycling drives in schools to using e-books or reducing paper waste,
every little effort counts. We are the next leaders, policy-makers, and educators. Thus, by championing eco-friendly habits today,
we ensure a brighter and greener future for all.
Remember, every effort counts, and together, we can make a world of difference.

miscellaneous Exercise
Note: All details presented in the questions are imaginary and created for assessment purposes.

Practise and create the following invitations in about 120-150 3. You are Arjun Verma of class XII-C. Pen an article for your
words. (5 m) school newsletter discussing the advantages and relevance
1. You are Priya/Piyush. You find illiteracy as the biggest of digital learning in the current educational scenario. Use
impediment to the development of a nation. You strongly the prompts provided along with your individual insights to
believe that education can play a very important role in the frame this article.
progress of a country. Write an article in 120-150 words on
‘Education—The Biggest Tool of Progress’. Use the given Current global situation prompting digital education
cues along with your own ideas to compose this article. Benefits:
• Importance of education For students: flexibility, vast resources
• Role in development of a country or society For teachers: real-time feedback, varied tools
• How does it make the society progressive Challenges and their solutions
2. You are Sohail Hassan of class XII-B. Write an article for your
4. With the rise of social media platforms and the ease with
school magazine, sharing the importance of young adults, as
volunteers in one’s local community, the need to do so and the which information can be shared online, false information
benefits involved. Use the given cues along with your own can have m dire consequences. Write an article for the
ideas to compose this article. (CBSE SQP, 2023) local daily, in 120-150 words, highlighting the problem
of the growing menace of fake news and misinformation,
Importance for personal growth + community development analysing the issue and provide suggestions to tackle it. You
Benefits - For self: new skills, experience, a sense of purpose are Kavita Bannerji, from Siliguri, West Bengal. Use the
For community: positive impact given cues, along with your own ideas to create the article.
What are the ways one could get involved? (CBSE APQ, 2023)

P
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 1. What is fake news, and why has it become a significant
8. Draft an article in 120-150 words to be published in a
concern in the age of social media? prominent journal on the problem of stray dogs in your
2. Analysing the Issue – colony. Give suggestions to overcome this menace. You are
• How
 does misinformation affect public opinion and Prerna/Prashant. Use the given cues along with your ideas
behaviour? to compose the article: (CBSE, 2023)
• What
 challenges do individuals and society face in
combatting the spread of false information? agressive, uncontrollable *ADOPT
3. Suggestions to Tackle Fake News - behaviour *BEFRIEND
• What
 measures can social media platforms take to Litter streets *VACCINATE
address the issue of fake news? by overturning
• How
 can individuals critically evaluate the information dustbins
Problem of
they encounter online?
stray Dogs
4. Local Perspective -
• How
 does the issue of fake news affect the local risk of diseases
community?
• What
 local resources or organisations can help address barking and howling
this problem? nuisance to the community

5. Hard work and punctuality are essential for a happy and 9. You are Priya Sharma of class XII-A. Write an article for
successful life. They help in meeting the desired targets your school magazine, emphasizing the significance of
of our life. You are Naira/Rishi. Write an article in environmental conservation, particularly by young adults,
120-150 words highlighting the importance of hard work the urgency of the matter, and the advantages it offers. Use
and punctuality in a student’s life.
the provided cues along with your personal insights to draft
6. You are Sneha Hassan of Class XII. Write an article this article.
for your school magazine on the Importance of Active
Listening. Expand on how good listening skills help one to Importance of environmental conservation for a sustainable
gain multiple perspectives, be an effective team player, and future
contribute to an overall atmosphere of empathy and care. Benefits:
Use the given cues along with your own ideas to compose
For self: Gaining knowledge about nature, physical activity,
this article. (CBSE APQ, 2023) mental well-being
Listening skills: listening and comprehending, paraphrasing, For community: Cleaner surroundings, reduced carbon
active body language like nodding, changing facial expressions, footprint, fostering a culture of sustainability
and sitting straight. How can young adults play a role?
Effect on others: we feel heard and acknowledged, builds
trust, others more likely to hear us, leads to more ideas being 10. Owning a car has become a status symbol these days.
shared. However, an increase in the number of cars has added to
7. To have a fair complexion is an obsession in our society. various types of pollution and other problems. Write an
Demand for fair brides in matrimonial columns and sale article in not more than 120-150 words highlighting the
of fairness creams are evidence enough. Write an article urgent need for reducing these man-made problems, giving
in 120-150 words giving your views on ‘Beautiful mind, suitable suggestions. You are Vijay/Vidhi.
better than a fair complexion.’ You are Reshma/Ritesh.
•  millions of vehicles
Use the following clues:
•  traffic congestion
• In the West, people like to be tanned
•  respiratory diseases
• fair complexion, only skin deep
•  man-made traffic hazards
• beautiful mind
• attitude to life
• behaviour in society, etc.

129 Article Writing P

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 Hints & Explanations
1. Education - The Biggest Tool of Progress home, ranging from video lectures to interactive quizzes,
By Piyush tailored for varied learning styles. For educators, digital
tools offer real-time feedback and content customization
Education is not just learning from books; it is the
options.
cornerstone of a nation's progress and development. It
equips individuals with the necessary skills and knowledge However, digital education isn’t without hurdles. Issues like
to face real-world challenges. A well-educated society is unequal access and internet affordability pose challenges.
a powerhouse of innovation and creativity, which leads Nevertheless, collaborative efforts between institutions and
to the development of new technologies and solutions for governments are ensuring quality education and structured
online curricula. With the world moving towards universal
pressing issues.
internet access and improved digital literacy, the potential
Furthermore, education fosters an understanding of rights of online learning is just beginning to be realized.
and responsibilities, encouraging individuals to participate
In essence, digital learning isn’t merely about technology;
actively in democratic processes and civic duties. It it’s about making quality education accessible to all,
reduces inequalities by offering equal opportunities to all, signifying a brighter educational future.
regardless of their background, promoting social harmony
4. The Threat of Fake News in Social Media
and inclusivity.
By Kavita Bannerji, Siliguri
Moreover, education contributes to economic growth
Fake news—false or misleading information passed off as
by enhancing productivity and efficiency. An educated
real news—has become a major concern in the age of social
workforce attracts investments and industries, which boosts
media. The rapid spread of misinformation can impact
a country’s economy. Thus, education truly acts as the
public opinion and influence behaviour, causing confusion,
biggest tool of progress, transforming not just individuals fear, and even harm.
but entire societies, paving the way for a brighter and more
Fake news can influence elections, spread false health
prosperous future.
information, and encourage violence. The difficulty is that
2. The Power of Youth in Community Volunteering it spreads rapidly and is difficult to manage. To combat this,
By Sohail Hassan, XII-B social media platforms should improve their fact-checking
Young adults have the power to drive change by volunteering tools and content moderation.
locally, which significantly aids personal growth and Individuals can assist by verifying facts before sharing
community enhancement. When we volunteer, we don't just and relying on reputable sources. Fake news in Siliguri
lend a hand; we develop critical skills and gain experiences has the potential to cause chaos and disrupt the harmony
that shape our future. It brings a sense of achievement and in our society. Local organisations and schools may assist
purpose to our lives. promote awareness and identify fake news in order to keep
our community informed and secure.
For our community, the effects are just as beneficial. We
5. Hard Work and Punctuality in a Student’s Life
spark positive changes that can grow over time, creating
stronger, more united neighbourhoods. Volunteering might By Naira
involve tutoring younger students, planting trees, or helping To achieve success as a student, hard work and punctuality
at community kitchens. are required. Students ought to develop this behaviour from
the very beginning of their lives. They can achieve success
The benefits are twofold: we grow as individuals, gaining
and happiness in life if they are punctual and hardworking.
new perspectives, while our actions contribute to the well-
As a result, punctuality and hard work are required in a
being of those around us. Let's embrace the opportunity to
student’s life in order to achieve his or her goals.
make a difference and discover our potential for leadership
Consider the following examples. When a student is late
and empathy. Together, our small efforts can lead to big
for class, he misses the lesson and annoys the teacher.
transformations.
Unpredictability is viewed as a sign of rudeness or an
3. Digital Learning: The Future of Education uncivilized temperament. Napoleon lost the battle of
By Arjun Verma, XII-C Waterloo because his marshal arrived too late. Nelson once
The landscape of education is undergoing a transformative stated that punctuality was the key to his success in life.
change, with digital learning emerging as a dominant force. As a result, diligence and punctuality drive a student to
This shift, driven by technological advancements and success. As a result, all students are expected to study hard
global challenges, has made learning flexible and diverse. and complete their given homework on time. Teachers will
Today’s students can access a plethora of resources from adore them, and they will achieve great success.

P
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6. Importance of Active Listening 9. Young Adults: The Torchbearers of Environmental
By Sneha Hassan, Class XII Conservation
Listening is an art that goes beyond just hearing words. By Priya Sharma, XII-A
It's about understanding, engaging, and responding with The environment is not just a backdrop to our lives; it is the
care. Active listening is not passive; it's an active process very essence that sustains us. In today’s age, with rising global
where you show you're engaged with nods and attentive
temperatures and depleting green cover, environmental
expressions. It means paraphrasing to show you've grasped
conservation is not just important – it’s imperative. For
the essence, maintaining eye contact, and sitting straight to
convey your involvement. young adults, taking part in conservation activities offers a
plethora of benefits. Personally, it provides an opportunity
Why is this important? Because when we listen actively, we
gain varied perspectives and become effective collaborators. to learn about our natural world, promotes physical health
It creates a space where everyone feels heard, valued, and through outdoor activities, and ensures mental well-
trusted. This mutual respect paves the way for a more being by connecting with nature. For the community, our
empathetic and caring environment. involvement ensures cleaner surroundings, reduces the
Good listeners are like the roots of a tree, keeping the collective carbon footprint, and most importantly, instills a
soil of our school community nourished and strong. Let's culture of sustainability. So, how can we, the young adults,
practise active listening; it's where understanding begins, become environmental champions? Simple. Participate in
and without it, communication just falls apart. tree plantation drives, advocate for sustainable practices
7. Beautiful Mind, Better Than a Fair Complexion at school and home, organize or join clean-up drives, and
By Reshma most importantly, educate ourselves and others about the
In contemporary society, a fair complexion is frequently importance of a balanced ecosystem. Let’s pledge to leave
associated with beauty, leading to a misguided sense of self- the world better than we found it.
worth. In contrast, in Western civilizations, people spend
hours sunbathing to obtain a tanned appearance. Isn’t it
fascinating how beauty standards differ between cultures? Nailing the Right Answer
However, the colour of our skin is only superficial. A • Structure
  your article effectively: Begin with a
beautiful mind that is rich in information, empathy, and captivating introduction that highlights the significance
a good attitude outperforms the assumed beauty of a fair
of environmental conservation and its relevance in
complexion. A person’s attitude toward life and behaviour
today’s world.
in society genuinely define them.
• Conclude
  with a pledge or call to action, encouraging
A fair skin may draw notice at first, but it is a beautiful
others to join the efforts in preserving the environment.
intellect that lasts. It’s past time to value the glow of
knowledge and kindness over skin fairness. True beauty,
after all, radiates from inside. 10. Cars: Convenience at a Cost
8. Panic Created by Stray Dogs By Vijay
By Prerna In today's world, owning a car is often seen as a symbol
Stray dogs have become frequent in most Indian Colonies, of success. However, the surge in the number of vehicles,
posing a serious threat to the people’s safety and health. reaching millions, is leading to severe environmental
Stray dogs have become a serious problem in our colony, and health issues. Our cities face relentless traffic
and their aggressive and uncontrollable behaviour has congestion that not only wastes time but also contributes
resulted in multiple incidents of attacks on people. The significantly to air pollution. This rise in pollutants in
dogs’ continuous barking and howling have also become the air is causing an alarming increase in respiratory
an annoyance to the neighbourhood. diseases among city dwellers, affecting the quality of life.
Furthermore, stray dogs litter the streets by topping trash The hazards created by such heavy traffic are not just
cans, causing unhygienic conditions that could contribute limited to health but extend to daily commuting challenges
to the spread of disease. The crisis requires immediate as well. Accidents and road blockages are common, causing
attention, and we can no longer afford to ignore it.
further chaos. We must address these man-made problems
One possible option is to launch a vaccination programme promptly. Encouraging public transport, promoting
to reduce illness transmission among dogs. We may also carpooling, and investing in eco-friendly vehicles are crucial
adopt these dogs and provide them with a safe and secure
steps we can take. Additionally, stricter regulations on
home. We may also try to befriend these dogs and train
them to be less violent. It is critical to recognise that the vehicle emissions can significantly reduce pollution levels.
problem of stray dogs can only be solved by collaborative As responsible citizens, it is our duty to adopt these
community initiatives. We must all work together to protect measures and contribute to making our cities cleaner and
our safety and the well-being of these animals. Let us take healthier places to live. Let’s drive change by changing how
the initiative to address the issue. we drive.

131 Article Writing P

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Report Writing
5
Preview
A report offers a concise summary of a past event or situation. Serving as a documented
record of notable occurrences from our daily lives, it delivers firsthand details about specific
incidents. While capturing the essence of the event, a report can incorporate the writer’s
thoughts, views, and interpretations. Essentially, it’s an organized, factual account based
on observation, research, or personal experience. Reports play a vital role in academic and
professional settings, allowing for both reflection on past events and a means of evaluating
one’s comprehension and analytical skills.

Steps to attempt Report Writing

 U
nderstand the Format: Familiarize yourself with the standard format of report
writing. Typically, it includes a title, the name of the writer, date, and the content of
the report.
 C
 lear and Concise Title: The title should be to-the-point and give a clear idea about
the content of the report. It should be centered and written in bold.
 O
 bjective Tone: Always maintain a formal and neutral tone. Avoid using slang or overly
complex language. Stick to third-person narration.
 O
 rganized Structure: Begin with an introduction that outlines the purpose of the
report. Follow this with the main content, detailing the event or situation, and conclude
with a summary or recommendation.
 5
 W’s and 1 H: Ensure that you answer the “5 W’s and 1 H” - What, Where, When,
Why, Who, and How. These are crucial for a comprehensive report.
 B
 e factually correct: Always ensure that the data, statistics, or facts you include in
the report are accurate. Avoid assumptions or hearsay.
 P
 aragraphing: Organize your content into clear, well-structured paragraphs. Each
paragraph should focus on a specific point or detail.
 U
 se Passive Voice: Typically, reports are written in passive voice. For instance, “A
seminar was organized by the school” rather than “The school organized a seminar.”
 P
 roofread: After writing, take a few minutes to review your report. Check for any
grammatical errors, unclear sentences, or information that might be missing.
 P
 ractice Regularly: Before the exam, practice writing reports on various topics. This will
help you get comfortable with the format and improve your speed.
 SOLVED EXAMPLE
1. You happened to witness a road accident in which ten people were killed while visiting a temple. As a reporter of ‘The Times
of India’, write a report in about 120-150 words.
Ans. Ten People Killed in Road Accident
By TOI Reporter
Baroda, 31 October, 2023: Ten people from Rajasthan were killed and four others were injured after their vehicle had been
struck from behind by a loaded truck on the Vadodara-Ahmedabad Expressway around 2.30 p.m. on Saturday. The injured
were transferred to Ahmedabad Civil Hospital. According to eyewitnesses, a Force Traveler carrying 14 passengers attempted to
overtake a vehicle but applied brakes after noticing another vehicle approaching from the opposite way and came to a stop. The
cement truck smashed into the tempo traveller from behind, killing ten people on the spot, including two children aged seven and
eight years. Police had to bring a crane to remove the van and recover the bodies. A three-year old daughter somehow escaped
uninjured. A case has been filed against the truck driver, and cops are examining whether the driver was driving recklessly or
whether the tempo traveller pulled the brakes too soon, causing the accident.

miscellaneous Exercise
Note: All details presented in the questions in imaginary and created for assessment purpose.

Practise and write the following reports in about 120-150 3. Your school recently organised a workshop on Active
words. (5 m) Listening for students of Classes XI and XII on 21 January,
1. Vidya Bhawan Senior Secondary School, Noida, recently 2023. As Simran Kaur, a member of the organising
organised a 3-day environment awareness camp for the committee, draft a report covering the event for your
students to be prefects/monitors. Write a report about the
school’s newspaper. Support your ideas with the outline
same in 120-150 words for the school magazine. Support
your ideas with outline cues given below, to craft your cues given below to craft your report.
report. (CBSE APQ, 2023)

• About awareness camp • Purpose of the workshop

• Activities and posters • Who attended the workshop
•  How to encourage everyone to be part of environment camp • Activities that took place
• About importance of conservation of environment • Key messages or takeways from the workshop
• Post workshop resources and information given to attendees
2. The Local Sports Club, Krishnapuram-IV, initiated a
community sports outreach program on 5th February • What
  impact the workshop would have on the students and
their environment
2024 to engage young adults in various sports activities.
As Rahul D., the sports editor for the local community
4. The Eco Club of Harmony Society initiated a tree-plantation
magazine, write a report covering this initiative. Use the
cues provided below to frame your report. drive on 5th March, 2024 to combat the rising air pollution
in the locality. As Rajan Mathur, the secretary of the club,
•  Which sports outreach program? - Purpose of the initiative write a report on the event, weaving in the details provided
- its significance
in the cues.
•  Sports and activities introduced - Who participated?
•  Any notable personalities present? • Which tree-plantation drive? its objective and significance
• How were young adults motivated to be part of the program? • Involvement of residents-age groups, number of participants.
•  Any kits or information handed out to participants? •  Number of saplings planted-types of trees
•  Any details on future matches or tournaments? • Keynote speech by an environmentalist
•  What positive outcomes are anticipated in the community? • Future plans of the Eco Club

133 Report Writing P

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 5. Every Sunday morning, you have observed senior residents
Cues:
of the colony teaching children from the neighbouring
1. What is the initiative?
slums. You are impressed by this spirit of giving back to
2. What
 are the key components and main objective of the
the community. You are Shalu/Shalin, news reporter for
initiative?
a national daily. Write a report in 120-150 words on a
3. How often are the visits?
community initiative. You may use following cues:
4. About
 the summer camp - specific activities, events and
(CBSE, 2023) benefits
Spirit of Community 5. Positive changes – village and students
• Senior Residents – volunteerism 6. Future plans
• Teach underprivileged children
8. The Annual Prize Distribution Function was held in your
• Be a part of fundraising school last Friday. Write a report to be published in your
• Help with animal shelters school magazine. You are Saket/Sudipti, Student Editor of
• Join groups for clean-up drives the magazine.
9. The R.W.A (Resident Welfare Association), Nandipura -II,
6. The Science Club of Stellar Heights hosted a webinar on launched a volunteer programme for the young adults in
the theme “Innovations in Renewable Energy” on 15th July, and around the neighbourhood, on 18 January 2023. As
2024. As Neha Iyer, the club’s president, draft a report Sunitha. J, the local correspondent of the neighbourhood
detailing the highlights of this virtual event, incorporating newsletter, write a report, covering this event. Support your
the points provided. ideas with outline cues given below, to craft your report.
(CBSE SQP, 2023)
• Objective of the webinar
• Number and profile of attendees • which
  volunteer programme? - purpose of the launch event
- it’s importance
• Guest speakers and their key points
•  activities that took place - who attended?
• Interactive Q & A session
•  any key messages delivered?
• Clubs's future plans for community awareness
• how
  were young adults encouraged to join the volunteer
7. You are Riddhi Sikharwar, studying in Azad School, proram?
Bhopal, M.P. Your school has adopted a village, as a social •  any resources and information provided to attendees?
responsibility initiative. Every month students visit the •  any insight into follow up activities planned?
village and carry out literacy and numeracy programmes, •  what impact is expected to be achieved in the community?
including a fun-filled summer Mela* for adults and children.
10. You are Jay/Jaya, studying in Shyamala Memorial Hall,
Use the given cues (on the next page), along with your own
Chandigarh. Some students of our school attend a first aid
ideas, to write a report in about 120-150 words, for your
training camp for a week at Red Cross Headquarter of our
school magazine, on the initiative and its positive impact. state. Write a report for your school magazine in 120-150
[*fair/ carnival] words on the events of the campo and your participation
(CBSE APQ, 2023) and performance.

P
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 Hints & Explanations
1. Environment Awareness Camp for Prefects and Monitors XI and XII. The event aimed to enhance students'
By Suhita Tiwari communication skills, a crucial aspect for their academic
April 24, 2024: Vidya Bhawan Senior Secondary School, and personal growth.
Noida, recently hosted a 3-day environment awareness camp The workshop was attended by over 100 students and
for students aspiring to be prefects or monitors. The camp several faculty members. Interactive activities like listening
aimed to foster environmental consciousness and encourage drills, role-playing, and group discussions were conducted,
students to take the lead in promoting sustainability. emphasizing the importance of attentive listening and
During the camp, various activities were conducted to effective response strategies.
engage the participants. There were interactive workshops Key takeaways included techniques for better engagement
on waste segregation, recycling, and composting. Students in conversations and understanding the speaker’s
also designed eye-catching posters with powerful messages perspective. Post-workshop, participants received handouts
about the importance of conserving the environment. A summarizing the skills discussed, along with access to
tree-planting drive was organized, allowing students to online resources for further learning.
actively participate in greening their surroundings. This initiative is expected to significantly improve
The camp focused on encouraging everyone to contribute interpersonal interactions among students, fostering a more
to a healthier planet. The participants were urged to spread cooperative and supportive school environment.
awareness among their peers and families, highlighting 4. Harmony Society’s Drive towards a
the role each person plays in preserving natural resources. Greener Tomorrow
Through discussions and teamwork, the camp reinforced
the idea that even small actions can make a significant By Rajan Mathur, Secretary, Eco Club
difference in environmental conservation. March 05, 2024: The Eco Club of Harmony Society took
a commendable initiative to combat rising air pollution
by organizing a tree-plantation drive. Aimed at enhancing
green coverage in our locality, the event witnessed
Mistakes 101 : What not to do!
enthusiastic participation from residents, with over 150
• Students
  might fail to emphasize the importance of the attendees ranging from young children to senior citizens.
camp's objectives, such as environmental conservation A remarkable count of 500 saplings, including neem,
and how the camp encourages participants to be proactive. banyan, and peepal, were planted during the drive. Mrs.
Sunita Rao, a renowned environmentalist, graced the
2. Community Sports Outreach Program Launched at occasion with her insightful keynote address, emphasizing
Krishnapuram-IV the importance of trees in maintaining ecological balance.
By Rahul D., Sports Editor The Eco Club has also laid out future endeavors, including
Krishnapuram-IV, February 06, 2024: The Local Sports setting up regular maintenance schedules for the newly
Club of Krishnapuram unveiled a commendable initiative planted saplings and organizing awareness workshops.
yesterday - a Community Sports Outreach Program, aiming Together, we hope to carve a sustainable and greener path
to involve young adults in an array of sports activities. for our community.
The key sports introduced were football, basketball, and 5. Community Initiative: Teaching the Spirit of Giving
badminton. The event saw enthusiastic participation from
By Shalu, National Daily
young adults of the community and neighbouring areas.
Notably, national-level badminton player, Ananya Singh, Every Sunday morning, the quiet of our local park is
graced the event, sharing her journey and motivating the broken by the gentle buzz of learning and laughter. Senior
youth to embrace sports. To ensure effective participation, residents from our community gather under the shade of
the club distributed sports kits and informational brochures old trees, imparting knowledge and skills to children from
about each sport. The highlight was the announcement of neighbouring slums. This initiative, powered by the spirit
intercommunity matches and a mini-tournament planned for of giving back, shows the immense value of volunteerism.
the summer. The organizers are hopeful that this initiative These generous elders not only focus on academic tutoring
will not only promote physical fitness among the youth but but are also actively involved in various community
also foster community bonding and teamwork. With such betterment activities like fundraising for educational needs,
dedicated efforts, the future of sports in Krishnapuram-IV helping at animal shelters, and organizing clean-up drives.
looks promising. This blend of education and social responsibility fosters a
3. Workshop on Active Listening Engages robust community spirit and teaches young minds about the
Senior Students importance of service.
By Simran Kaur, Organizing Committee Member The initiative is a beautiful testament to the power of
On January 21, 2023, our school hosted an insightful community engagement and the impact it can have on the
workshop on Active Listening for the students of Classes lives of underprivileged children. It's not just about lessons in

135 Report Writing P

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 textbooks but lessons in life, compassion, and coexistence. was the state’s Education Minister. It was a source of
This movement is gaining momentum, encouraging more immense pride for us. He arrived at 10:00 a.m. Our Principal
residents to join in and support the cause, truly making it a and a few educators greeted him with a flower bouquet. The
communal effort towards a brighter future. parents of the pupils, as well as other prominent citizens
6. Illuminating Minds: Webinar on Renewable Energy of the city, were invited to grace the occasion. The event
began exactly at 10:00 a.m. The ceremony began with a
By Neha Iyer, President, Science Club
welcome song. The lamp was lit in front of the portrait of
July 15, 2024: ‘Stellar Heights’ Science Club successfully Goddess Saraswati by the Principal and the Chief 69 Guest.
conducted a webinar titled “Innovations in Renewable The students of our school brought cultural items. People
Energy”. The primary objective was to enlighten students applauded enthusiastically during this Cultural Program.
about the latest advancements in green energy solutions. Following that, our Principal presented the Annual Report
The virtual seminar saw over 300 attendees, comprising of the school. The Prize Distribution continued immediately
students, educators, and local community members. after that. The awards were distributed by the Chief Guest
Esteemed guest speakers, Dr. Ramesh Kumar and Ms. to the winners of the various fields. Then he delivered a very
Simran Kaur, shared their insights on solar innovations and inspiring speech. The event concluded with the National
the potential of wind energy, respectively. The highlight
Anthem.
was an interactive Q&A session, where participants
raised pertinent queries, reflecting deep engagement with 9. R.W.A Launches Youth Volunteer Program
the subject. Inspired by the overwhelming response, the By Sunitha J., Local Correspondent
Science Club is planning more such enlightening sessions, On January 18, 2023, the Resident Welfare Association of
focusing on community awareness and active participation Nandipura-II kicked off a new volunteer program aimed at
in adopting renewable energy solutions. We are optimistic engaging young adults in community service. The launch
about shaping an informed and eco-conscious community. event was a vibrant gathering, marked by the presence of
7. Empowering Through Education: Azad School's local dignitaries and enthusiastic youths ready to make a
Village Adoption Initiative difference.
By Riddhi Sikharwar, Azad School, Bhopal The event spotlighted the importance of volunteerism, with
Azad School has undertaken a great journey of social speakers emphasizing how such initiatives can transform
responsibility by adopting a village. This initiative involves our neighborhood. Key activities included workshops on
monthly visits by students to participate in literacy and community development and interactive sessions designed
numeracy programmes aimed at empowering rural residents. to inspire and recruit volunteers. Attendees received
A highlight of this initiative is the summer Mela, a carnival materials detailing the program's objectives and upcoming
filled with educational games, reading sessions, and events, like a neighborhood clean-up drive planned as a
numeracy activities. It not only entertains but also educates, follow-up activity.
strengthening community bonds. This program is expected to foster a strong sense
These interactions have led to significant positive effects. of community and contribute significantly to local
Villagers are more eager to study, and students have gained development, encouraging a new generation to step forward
a stronger sense of empathy and responsibility. The mutual and lead by example.
exchange of respect and knowledge was transformative. 10. First Aid Training Camp at Red Cross Headquarters
Future goals include expanding educational programmes By Jaya, Class XII
and introducing vocational training for adults. This April 28, 2024: Shyamala Memorial Hall witnessed an
ongoing commitment is aimed to give hope and inspire enriching experience for its students when a group, including
future development in the village, demonstrating the myself, attended a week-long first aid training camp at the
transformative potential of community engagement. State's Red cross Headquarters. The camp commenced with
an inaugural speech, emphasizing the importance of first aid
in everyday life. Over the week, expert trainers imported
Nailing the Right Answer knowledge on various first aid techniques from simple
wound dressing to CPR. Demonstrations were followed by
• Explain
  what activities are involved, such as literacy
hands-on practice sessions, ensuring that each participant
and numeracy programmes, and how often students
was equipped to handle emergencies.
visit the village.
One of the highlights was the real-life simulation exercise,
•  Include details about the summer Mela and the specific
where participants had to provide first aid in a staged
activities conducted during the fair or carnival. accident scenario, my participation in this exercise was
commendable. I successfully managed a "burn injury"
8. Annual Prize Distribution Function
earning appreciation from the trainers. The camp concluded
By Saket, Student Editor with a closing ceremony where participants received
April 29, 2024: Last Friday, our school hosted the Annual
certificates. This training not only provided us with life
Prize Distribution Function. The Chief Guest of the function saving abilities, but also built in us a sense of responsibility.

P
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 THE LAST LESSON
—Alphonse Daudet
1
Narrative Sketch
Theme
 T
 he story is set during a tough time when France lost two regions, Alsace "The Last Lesson" shows
how losing the right to use
and Lorraine, to Prussia because of a war. This change means that students one's own language due to
in these regions, including a young schoolboy Franz, can no longer learn foreign control causes deep
French at school; they must learn German instead. emotional pain. It underlines
the bond between language
 O
n the morning the story takes place, Franz is really nervous about going
and identity, and how
to school because he hasn’t prepared for his French grammar lesson. He’s language connects us to our
afraid of being scolded and even thinks about skipping school to enjoy the culture and freedom. This
beautiful day outside. story illustrates the value of
our native language and the
 W
hen Franz arrives at school, it’s unusually quiet—like a Sunday morning. regret of not appreciating it
The classroom, usually bustling with the noise of recitations and lessons, is until it's nearly lost.
unusually quiet, with villagers, including old Hauser and the former mayor,
Important Characters’ Sketch
attending the class as a mark of respect for their language and M. Hamel's
 Franz: The young
dedication. protagonist, Franz,
he teacher, M. Hamel, who has served the village for forty years, delivers
 T symbolizes innocence and
this news with a heavy heart. From tomorrow, they'll only learn German the everyday person's
because of new rules from the government. This news hits everyone hard ignorance about the
importance of their
because it means losing a part of who they are. cultural heritage. Initially
 T
he presence of the village elders in the classroom is a sign of their regret for reluctant and uninterested
not appreciating the value of their in learning, he undergoes a
transformation, realizing the
language and education sooner. It’s
value of his language and
a touching moment showing that culture as he faces the loss
they, too, feel the pain of losing of his right to learn it.
their language.  M. Hamel: M. Hamel is
 M
. Hamel talks about how beautiful a dedicated French teacher
in a small village, deeply
and important the French language connected to his language
is. He tells the students that and culture. Known for
holding onto their language is like being strict, he wears his
holding onto a key that can keep best clothes for his final
them free, even when times are lesson, showing respect for
the occasion. Facing the
tough. loss of his teaching role,
 M
. Hamel makes the most of their he becomes gentle and
final class by teaching them about grammar and writing. He even writes "Vive patient, regretfully sharing
his knowledge one last
La France!" on the board, which is his way of saying that their language and
time, and inspiring a sense
culture are still worth fighting for. of national pride in his
Contd... students.
  F
ranz regrets not paying attention in class before and now feels a strong connection to his teacher and his
language. It’s a big wake-up call for him about what truly matters.
 T
he story shows us how our language tells a lot about who we are. It also tells us that sometimes we don’t
realize how important something is until we're about to lose it.
Despite the sadness, there’s a strong sense of community as everyone comes together to face this change.
 
 "
The Last Lesson" is a heart-touching story about a small village facing big changes. Through Franz’s eyes,
we learn how important it is to cherish our culture and education. The story leaves us thinking about what it
means to stand together and remember our roots, no matter what challenges come our way.

NCERT EXERCISE
Ans. Franz's feelings changed from seeing school and M. Hamel
Think as you read as a chore and a source of fear, respectively, to valuing them
deeply. He regretted not learning more and appreciated
(NCERT, Pg 7) M. Hamel's efforts and the importance of his lessons.
1. What was Franz expected to be prepared with for school
that day?
Understanding the text
Ans. Franz was expected to be prepared with a lesson on
participles for school that day. He was worried because he
1. The people in this story suddenly realise how precious
did not know anything about them.
their language is to them. What shows you this? Why
2. What did Franz notice that was unusual about the school does this happen?
that day?
Ans. The people's attendance at the last French lesson, including
Ans. Franz noticed that the school was unusually quiet and villagers who hadn't attended school in years, shows how
solemn that day. He also saw villagers, including old Hauser precious their language is to them. This realization happens
and the former mayor, sitting in the back of the classroom,
because the order from Berlin to only teach German signifies
which was not normal.
a loss of their cultural identity and freedom, making them
3. What had been put up on the bulletin-board? value their native language more.
Ans. The bulletin-board had an order from Berlin announcing 2. Franz thinks, “Will they make them sing in German,
that only German would be taught in the schools of Alsace even the pigeons?” What could this mean?
and Lorraine. This meant that French lessons were to be

(There could be more than one answer.)
stopped, making it Franz's last French lesson.
Ans. Franz wonders if even the pigeons will have to sing
(NCERT, Pg 8)
in German. This shows how deeply the change affects
1. What changes did the order from Berlin cause in school everything. He feels like the new rules will change every
that day? part of life, even the natural sounds of birds.
Ans. The order from Berlin caused the school to have its last This thought also shows Franz's worry about losing
French lesson, bringing a solemn atmosphere. Villagers his own culture and language. It's like he's saying that
attended the class to show respect for their language and M. everything familiar, even birds' songs, might be taken
Hamel's dedication. over by the foreign rule, making his world unrecognizable.
2. How did Franz's feelings about M. Hamel and school The language was as natural to them as cooing is to the
change? pigeons..
P
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 miscellaneous Exercise
1. Which of the following can be attributed to M. Hamel’s
Extract Based Questions declaration about the French language?
(a) subject expertise (b) nostalgic pride
I. Read the following extract and answer the questions
that follow: (6 M) (c) factual accuracy (d) patriotic magnification
Usually, when school began, there was a great bustle, which 2. Read the quotes given below.
could be heard out in the street, the opening and closing Choose the option that might best describe M. Hamel's
of desks, lessons repeated in unison, very loud, with our viewpoint.
hands over our ears to understand better, and the teacher’s
(i) Those who know nothing of
great ruler rapping on the table. But now it was all so still!
foreign languages know nothing of
I had counted on the commotion to get to my desk without
their own.
being seen; but, of course, that day everything had to be as
quiet as Sunday morning. Through the window I saw my —Johann Wolfgang von Goethe
classmates, already in their places, and M. Hamel walking
(ii) Language is the road map of a
up and down with his terrible iron ruler under his arm. I culture. It tells you where its people
had to open the door and go in before everybody. You can come from and where they are going
imagine how I blushed and how frightened I was. —Rita Mae Brown
(CBSE SQP, 2023)
1. List any two sensory details present in this extract. (iii) A poor man is like a foreigner in
his own country.
2. Why does the protagonist feel anxious about entering the
classroom on this particular day? —Ali Ibn Abi Talib
(a) The classmates have started the lesson (iv) The greatest propaganda in the
(b) The teacher is in a bad mood world is our mother tongue, that is
(c) The classroom is too quiet what we learn as children, and which
we learn unconsciously. That shapes
(d) The protagonist is running late our perceptions for life.
3. Complete the sentence appropriately. —Marshal McLuhan
The phrase “as quiet as Sunday morning” suggests that
____________. (a) Option (i) (b) Option (ii)
4. Pick evidence from the extract that helps one infer that this (c) Option (iii) (d) Option (iv)
was not the protagonist’s first time being late to school. 3. “I was amazed to see how well I understood it.”
5. What does the term ‘terrible iron ruler’ indicate about M. Select the option that does NOT explain why Franz found
Hamel? the grammar lesson “easy”.
6. Which of the following headlines best suggests the central (a) Franz was paying careful attention in class this time.
idea of the extract? (b) M. Hamel was being extremely patient and calm in his
(a) The Fears of a Latecomer teaching.
(b) The Importance of Punctuality (c) Franz was inspired and had found a new meaning and
(c) The Rigidity of the School System purpose to learning.
(d) The Anxiety of a Young Student (d) Franz had realized that French was the clearest and most
II. Read the following extract and answer the questions logical language.
that follow: (4 M) 4. Franz was able to understand the grammar lesson easily
M. Hamel went on to talk of the French language, saying because he was
that it was the most beautiful language in the world — the (a) receptive (b) appreciative
clearest, the most logical; that we must guard it among us
(c) introspective (d) competitive
and never forget it, because when a people are enslaved, as
long as they hold fast to their language it is as if they had III. Read the following extract and answer the questions
the key to their prison. Then he opened a grammar book that follow: (6 M)
and read us our lesson. I was amazed to see how well I My last French lesson! Why? I hardly knew how to write!
understood it. All he said seemed so easy, so easy! I should never learn anymore! I must stop there, then! Oh,
(CBSE QB) how sorry I was for not learning my lessons, for seeking

141 The Last Lesson P

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 birds’ eggs, or going sliding on the Saar! My books, that V. Read the following extract and answer the questions
had seemed such a nuisance a while ago, so heavy to carry, that follow: (4 M)
my grammar, and my history of the saints, were old friends Poor man! It was in honour of this last lesson that he had
now that I couldn’t give up. And M Hamel, too; the idea put on his fine Sunday clothes, and now I understood why
that he was going away, that I should never see him again,
the old men of the village were sitting there in the back of
made me forget all about his ruler and how cranky he was.
the room. It was because they were sorry, too, that they had
1. Select the proverb that best outlines the moral in the above not gone to school more. It was their way of thanking our
extract. master for his forty years of faithful service and of showing
(a) Procrastination is the thief of time. their respect for the country that was theirs no more.
(b) ‘Both positive and negative thinking are contagious.’ (CBSE QB)
(c) An ounce of prevention is worth a pound of cure. 1. Why does the narrator refer to M. Hamel as ‘Poor man!’?
(d) The grass is always green on the other side of the fence. (a) He empathizes with M. Hamel as he had to leave the
2. List the objects which seem intimidating to Franz. village.
3. What does the narrator regret most about his past actions in (b) He believes that M. Hamel’s “fine Sunday clothes”
the context of the extract? clearly reflected that he was not rich.
4. According to the extract, what changed the narrator’s (c) He feels sorry for M. Hamel as it was his last French
perspective towards his books and lessons? lesson.
5. Which of these uses ‘heavy’ in the same way as the (d) He thinks that M. Hamel’s patriotism and sense of duty
following expression in the extract?
resulted in his poverty.
‘my books... so heavy to carry’
2. Which of the following idioms might describe the villagers’
(a) The heavy rain prevented us from going out. act of attending the last lesson most accurately?
(b) The topic was heavy and complex for the students.
(a) ‘Too good to miss’
(c) The heavy load was hard to lift.
(b) ‘Too little, too late’
(d) The atmosphere in the room was heavy with tension.
(c) ‘Too many cooks spoil the broth’
6. Complete the given sentence appropriately.
(d) ‘Too cool for school’
Upon realizing that it was his last French lesson, the narrator
felt __________. 3. Choose the option that might raise a question about
M. Hamel’s “faithful service”.
IV. Read the following extract and answer the questions
that follow: (6 M) (a) When Franz came late, M. Hamel told him that he was
I jumped over the bench and sat down at my desk. Not till about to begin class without him.
then, when I had got a little over my fright, did I see that (b) Franz mentioned how cranky M. Hamel was and his
our teacher had on his beautiful green coat, his frilled shirt “great ruler rapping on the table”.
and the little black silk cap, all embroidered, that he never (c) M. Hamel often sent students to water his flowers, and
wore except on inspection and prize days.
gave a holiday when he wanted to go fishing.
1. What was the reason for Franz’s fright in the above extract?
(d) M. Hamel permitted villagers put their children “to work
2. ‘I had got over a little over my fright’ means: on a farm or at the mills” for some extra money.
(a) A bit afraid (b) A bit settled
4. Choose the option that most appropriately fills in the blanks,
(c) Being cautious (d) All of these for the following description of the given extract.
3. What change did Franz find in the teacher’s appearance? The villagers and their children sat in class, forging with
(a) He was dressed for a special occasion their old master a (i) _____ togetherness. In that moment,
(b) He wore a hat the class room stood (ii) _____. It was France itself, and the
(c) He wore a black coat last French lesson a desperate hope to (iii) ______ to the
(d) None of these remnants of what they had known and taken for granted.
Their own (iv) _______.
4. Identify the line from the text that bears evidence to the fact
that on entering the class, Franz noticed a change? (a) (i) graceful; (ii) still; (iii) hang on; (iv) country
5. Complete the sentence appropriately. (b) (i) bygone; (ii) up; (iii) keep on; (iv) education
M. Hamel was dressed so exquisitely because __________. (c) (i) beautiful; (ii) mesmerised; (iii) carry on; (iv) unity
6. What does the phrase “Jumped over the desk” means? (d) (i) forgotten; (ii) transformed; (iii) hold on; (iv) identity
P
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 7. Why was M. Hamel dressed in formal clothes in school?
Short Answer Type Questions (CBSE 2023)
8. What impact does the last lesson in French have on the
Answer the following questions in about 40-50 words. (2 M)
students in the story ‘The Last Lesson’?
1. A crisis brings out our true capacity for compassion and
kindness. 9. In ‘The Last Lesson’, how does the author emphasize the
significance and value of one’s native language?
Support the above statement using M. Hamel from ‘The
Last Lesson’ as an example. State any one detail from the 10. If this had been M. Hamel’s first lesson, how do you think
text. (CBSE APQ, 2023) the school experience of the students might have been
impacted? (CBSE QB)
2. In “The Last Lesson,” M. Hamel’s attitude towards
teaching and his students undergoes a change. How does 11. Little Franz is the narrator of the story. The name ‘Franz’
his blackboard message highlight this shift? means ‘from France’. In what way does the story being told
3. In “The Last Lesson,” what does the character of M. as a first-person narrative of Franz impact your reading and
Hamel symbolize in the context of cultural erosion and the understanding of the story? Provide at least one evidence
importance of one’s native language? from the text to support your opinion. (CBSE QB)
4. "Our native language is part of our culture and we are proud of 12. At the end of his last lesson, M. Hamel decides to leave a
it." How does the presence of village elders in the classroom little note for each of his students for them to find the next
in M Hamel's last lesson show their love for French? day at their desks. Based on your reading of the story, what
5. “You realise the true value of a thing only on losing it.” might his note to Franz read?  (CBSE QB)
Comment on this statement in the light of the story. You may begin like this:
6. In ‘The Last Lesson’, how does M. Hamel’s demeanor
Dear Franz, I know you have always preferred to run in the
change from his usual strictness on the last day of class? open fields …

Hints & Explanations

[The extract revolves around the theme of punctuality and
Extract-Based Questions the consequences of being late. The protagonist is anxious
and embarrassed about being late to school, highlighting
Extract-I
the importance of being on time. The other options, such as
1. (Any two) A, C and D touch upon some of the themes in the extract
• the sounds of the opening and closing of desks but do not fully capture the central idea.]
• the loud lessons repeated in unison
• the teacher’s ruler rapping on the table
2. (a) The classmates have started the lesson. Nailing the Right Answer
[The protagonist sees through the window that the other Always read the extract closely to catch specific details
students are already in their places, and this makes him feel that directly answer the question. For instance, if asked
anxious as he is trying to avoid being seen and is worried about sensory details, scan the text for descriptions that
about being scolded by the teacher and embarrassed in appeal to the senses, such as sounds, sights, or physical
front of his classmates. The fact that the other students have sensations, which can help provide accurate answers.
already started the lesson adds to his anxiety.]
3. ...the school was unusually quiet and still, as if it were a day
of rest/holiday/school off, rather than a bustling school day. Extract-II
4. The protagonist seems to have a plan for how to sneak into 1. (d) patriotic magnification
class without being noticed, suggesting that they may have [M. Hamel spoke about the French language with a lot of
been in similar situations before. love and pride, especially because their area was taken over
[The protagonist states: “I had counted on the commotion by another country. He wanted everyone to remember their
to get to my desk without being seen,” which implies that
language because it was a part of who they are.]
they have been late before and have developed a strategy
for avoiding punishment.] 2. (b) Option (ii)
5. The strictness and severity of M. Hamel’s discipline. 3. (d) Franz had realized that French was the clearest and most
6. (b) The Importance of Punctuality logical language.

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 [Franz found the grammar easy not because French was Extract-V
the easiest or clearest language, but for other reasons like 1. (c) He feels sorry for M. Hamel as it was his last French
paying attention.] lesson.
4. (a) receptive [The term "poor man" reflects a sense of pity for M. Hamel's
[Franz could understand the lesson well because he was situation—having to say goodbye to his students and his
really open to learning that day.] role due to the political change.]
2. (b) ‘Too little, too late’
Extract-III
[This implies that while the villagers show regret and a
1. (a) Procrastination is the thief of time.
desire to honor M. Hamel, their action comes at a point
[The narrator regrets not having paid attention to his lessons when it can no longer change the situation—the loss of their
earlier and realizes too late that his opportunity to learn is language in the education system.]
now gone.]
3. (c) M. Hamel often sent students to water his flowers, and
2. The objects that seem intimidating to Franz are likely the gave a holiday when he wanted to go fishing.
books and his grammar, as well as the history of the saints,
[This indicates that M. Hamel may have sometimes
which he previously found burdensome.
prioritized his personal activities over teaching, which
3. The narrator regrets most not learning his lessons and could be seen as a lapse in his otherwise faithful service.]
wasting time on other activities like looking for bird’s eggs
4. (d) (i) forgotten; (ii) transformed; (iii) hold on; (iv) identity
or going sliding on the Saar.
[This option captures the themes of the extract, emphasizing
4. The narrator's perspective towards his books and lessons
the solemn and transformative nature of the last lesson and
changed because he realized that he was going to lose them
the collective effort to cling to their cultural identity in the
soon. They became valuable to him because they were a
face of change.]
connection to his language and his teacher, M. Hamel.
5. (c) The heavy load was hard to lift.
[This is because it refers to the actual, physical weight of an Mistakes 101 : What not to do!
object, similar to the weight of the books.]
A common error is to choose an answer based on a
6. regretful superficial understanding of the text without considering
the deeper emotional or symbolic implications. For
Extract-IV example, failing to recognize the significance of the
1. The reason for Franz’s fright in the above extract was likely villagers' presence and attire might lead to misinterpreting
because he was late for school and expected to be scolded, why the text describes the classroom or the nation in a
especially since there was a lesson on participles that he particular way.
was not prepared for.
2. (b) A bit settled
[This phrase indicates that Franz had begun to feel less Short Answer Type Questions
afraid after his initial fear upon entering the classroom.]
1. The crisis of losing France to the Prussians brings out M
3. (a) He was dressed for a special occasion
Hamel’s compassion. (Any one):
[M. Hamel was wearing his fine clothes that he only wore
• He
  stops being strict and talks to all the students in his
for inspections and prize days, indicating the special nature
class in a gentle and understanding manner.
of this day.]
4. The line from the text that bears evidence to the fact • He
  does not use the ruler in class to beat the students, but
that on entering the class, Franz noticed a change is the triesto impart as much knowledge as he can in the last
observation of M. Hamel’s attire: "I see that our teacher French lesson.
had on his beautiful green coat, his frilled shirt and the little • He
  admits his own fault in why many students never
black silk cap, all embroidered, that he never wore except learnt the language and feels apologetic for the choices
on inspection and prize days." he made as a teacher.
5. it was the last French lesson he was giving, which was a • He
  does not berate Franz when Franz is unable to recite
significant and solemn occasion. in French. Instead, he understands why Franz is behind
6. The phrase “Jumped over the bench” means that Franz due to his family’s limitations and being given chores by
quickly and somewhat abruptly entered his seating area in M Hamel.
the classroom, indicating his haste and the urgency of the • M.
  Hamel delivers the last lesson with patience that
situation. It's a way to describe moving quickly to one's generates a lot of interest. He speaks about everything
seat, not literally jumping over a desk. that he loves about the French language.
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 • M.
  Hamel chokes up by the end of the lesson and is the importance of their language and heritage, sensing
unable to speak. He fixes his eyes on everyone in the a profound loss. The atmosphere becomes heavy with
class feeling extremely sentimental and with a kind gaze. emotions, and even the usually restless young children pay
2. In “The Last Lesson,” M. Hamel writes “Vive La France!” attention to the lesson. The lesson becomes a wake-up call
on the blackboard. This action captures his deep love for for Franz and his classmates, making them appreciate their
his country and language. It symbolizes the weight of language and culture.
the realization that this might be the last French lesson in 9. The author emphasizes the significance of one's native
Alsace. By writing these words, M. Hamel emphasizes the language by showing how the villagers react to losing the
importance of language and identity. It serves as a reminder right to learn French. M. Hamel speaks about the beauty
to the students that their French language and culture are and importance of the French language, highlighting it as a
worth preserving, even in challenging times. key to freedom and identity in the face of oppression.
3. M. Hamel symbolizes the cultural heritage and the
10. If this had been M. Hamel's first lesson, the students would
emotional connection one has with their native language.
have experienced a more typical, less emotional class
His last lesson is a poignant reminder of the loss of identity
focused mainly on regular studies without the deep sense
and cultural erosion that comes with the imposition of a
of loss and urgency that marked his last lesson. The lesson
foreign language.
would have likely been stricter and less reflective on the
4. The village elders' presence in the classroom during M. value of their language.
Hamel's last French lesson symbolizes their respect for
the French language and their cultural heritage. It shows 11. Franz narrating the story adds a personal and emotional
their regret for not attending school more often and their layer, making the impact of the events more immediate and
heartfelt. For example, his regret for not paying attention in
solidarity in facing the loss of their language rights.
class earlier and his sudden realization of the importance of
5. The statement reflects the villagers' sudden appreciation his lessons make the story relatable and poignant (deeply
for the French language when faced with the order to only affecting).
teach German. This realization of value upon the threat of
loss highlights a common human tendency to appreciate
something fully only when it is at risk of being taken away.
Nailing the Right Answer
6. On the last day of class, M. Hamel's demeanor changes
from his usual strictness to a more gentle and reflective Provide textual evidence: Support your opinions with
specific examples or quotes from the text.
mood. He expresses patience and understanding, wanting
to impart as much knowledge as possible, recognizing the 12. Dear Franz,
gravity and finality of the situation.
I know you have always preferred to run in the open fields,
7. M. Hamel dressed in formal clothes as a mark of respect for yet remember that knowledge also opens vast landscapes
the solemnity of the occasion. It was his last opportunity to to explore. Regret for wasted time is common, but take this
teach French, symbolizing honor both to his profession and as a lesson to cherish and utilize every moment forward,
to the French language. for time once passed never returns. Keep learning, keep
8. In “The Last Lesson”, the final session of French teaching growing.
leaves a profound impact on the students. They realize —M. Hamel

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