2PBDBEPS(Vijaya Patake)

BOARD MODEL PAPER - 1 - 2024-25 7. The direction of current induced in the loop ‘abc’ shown in the figure is
II PUC – PHYSICS - (33) (A) along ‘abc’ if I is increasing
Time: 3 hours Maximum marks: 70 (B) along ‘abc’ if I is decreasing
Instructions: (C) along ‘acb’ if I is increasing
1. All parts (A to D) are compulsory. PART - E is only for visually challenged students. (D) along ‘acb’ if I is constant
2. For Part - A questions, first written-answer will be considered for awarding marks.
3. Answers without relevant diagram / figure / circuit wherever necessary will not carry any marks. Ans (C)
4. Direct answers to numerical problems without relevant formula and detailed solutions will not carry any marks. 8. An ideal step-up transformer decreases _________.
(A) current (B) voltage (C) power (D) frequency
PART - A Ans (B)
I. Pick the correct option among the four given options for ALL of the following 9. The displacement current is due to
questions: [15 × 1 = 15] (A) flow of electrons (B) flow of protons
1. The S.I. unit of electric charge is (C) changing electric field (D) changing magnetic field
(A) coulomb metre (B) coulomb per metre Ans (C)
(C) coulomb (D) per coulomb 10. An object of finite height is placed in front of a concave mirror within its focus. It forms
Ans (C) (A) a real enlarged image (B) a real diminished image
2. The angle between equipotential surface and electric field is (C) a virtual enlarged image (D) a virtual diminished image
(A) 90º (B) 0º (C) 180º (D) 45º Ans (C)
Ans (A) 11. A beam of unpolarised light of intensity I0 is passed through a pair of polaroids with their pass-axes
3. Statement-I: The resistivity of metals increases with increase in temperature. inclined at an angle of θ. The intensity of emergent light is equal to
Statement-II: Increasing the temperature of metals causes more frequent collisions of electrons. I I
(A) I0 cos2   (B) I0 cos  (C) 0 cos  (D) 0 cos 2 
2 2
(A) both I and II are true and II is the correct explanation of I.
Ans (D)
(B) both I and II are true but II is not the correct explanation of I.
(C) I is true but II is false. 12. Emission of electrons from a metal surface by heating it is called
(D) both I and II are false. (A) photoelectric emission (B) thermionic emission
Ans (A) (C) field emission (D) secondary emission
Ans (B)
4. A moving coil galvanometer can be converted into a voltmeter by connecting
(A) a low resistance in parallel with galvanometer. 13. When alpha particles are passed through a thin gold foil, most of them go undeviated because
(B) a low resistance in series with galvanometer. (A) most of the region in an atom is empty space
(C) a high resistance in parallel with galvanometer. (B) alpha particles are positively charged particles
(D) a high resistance in series with galvanometer. (C) alpha particles are heavier particles
Ans (D) (D) alpha particles move with high energy
Ans (A)
5. When a bar magnet is suspended freely, it points in the direction of
(A) east-west (B) north-south (C) northeast-southeast (D) northwest-southwest 14. Nuclei with same atomic number are called
Ans (B) (A) isotopes (B) isobars (C) isomers (D) isotones
Ans (A)
6. The energy stored in an inductor of inductance L in establishing the current I in it is
1 1
(A) LI (B) LI2 (C) LI (D) LI 2
2 2
Ans (D)

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15. The column-I is the list of materials and the column-II, the list of energy band gaps 𝐸g. Identify the 23. A long air-core solenoid of 1000 turns per unit length carries a current of 2 A. Calculate the magnetic
correct match. field at the mid-point on its axis.
Column - I Column - II Solution
(i) conductors (a) Eg < 3 eV B  o nI
(ii) insulators (b) Eg = 0 eV
B  4 107 1000  2
(iii) semiconductors (c) Eg > 3 eV
(A) (i) - (a), (ii) - (b), (iii) - (c) (B) (i) - (b), (ii) - (a), (iii) - (c) B  2.512  103 T
(C) (i) - (c), (ii) - (a), (iii) - (b) (D) (i) - (b), (ii) - (c), (iii) - (a) 24. Give the principle of AC generator. Why is a current induced in an AC generator called alternating
Ans (D) current?
Solution
II. Fill in the blanks by choosing appropriate answer given in the bracket for ALL of
The principle of AC generator is electromagnetic induction.
the following questions [5 × 1 = 5]
The current induced in AC generator is called alternating current since the polarity of the induced current
(photon, polar, zero, infinite, phase, phasor)
reverses after every half rotation of the coil.
16. A molecule possessing permanent dipole moment is called ____________ molecule.
25. Write any two uses of ultraviolet radiations.
Ans polar
Solution
17. The net magnetic flux through any closed surface is _______________.
 UV rays are used to sterilize surgical equipment
Ans zero  UV rays are used to kill germs in water purifiers
18. A rotating vector used to represent alternating quantities is called __________. 26. Name the objective used in (a) refracting type telescope and (b) reflecting type telescope.
Ans phasor Solution
19. A wavefront is a surface of constant ___________. (a) Convex lens of large aperture and large focal length
Ans phase (b) Concave mirror

20. In interaction with matter, light behaves as if it is made up of packet of energy called __________. 27. Write the two conditions for the total internal reflection to occur.

Ans photon Solution

 Light ray must travel from denser medium to rarer medium.
PART - B  Angle of incidence must be greater than the critical angle.
III. Answer any FIVE of the following questions: [5 × 2 = 10] 28. Name the majority and the minority charge carriers in n-type semiconductor.
21. State and explain Gauss’s law in electrostatics. Solution
Solution Electrons are the majority charge carriers and holes are the minority charge carriers in n type
1 semiconductor.
The total electric flux through a closed surface in an electric field is equal to times the total charge
0
enclosed by it. PART - C
1 IV. Answer any FIVE of the following questions [5 × 3 = 15]
  qi
0
29. Write any three properties of electric field lines.
22. Define drift velocity and mobility of free electrons in conductors. Solution
Solution Electric field lines never intersect each other.
The drift velocity is the average velocity of free electrons in a conductor due to an external electric field. Electric field lines never form closed loops.
The- mobility of a charge carrier is defined as the drift velocity per unit electric field.
Electric field lines never pass through the conductors.

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30. Obtain the expression for the effective capacitance of two capacitors connected in parallel. 32. Write any three differences between diamagnetic and paramagnetic materials.
Solution Solution
Q1 -Q1 Diamagnetic substances Paramagnetic substances
(1) These are repelled by a magnet These are attracted by a magnet
Q Q (2) When diamagnetic substance is placed in a Magnetic field lines pass through this substance.
+ –
+  magnetic field, magnetic field lines do not
+  pass through substance.
+ 
CP (3) The relative permeability (r) of diamagnetic The relative permeability of a paramagnetic
=
substance is always less than 1 ; r < 1 substance is more than 1 ; r > 1
Q2 -Q2
Or Or
V Magnetic susceptibility is negative ( < 0) Magnetic susceptibility is positive ( > 0)

V Fig (b) (4) The magnetic susceptibility of a diamagnetic The magnetic susceptibility of a paramagnetic
Fig (a)
substances does not change with temperature substance is inversely proportional to absolute
Consider two capacitors of capacitances C1 and C2 connected in parallel between points A and B. temperature
Let V = potential difference across the combination. (5) Magnetic susceptibility is independent of Magnetic susceptibility depends on temperature
Q = total charge stored in the combination. temperature
The charge Q divides into Q1 and Q2 and stored separately in the two capacitors C1 and C2 respectively. (6) Magnetic field lines are expelled out when a Magnetic field lines flow inside when a
Hence, Q = Q1 + Q2 …(1) diamagnetic substance placed in an external paramagnetic substance placed in an external
Since potential difference across each capacitor remains same, magnetic field. magnetic field.
Q1 = C1V and Q2 = C2V  Q = C1V + C2V
33. Describe an experiment to demonstrate the phenomenon of electromagnetic induction using a bar magnet
Q = (C1 + C2)V …(2)
and a coil.
Q
The effective capacitance of the parallel combination is CP   Q  CPV …(3)
V Solution
Comparing Eq. (2) and Eq. (3), we get
CpV = (C1 + C2)V or Cp = C1 + C2 …(4)
31. What is Lorentz force? Write its expression and explain the terms.
Solution
Referring to figure, C is a large coil of several turns of a conductor connected to a sensitive galvanometer
The resultant force acting on a charged particle moving through a region of combined electric and
(G). Consider a bar magnet NS placed co-axially and close to the coil.
magnetic fields is called Lorentz force.
The expression for Lorentz force is Observations
  
F  Fele  Fmag  When the magnet is moved towards the coil, the galvanometer shows a momentary deflection.
     When the magnet is moved away from the coil, the galvanometer again shows deflection but in
F  qE  ( qv  B)
    opposite direction.
F  q  E  (v  B ) 
 Deflection is also observed when the magnet is kept stationary and the coil is moved towards or
away from the magnet.
 If the relative motion between the coil C and the magnet takes place faster, a large deflection is
observed in the galvanometer.
 No deflection is observed in the galvanometer if both the coil and magnet are at rest or if there is no
relative motion between them.

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Explanation 35. Write the three postulates of Bohr’s atom model.

 From the above observations, it can be concluded that whenever there is a relative motion between Solution
the coil and the magnet a momentary emf and hence a current are induced in the coil. 1. The Electrons revolve only in certain orbits called stationary orbits without emission of radiant
 During the relative motion, the magnetic flux linked with the coil changes continuously. This is the energy.
reason for induced emf. 2. Electrons revolve around nucleus only in such orbits for which the angular momentum is an integral
 The induced emf lasts as long as the change of flux takes place.
multiple of  h  . That is, mvr  nh
 If the change of flux takes place rapidly, the emf induced is larger.  2  2
34. Give any three results of experimental study of photoelectric effect. 3. An electron emits energy only when it jumps from higher orbit to lower orbit. If Ef and Ei are the
Solution energies of electron in higher orbit and lower orbit respectively, then frequency of emitted photon is
1. Threshold frequency and limiting wavelength: For a given photosensitive cathode, there is a given by hv = Ef  Ei
certain frequency of radiation, below which no photoelectric emission takes place, whatever be the 36. Find the energy equivalent of one atomic mass unit, first in joule and then in MeV.
intensity and whatever be the duration for which the radiation is incident on the surface. This Given: 1𝑢 = 1.6605 × 10–2 kg, e = 1.602  10–19 C and c = 2.9979  108 m s–1.
minimum frequency is called threshold frequency (v0) [Fig (a)]. The corresponding wavelength is Solution
called limiting wavelength (0). This is the maximum wavelength beyond which photoemission is According to Einstein’s mass energy relation E = mc2. It can be shown that the energy equivalent of
not possible. The threshold frequency is found to be different for different materials. 1u  931.5MeV
2. Photoelectric effect is almost an instantaneous process (even when the incident radiation is of 1u  1.6605  1027 kg and c  2.9979  108 m s 1
very low intensity). The time lag between the incidence of radiation and the emission of  E  1.6605  1027  (2.9979  108 ) 2 J
photoelectrons is of the order of 10−9 s or even less. 1.6605  1027  (2.9979  108 ) 2
 eV
3. (i) Effect of intensity of radiation on photocurrent: For a given frequency of radiation higher than 1.602  1019
~ 931.5  10 eV
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the threshold frequency and for a given photocathode, the photoelectric current (i.e. the number of
photoelectrons) is directly proportional to the intensity of incident radiation. The variation of 1u  931.5 MeV (Energy equivalent)
photoelectric current with the intensity of incident radiation is as shown in Fig. (b)
(ii) Saturation current: As the plate potential (V) is increased, the photoelectric current (I) PART - D
increases gradually upto a certain constant value called the saturation current (Is). V. Answer any THREE of the following questions [3 × 5 = 15]
The number of photoelectrons and hence the saturation current increases with the intensity of the
incident radiation as shown in Fig. (c). 37. Derive the expression for the electric field at a point on the axis of an electric dipole.
4. Effect of frequency of incident radiation on the kinetic energy of photoelectrons: When the Solution
frequency of radiation incident on a photoemissive surface is increased above the threshold Consider a point A on the axis of a dipole at a distance r from its centre O, as shown in the figure.
frequency, the kinetic energy of the photoelectrons increases linearly [Fig. (a)]. The electric field at A due to the charge +q is given by
 1 q
5. Stopping potential: If the plate potential is E q  pˆ 2a
made negative, current still flows, but 40 (r  a )2 A O
where p̂ is a unit vector along dipole axis. E+q E–q +q p –q
decreases as the negative potential is
increased. At a particular negative potential The electric field at A due to the charge –q is r
 1 q
(V ), the photoelectric current becomes
0
E q   ( pˆ )
almost zero. This potential is called stopping 40 (r  a ) 2
potential or retarding potential as shown in The total electric field at A is
Fig. (c) and Fig. (d).    q  1 1  q  (r  a )2  (r  a) 2 
E  E q  E q =  pˆ    pˆ
Figure (d) shows the variation of 40  (r  a) 2 (r  a ) 2  4 0  (r 2  a 2 ) 2 
photoelectric current with plate potential for Fig (d) q  (r 2  a 2  2ar )  (r 2  a 2  2ar ) 
   pˆ
radiation of different frequencies. 4 0  (r 2  a 2 ) 2 
q  4ar 
 pˆ
40  (r 2  a 2 )2 

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 1 2 pr Consider a current loop of radius r carrying current I in the direction as shown in the figure.
E pˆ …(2)
4 0 (r 2  a 2 )2 Let the plane of the loop be perpendicular to the plane of the paper. Let P be a point on its axis at a

For r >> a, a2 can be neglected with respect to r2. E can be approximated from eq. (2) as distance x from the centre O of the current loop.
 1 2p   1 4qa  The entire loop is assumed to be divided into a large number of small current elements each of length
E . p or E  p …(3)
4 0 r 3 40 r 3 dl.
 
E is directed along the dipole axis i.e., along p̂ . Consider one such current element CD  I dl of the loop.

1 2p According to the Biot-Savart’s law, the magnitude of magnetic field dB at the point P due this current
In magnitude, E  …(4)
40 r 3 element is
If r >> a, then the dipole is called short dipole.    I dl sin 90
dB   0 
38. Two cells of different emfs and different internal resistances are connected in series. Derive the  4  a2
expression for effective emf and effective internal resistance of the combination.    I dl
dB   0  2 along PM
 4  a
Solution
The magnetic field dB along PM can be resolved into its rectangular components
(E1, r1) (E2, r2) (Eeq, req)
 dB sin  along PY (perpendicular to the axis of loop)

A I B I I C A I I C  dB cos  along PX (parallel to the axis of loop)
V1 V2 V     I dl
V Similarly, magnitude of magnetic field dB at the point P due to current element CD is dB   0  2
 4  a
Consider two cells of emfs E1 and E2 having internal resistances r1 and r2 respectively, connected in along PN.

series between A and C. (as shown in the figure). The magnetic field dB along PN can be resolved into its rectangular components
Terminal potential difference across 1st cell is  dB sin  along PY (perpendicular to the axis of loop)
V1 = E1  I r1  dB cos  along PX (parallel to the axis of loop)
Terminal potential difference across 2nd cell is The components perpendicular to the axis of the loop will be equal and opposite and hence they will
V2 = E2  I r2 cancel out. While, their axial components will be equal and along the same direction and hence they get
Total terminal potential across the combination is added up.
V = V1 + V2 The magnetic field at the point P due to pair of diametrically opposite current elements
V = (E1 + E2)  I (r1 + r2) …(1) (CD and CD) = 2dB cos  along X direction
If Eeq and req are the equivalent emf and internal resistance of series combination of cells, then Magnetic field at point P due to one turn
V = Eeq  I req …(2) r
   I dl r
On comparing eq. (1) and eq. (2), we get B   2dB cos  But dB   0  2 and cos  
0  4  a a
Eeq  E1  E2 …(3) r
   I dl r
and req  r1  r2 …(4) B   2 0  2
0 
4  a a
r
39. Derive the expression for the magnetic field at a point on the axis of a circular current loop.    2I  r 
  0  2    dl
Solution  4  a  a  0
   2 Ir
1
1
  0  3  r  a3  a  a2 a3   r 2  x2  2  r 2  x 2 
a   r 2  x2  2  4  a
  (90  )    2I r
2
B 0 
sin   sin (90  )  3
 4   r 2  x2  2
r
cos   40. (a) Two coherent waves of a constant phase difference undergo interference. Obtain the expression for
a
the resultant displacement. [3]
(b) Write the conditions for constructive and destructive interference in terms of phase difference. [2]
Solution

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(a) Consider two coherent light sources S1 and S2 emitting waves of same amplitude and same 41. What is a rectifier? Explain the working of a full-wave rectifier using a neat circuit diagram. Draw its
frequency. The light waves from S1 and S2 reach the observation screen. Let O be the centre of the input-output waveforms.
screen and let it be equidistant from S1 and S2. Solution
It is a device which converts both the half cycles of AC into DC

Consider an arbitrary point P on the screen.

Let the phase difference between the waves coming from S1 and S2 at P be .
 N
M
Then, if the displacement ( E  vector) due to S1 at P is given by
y1 = a cos t …(1)
Then, the displacement produced by S2 will be
y2 = a cos(t + ) …(2)
The resultant displacement at point P due to S1 and S2 will be Fig. 14.23 Full-wave rectifier circuit
Fig. 14.24 Input and output waveform
y = y1 + y2 in a full-wave rectifier
y = a[cos t + cos(t + )] The circuit connections are made as shown. It contains ac source, step down centre tapped transformer,
   two diodes (D1 and D2) and a load resistance (RL).
y  2a cos   cos  t   …(3)
2  2
During positive half cycle of in put ac, the point A is positive with respect to the point B.
 Then D1 is forward biased and D2 is reverse biased. Hence the current I1 flows through RL from M to N.
From equation (3), the amplitude of the resultant displacement is 2a cos   .
2
During negative half cycle of ac, the point A is negative with respect to the point B. Then D1 is reverse
2
Therefore, the intensity at that point will be I  4 I 0 cos   … (4) biased and D2 is forward biased. Hence the current I2 flows from M to N.
2
Thus the diode allows both positive and negative half cycles of input ac through RL in the same direction
From equation (4), we see that the value of I depends on the phase difference  between y1 and y2.
(from M to N). Hence, the diode acts as full wave rectifier.

(b) Condition for constructive interference

VI. Answer any TWO of the following questions: [2 × 5 = 10]

I is maximum [Fig. (b)], when cos 2    1   = 0,  2,  4, …, Imax = 4I0 42. (a) Calculate the potential at point P due to a charge of 400 nC located 9 cm away. [2]
2
Phase difference,  = 2n, where n = 0, 1, 2, …(5a) (b) Obtain the work done in moving a charge of 2 nC from infinity to the point P. Does the answer
This corresponds to bright points on the screen. The bright points on the screen are called depend on the path along which the charge is moved? [3]
interference maxima. Interference at such points is called constructive interference Solution
Path difference,  = S1P ~ S2P = n where n = 0, 1, 2, …(5b) 1 q
(a) V 
 4 o r
(By using, path difference =  phase difference)
2 9  109  400  109
=
Condition for destructive interference 9  102
 V = 4  104 V
I is minimum when cos 2    0   =  ,  3,  5, …,
2 (b) W = qoV
Imin = 0 = 2  10–9  4  104
Phase difference,  = (2n – 1) where n = 1, 2, 3, …(6a) W = 8  0–5J
This corresponds to dark points on the screen. Interference at such points is called destructive
interference.

Path difference,  = S1 P ~ S 2 P  (2n  1) where, n = 1, 2, 3, …(6b)
2
The dark points on the screen are called interference minima.

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43. In the following network, find the current I3. = 2  3.14 50 1
XL = 314 
1 1
XC = 
c 2fL
1
=
2  3.14  50  90  106
XC = 35.38
Since XL > XC, voltage leads current
Solution X  X C 314  35.38
tan = L 
10  10  R 100
I1 I3 (I2 – I3) tan = 2.78
I 5 II  = tan–1 2.78
I2 (I2 + I3)
1A 45. An equilateral prism is made of glass of unknown refractive index. A parallel beam of light is incident
20  5 on a face of the prism. The angle of minimum deviation is 40º. Find the refractive index of the material
of the prism. If the prism is placed in water of refractive index 1.33, find the new angle of minimum
for Loop I, deviation of a parallel beam of light.
10I1 + 5 I3 – 20I2 = 0
Solution
For loop II,  2I1 + I3 – 4I2 = 0
AD
10 (I1 – I3) – 5 I3 – 5 (I2 + I3) = 0 sin  
n  2 
10I1 – 10I3 – 5 I3 – 5I2 – 5I3 = 0 A
sin  
10I1 – 20I3 – 5I2 = 0 2
2I1 – 4I3 – I2 = 0 … (2)  60  40 
sin  
But I1 + I2 = 1  2 
=
I2 = 1 – I1  60 
sin  
(1)  2I1 + I3 – 4 (1 – I1) = 0  2 
2I1 + I3 – 4 + 4I1= 0 sin 50o
n=
6I1 + I3 = 4 ... (3) sin 30o
(2)  2I1 – I3 – (1 – I1) = 0 0.766
n=
2I1 – 4I3 – 1 + I1 = 0 0.5
3I1 – 4I2 = 1 … (4) n = 1.532
6I1 – I3 = 4 if prism is placed in water then
(3I1 – 4I3 = 1)  2 AD
sin  
 2 
6I1 – I3 = 4  ng 
A
6I1 – 8I3 = 2 sin  
2
9 I3 = 2
 60  D 
2 sin  
I3 = A n g 1.532  2 
9  
n  1.33  
60
I3 = 0.22A sin  
 2 
44. An AC source of frequency 50 Hz is connected in series with an inductor of 1 H, a capacitor of 90 µF
 60  D 
and a resistor of 100 Ω. Does the current leads or lags the voltage? Calculate the phase difference 1.15  sin30 sin   
 2 
between the current and the voltage.
 60  D 
 sin  
Solution  2 
XL = L = 2fL

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 2PBDBEPS(Vijaya Patake) 2PBDBEPS(Vijaya Patake)

60  D BOARD MODEL PAPER - 2 - 2024-25

 sin 1 (0.575)
2
II PUC – PHYSICS - (33)
60 + D = 2 
Time: 3 hours Maximum marks: 70
D = 70 – 60
Instructions:
D = 10 1. All parts (A TO D) are compulsory. PART-E is only for visually challenged students.
2. For Part – A questions, first written-answer will be considered for awarding marks.
3. Answers without relevant diagram / figure / circuit wherever necessary will not carry any marks.
PART - E 4. Direct answers to numerical problems without relevant formula and detailed solutions will not carry any marks
(For Visually Challenged Students only)
7. A circular conducting loop is placed in the plane of the paper to the right of a long straight conductor PART - A
carrying current I in the upward direction. The direction of current induced in the loop is I. Pick the correct option among the four given options for ALL of the following
(A) clockwise if I is increasing (B) clockwise if I is decreasing questions: [15 × 1 = 15]
(C) anti clockwise if I is increasing (D) anti clockwise if I is constant 1
1. ‘The total electric flux through a closed surface in air is equal to times the total charge enclosed by
43. In a Wheatstone bridge, AB = 10 Ω, BC = 10 Ω, CD = 5 Ω and DA = 20 Ω are connected in cyclic order. 0
A galvanometer of 2 Ω is connected between B and D. A current of 1 A enters at A and leaves the that surface’. This is the statement of
network at C. Find the current through the galvanometer. (A) Coulomb’s law in electrostatics (B) Gauss’s law in magnetism
(C) Gauss’s law in electrostatics (D) Ampere’s circuital law
Ans (A)
2. The electric potential due to a negative point charge at a distance ‘r’ is
1 1
(A) positive and it varies as 2 (B) positive and it varies as
r r
1 1
(C) negative and it varies as 2 (D) negative and it varies as
r r
Ans (D)
3. Identify the WRONG statement from the following
(A) The drift speed acquired by free electrons per unit electric field is called mobility.
(B) The conductivity of semiconductors decreases with increase in temperature.
(C) The conductivity of conductors decreases with increase in temperature.
(D) Alloys are widely used in the construction of standard resistors.
Ans (B)
4. The physical quantities related to magnetism are listed in column I and the dimensions are listed in
column II. Identify the correct match
Column I Column II
(i) Magnetic field (a) [MLT2A–2]
(ii) Magnetic permeability (b) [L2A]
(iii) Magnetic moment (c) [M T–2 A–1]
(A) (i) - (b), (ii) - (c), (iii) - (a) (B) (i) - (c), (ii) - (b), (iii) - (a)
(C) (i) - (a), (ii) - (b), (iii) - (c) (D) (i) - (c), (ii) - (a), (iii) - (b)
Ans (D)
5. The ferromagnetic material among the following is
(A) copper (B) nickel (C) lead (D) calcium
Ans (B)

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6. The following are the statements related to self-inductance: 14. The radioactive decay in which very high energy photons are emitted is called __________.
(i) The self-inductance of a coil depends on its geometry and on the permeability of the medium inside it. (A) gamma decay (B) alpha decay (C) negative β decay (D) positive β decay
(ii) The self-inductance is a measure of electrical inertia and opposes the change in current in the coil. Ans (A)
(A) Both the statements are wrong (B) Only statement (i) is correct 15. When a forward bias is applied to a p-n junction, it
(C) Both the statements are correct (D) Only statement (ii) is correct (A) raises the potential barrier. (B) reduces the majority carrier current to zero.
Ans (C) (C) lowers the potential barrier. (D) raises the width of depletion region.
7. In a transformer, NP and NS are the number of turns present in its primary and secondary coils Ans (C)
respectively. The transformer is said to be a step-up transformer if
(A) NP < NS (B) NP > NS (C) NP = NS (D) NP >> NS II. Fill in the blanks by choosing appropriate answer given in the bracket for ALL of
Ans (A) the following questions: [5 × 1 = 5]
8. The expression for displacement current id is (zero, paramagnetic, transverse, ac generator, one, diamagnetic)
d d d E d E 16. The magnetic susceptibility is negative for ____________ materials.
(A) id  02 E (B) id   0 0 E (C) id   0 (D) id  0
dt dt dt dt Ans diamagnetic
Ans (D)
17. The device which works on the principle of electromagnetic induction is _______________.
9. Identify the statement which is true for a compound microscope from the following.
Ans ac generator
(A) Its objective is a convex lens of greater aperture.
(B) Its eyepiece is a convex lens of smaller aperture. 18. The power factor of an AC circuit containing pure resistor is __________.
(C) The image formed by its objective is real and inverted. Ans one
(D) Its eyepiece produces the final image, which is virtual and diminished. 19. The light waves are __________ in nature.
Ans (C) Ans transverse
10. Diffraction effect is exhibited by ________. 20. The charge of a photon is __________.
(A) only sound waves (B) only light waves (C) only matter waves (D) all types of waves
Ans zero
Ans (D)
11. In photoelectric effect experiment if only the frequency of incident radiation is increased, then PART - B
(A) the maximum kinetic energy of photoelectrons decreases. III. Answer any FIVE of the following questions: [5 × 2 = 10]
(B) the stopping potential increases.
21. ‘The charges are additive in nature’. Explain.
(C) the photoelectric current increases.
(D) the photoelectric current decreases. Solution
Ans (B) The total charge of a system is the algebraic sum of all the individual charges contained in the system.
If a system contains charges q1 , q2 , q3 .......qn , the total charge of the system is
12. The impact parameter is minimum in alpha (α) - scattering experiment for the scattering angle of
(A) 180 (B) 0 (C) 120 (D) 90 q  q1  ( q2 )  q3  .....  qn . where q represents the algebraic summation.
Ans (A) 22. What is an equipotential surface? What will be the shape of equipotential surfaces corresponding to a
13. The standing wave pattern of matter waves associated with an electron revolving in a stable orbit is single point charge?
shown in the diagram. The principal quantum number (n) and radius (rn) of the orbit are respectively Solution
4
(A) 8 and An equipotential surface is a surface with a constant value of potential at all points on it.

4 The shape of equipotential surfaces corresponding to a single point charge is a sphere.
(B) 4 and
 23. Give any two differences between current and current density.
2 Solution
(C) 8 and

2 Current Current Density
(D) 4 and
 Current is a scalar. Current density is a vector.
Ans (D) Current is a macroscopic quantity. Current density is a microscopic quantity.

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24. A moving coil galvanometer gives a deflection of 10 divisions when 200 µA of current is passed through PART - C
it. Find the current sensitivity of the galvanometer. IV. Answer any FIVE of the following questions: [5 × 3 = 15]
Solution 29. What is an electric dipole? Define electric dipole moment. Give its direction.
 Solution
SI 
I A system of two equal and opposite point charges separated by a small distance is called electric dipole.
10 The dipole moment of an electric dipole is defined as the product of the magnitude of either charge of a
SI   5 10 4 div A1
200  106 dipole and the separation between them.

25. State Faraday’s law and Lenz’s law of electromagnetic induction. Mathematically, dipole moment p  q  2a pˆ
Solution where p̂ is a unit vector along the dipole axis directed from (q) to (+q).
Faraday’s law: 30. What is a capacitor? Mention any two factors on which the capacitance of a parallel plate capacitor
The magnitude of the induced emf (e) in a conductor at any time instant is equal to the rate of change of depends.
magnetic flux () linked with the conductor at that time instant. Solution
Lenz’s law:
A capacitor is an arrangement having a pair of close lying conductors of any shape separated by a
The polarity of the induced emf in a coil (conductor) is always such that it tends to produce a current dielectric, generally employed for storing charge.
which opposes the change in magnetic flux that produced it. Factors affecting capacitance of a parallel plate air capacitor
26. Name the electromagnetic waves used for the following applications. (i) Area of each plate: C  A for given separation d
(a) The radar systems used in aircraft navigation. 1
(ii) Separation between the plates: C  for given plate area A
(b) The remote switches of household electronic systems such as TV. d

Solution 31. Derive an expression for angular frequency of revolution for a charged particle moving perpendicular to
a uniform magnetic field.
(a) Microwaves
(b) Infrared rays Solution
The period of revolution of the charged particle is
27. How is total energy of an electron revolving in an orbit of hydrogen atom related to the principal
2r mv
quantum number of the orbit? What is the significance of the negative sign in the expression for total T WKT r 
v qB
energy of electron in a hydrogen atom?
 mv 
Solution 2  
qB 
1 Substituting for r, we get T 
En  v
n2 2m
T …(2)
Negative sign indicates that the electron always bound to the nucleus. qB

28. What are intrinsic and extrinsic semiconductors? Angular frequency is given by
2
Solution 
T
Intrinsic semiconductors:
2
The pure semiconductors (impurity less than 1 part in 1010) are called Intrinsic semiconductors. 
 2 m 
Extrinsic semiconductors  qB 
A semiconductor which is doped with impurity atoms is called an extrinsic semiconductor.  
qB

m
32. Mention any three properties of magnetic field lines.
Solution
 The magnetic field lines of a magnet (or a solenoid) form continuous closed loops.
 The tangent to the field line at a given point represents the direction of the net magnetic field at that
point.

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 The magnetic field lines do not intersect. If they intersect, at the point of intersection, the direction of He proposed that when a photon interacts with a free electron of a metal, it imparts its energy (E)
the magnetic field would not be unique. completely to that electron. Part of this energy is utilized by the electron to just come out of the metal
33. A horizontal straight wire 10 m long is falling with a speed of 5.0 m s–1, at right angles to a magnetic surface.
field, 0.30 × 10–4 Wb m–2. Find the instantaneous value of the emf induced in the wire. The minimum energy required to do so is called the photoelectric workfunction (0) of that metal. It is a
Solution characteristic of that metal. The remaining energy appears as the kinetic energy of the photoelectron with
  Blv sin  which it is emitted.
  0.30  104  10  5.0  sin 90  15 104 V Hence, the energy equation for photoelectron with minimum kinetic energy can be written as,
 Energy of an   minimum energy used by an electron   maximum kinetic energy 
34. Derive the relation between radius of curvature and focal length in case of a concave mirror. incident photon    to come out of the metal    of the photoelectron 
     
Solution
 (h)   (0  workfunction)   ( K max ) 
Consider a concave mirror of focal length f and of radius of curvature R. Let P represent its pole, F
Thus, h  f 0  (Kinetic energy)max
represent its principal focus and C its centre of curvature. A ray of light parallel to the principal axis of E h
the concave mirror (of small aperture) is incident at M. The ray is reflected along MF. m 2  2 …(1)
c c
h h h
where     mc  ...(2)
AM – incident ray p mc 
MF – reflected ray Work functions also denoted by W.
C – Centre of curvature This is called Einstein’s photoelectric equation.
F – Principal focus
36. Define ‘binding energy’ and ‘mass defect’. Write the relation between them.
P – Pole
Solution
CM – Normal to mirror at M.
The difference between the sum of the masses of the constituent particles and the actual mass of a
Reflection from a concave mirror nucleus is called mass defect.
Thus, mass defect m  [ Zmp  ( A  Z )mn ]  M
From the law of reflection, angle of reflection = angle of incidence.
 CMF  AMC   The binding energy of a nucleus is defined as the energy required to be provided from external agency to
From the above figure MCF  AMC   (alternate angles) separate the nucleons to an infinite distance apart from the nucleus, so that they may not interact with
Also, MFP  CMF  MCF (Exterior angle = sum of the interior opposite angles) one another or energy with which nucleons are bound in the nucleus.
The binding energy Eb is related to mass defect (M) by Einstein’s mass-energy relation (E = mc2)
As the aperture is small and angle  is small, arc PM may be considered as a straight segment.
Eb (in MeV) = M (in MeV/c2)c2
PC = R, PF = f (by sign convention)
  tan   
From le PMC, tan   PM    PM  lim    1  tan   ,  is in radian  . PART - D
PC R  0    
V. Answer any THREE of the following questions: [3 × 5 = 15]
PM PM
From  PMF, tan 2 
le
 2  ,
PF f 37. (a) Obtain an expression for potential energy of an electric dipole placed in a uniform electric field. [3]
 PM  (b) Define energy density of a charged capacitor. How is the energy density related to electric field
  R
 R present between the plates of capacitor? [2]
Hence,   or f 
2  PM  2 Solution
   
 f  (a) Consider an electric dipole of dipole moment p held in a uniform electric field E as shown in the
 
35. Give Einstein’s explanation of photoelectric effect and write Einstein’s photoelectric equation. figure. Let  be the angle between p and E . F1
Net force on the dipole is zero.  +q
Solution     E 2a  B
To account for the experimental observations of photoelectric emission, Albert Einstein, (1905), made a The dipole experiences a torque  due to the field given by    p  E . 

simple but remarkable assumption, based on Planck’s quantum theory of radiation. Einstein assumed that This torque tends to rotate the dipole. –q
C
The magnitude of torque is given by  = pE sin. …(1) A
a beam of light is composed of discrete packets of energy called quanta of energy of radiation or  F2
photons. Each photon has energy h, where h = Planck’s constant, and  = frequency of radiation. Let an external torque ext be applied on the dipole such that it is equal and

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 E1  V E2  V
opposite to  . I 
As a result, the angular speed will be negligible and the rotation takes place without angular r1 r2
acceleration. E E  1 1
 I   1  2  V   
Work done by ext in rotating the dipole through a small angle d is  r1 r2   r1 r2 
dW   d  pE sin  d . …(2) E r  E2 r1 r r 
I  1 2 V  1 2 
Total work done in rotating the dipole from the orientation 1 to 2 is given by r1 r 2  r1 r2 
2
E1 r2  E2 r1  rr 
W   dW   pE sin  d   pE (cos  2  cos 1 ) …(3) V  I  1 2  …(2)
1 r1  r 2  r1  r2 
The work done is equal to the increase in the potential energy of the system. If Eeq and req are the equivalent emf and internal resistance of parallel combination of cells,
U2  U 1  W then
 U2  U1   pE  cos 2  cos 1  …(4) V = Eeq  I req ...(3)
On comparing eq. (2) and eq. (3) we get,
To find the potential energy in any position, a reference level of zero potential energy is required.
E r  E2 r1
For convenience, Eeq  1 2 …(4)
r1  r2
we take U = 0 at the position 1 = 90 (where cos 90 = 0).
rr
  req  1 2 …(5)
Then U ( )  U (90 )   pE  cos   cos    pE cos r1  r2
 2
Here, U (90)  0 and U ( )  U (say) 39. Derive an expression for force per unit length on two infinitely long thin parallel straight conductors
Potential energy of an electric dipole in an external electric field is U   pE cos …(5) carrying currents and hence define ‘ampere’.
  Solution
Eq. (5) can also be written as U   p  E …(6)
(b) The energy stored per unit volume in a region of space where electric field exists is called energy
density of the electric field.
1
u  ε0 E 2
2
u  E2
38. Derive an expression for effective emf and effective internal resistance of two cells of different emfs and
internal resistances connected in parallel.
Solution
E1, r1
Two long straight parallel conductors carrying steady currents Ia and Ib and separated
I1 by a distance d. Ba is the magnetic field set up by conductor ‘a’ at conductor ‘b’
I1
(Eeq, req)
Consider two very long straight conductors ‘a’ and ‘b’ placed parallel to each other in free space.
A 
I I C A I I C Let d = separation between the conductors
I2 I2
V Ia and Ib = currents in a and b respectively.
E2, r2 
The current Ia produces a magnetic field Ba at all points on the conductor ‘b.’
V
 I
Ba  0 a …(1)
Consider two cells of emfs E1 and E2 having internal resistance r1 and r2 respectively, connected in 2d

parallel between A and C. (as shown in the figure). According to right-hand clasp rule, the field Ba is directed normal to the length of the conductor ‘b.’

Let I1 and I2 be the currents drawn from the individual cells. (in figure Ba acts downwards)
E V E V 
I1  1 I2  2 The field Ba exerts magnetic force on the conductor ‘b’ carrying a current Ib.
r1 r2
The magnetic force on a segment of length L of conductor ‘b’ is given by the relation
Total current drawn from the grouping,
Fba = IbLBa …(2)
I = I1 + I2 …(1)
Substituting for Ba from eq. (1) in eq. (2), we get

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0 I a 41. What is a half-wave rectifier? Explain the working of a half-wave rectifier using neat circuit diagram.
Fba  I b L 
2d Also draw input-output waveforms corresponding to it.
 0 I a Ib 
Fba   L …(3) Solution
 2d 
 It is a device which converts only half cycles of AC into DC
The direction of the force Fba acting on the conductor ‘b’ can be determined by Fleming’s left hand rule. M
 1
Fba is directed towards the conductor ‘a’.
It can be shown that the force on conductor ‘a’ due to the current Ib in conductor ‘b’ is
 I I  N
(i) Fab   0 a b  L (in magnitude) …(4) and
 2d  Fig. 14.21 Half-wave rectifier circuit

(ii) directed towards the conductor ‘b’

The forces constitute action reaction pair obeying Newton’s third law.
 
Fba   Fab
The force per unit length of each conductor is given by
 I I
f ba  f ab  0 a b (in magnitude) …(5)
2d
Consider the expression f  0 a b  I I
2d
In SI, 0 = 4  107 H m1
Let d = 1 m and Ia = Ib = 1 A
7
Then f  4  10  1  1  2  107 N m1 Fig. 14.22 Input and output waveforms in a half-wave rectifier
2  1
The circuit connections are made as shown.
If two infinitely long thin straight conductors carrying equal currents and placed parallel to each other at
It contains ac source, step down transformer (T), diode (D) and a load resistance (RL).
separation of one meter in free space exert a force 2  107 N m1 on each other, then the current in each
During positive half cycle of input ac, the point A is positive with respect to the point B. Then the diode
is said to be one ampere.
is forward biased. Hence the current flows through RL from M to N.
40. (a) What is a wavefront? [1] During negative half cycle of input ac, the point A is negative with respect to the point B. Then the diode
(b) Explain the refraction of a parallel plane wave through a thin prism with a neat diagram. [2] is reverse biased. Hence current does not flow through RL.
(c) Give any two differences between constructive and destructive interferences of light. [2] The diode allows only positive half cycles of ac through RL from M to N.
Solution Hence it acts as half wave rectifier.
(a) The locus of all such points which have the same amplitude, vibrate in the same phase is called a
wavefront. VI. Answer any TWO of the following questions: [2 × 5 = 10]
42. Two points charges of +4 nC and +8 nC are placed at the points A and B respectively separated by a
(b)
distance 0.2 m in air. Find the magnitude of the resultant electric field at the midpoint ‘O’ of the line
joining A and B. What will be the magnitude of resultant electric field at ‘O’ if +4 nC is replaced by
another +8 nC charge?
(c) Solution
Constructive Interference Destructive Interference 1 qA
EA =
Bright fringe is obtained. Dark fringe is obtained. 40 AD 2
Phase difference is 2n ( n  0,1, 2, 3,...) Phase difference is (2n  1) ( n  1, 2,3,...) 4  109
= 9  109 
  (0.1)2
Path difference is 2n (n  0,1, 2, 3,...) Path difference is (2n  1) (n  1, 2,3,...)
2 2 EA = 3600 NC–1
1 qB
EB =
40 BD 2

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9
8  10 = 402  (1570  159.23) 2
= 9  109 
(0.1) 2
= 1600  1990271.99
EB = 7200NC–1 = 1991871.99
 Resultant electric field is
Z = 1411.33
E = EB – EA 1
= 7200 – 3600 fo =
2 LC
E = 3600 NC–1 1
If + 4nC is replaced by + 8nC, then net field at mid point is zero since charges are identical. =
23.14 5  20  106
43. Find the currents I1 and I2 in the given electrical network. fo = 15.92 Hz
45. An object is placed at a distance 0.3 m from a convex lens of focal length 0.2 m. Find the position and
nature of the image formed. Also find the distance through which the object should be moved to get an
image of linear magnification ‘1’.
Solution
Solution 1 1 1
 
I3 = I1 + I2 v u f
For loop 1, 1 1 1
 
6I1 + 10 (I1 + I2) + 4I2 + 1I2 = 2 v 0.3 0.2
1 1 1
10I1 + 21I2 = 2 … (2)  
22 10 v 0.2 0.3
  362 1 3 2 1
10 21  
v 0.6 0.6
4 10
1   64 V = + 0.6 m
2 21
Since, v is positive image is real and inverted magnification becomes 1 only when u = 2f
22 4  object has to be move 0.1 m away from the ……..to get magnification –1.
2  4
10 2
 64
I1 = 1   0.176A PART - E
 362
(For Visually Challenged Students only)
2 4
I2 =   0.011A
 362 13. The standing wave pattern of matter waves associated with an electron revolving in a stable orbit is
44. A series LCR circuit contains a pure inductor of inductance 5 H, a capacitor of capacitance 20 µF and containing 4 complete waves. The principal quantum number (n) and radius (rn) of the orbit are
resistor of resistance 40 Ω. If the AC source of 200 V, 50 Hz is present in the circuit, find the impedance. respectively
4 4 2 2
Also find the resonant frequency of the circuit. (A) 8 and (B) 4 and (C) 8 and (D) 4 and
   
Solution
XL = 2fL 43. The positive terminals of two cells of emfs 4 V and 2 V with internal resistances 2 Ω and 1 Ω are
= 2  3.14  50  5 connected by a uniform wire of resistance 12 Ω. Their negative terminals are connected by a second
XL = 1570  uniform wire of resistance 8 Ω. The mid points of these two wires are connected by a third uniform wire
1 of resistance 10 Ω. Find the current through 4 V cell.
XC =
2fc
1
=
2  3.14  50  20  106
XC = 159.23
R 2   X L  XC 
2
Z=

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BOARD MODEL PAPER - 3 - 2024-25 6. Below are the two statements related to magnetic flux and magnetic field lines.
II PUC – PHYSICS - (33) Statement-I: The net magnetic flux through any closed surface is zero.
Time: 3 hours Maximum marks: 70 Statement-II: The number of magnetic field lines leaving the surface is balanced by the number of lines
Instructions: entering it.
1. All PARTS (A to D) are compulsory. PART-E is only for visually challenged students. (A) Both the statements I and II are correct and II is the correct explanation for I.
2. For PART – A questions, first written-answer will be considered for awarding marks. (B) Both the statements I and II are correct and II is not the correct explanation for I.
3. Answers without relevant diagram / figure / circuit wherever necessary will not carry any marks. (C) Statement I is wrong but the statement II is correct.
4. Direct answers to numerical problems without relevant formula and detailed solution will not carry any marks. (D) Statement I is correct but the statement II is wrong.
Ans (A)
PART - A 7. The polarity of induced emf in a coil is given by _________
I. Pick the correct option among the four given options for ALL of the following (A) Lenz’s law (B) Faraday’s law
questions: [15 × 1 = 15] (C) Gauss’s law in magnetism (D) Ampere’s circuital law
1. The SI unit of surface charge density is __________. Ans (A)
(A) C m–1 (B) C m–2 (C) C m–3 (D) kg m–3 8. In a transformer, the windings of the primary and secondary coils are wound one over the other to reduce
Ans (B) the energy loss due to _________
2. The values of electric field (E) and electric potential (V) at any point on the equatorial plane of an (A) flux leakage (B) resistance of the windings
(C) eddy currents (D) hysteresis
electric dipole are such that
Ans (A)
(A) E = 0, V = 0 (B) E = 0, V ≠ 0 (C) E ≠ 0, V = 0 (D) E ≠ 0, V ≠ 0
Ans (C) 9. The electromagnetic waves suitable for RADAR systems used in aircraft navigation are
(A) Gamma rays (B) Ultraviolet rays (C) Microwaves (D) Infrared waves
3. If the potential difference across a capacitor is doubled, then the energy stored in it
Ans (C)
(A) is doubled (B) is quadrupled (C) is halved (D) remains same
Ans (B) 10. A ray of light is incident on glass-air interface at an angle greater than the critical angle for the pair of
media. Then the ray undergoes
4. A wire has a non-uniform cross-sectional area as shown in the figure. A steady current I flows through it.
(A) refraction only
Which one of the following statements is correct?
(A) The drift speed of electron is constant. (B) partial reflection and partial refraction
(C) total internal reflection
(B) The drift speed of electron increases while moving from A to B.
(D) grazes the surface at the interface of the two media.
(C) The drift speed of electron decreases while moving from A to B.
(D) The drift speed of electron varies randomly. Ans (C)
Ans (C) 11. To observe sustained interference pattern on a screen placed at a suitable distance in Young’s double slit
experiment, which of the following condition/s is/are necessary?
5. A charged particle of charge q is moving in a uniform magnetic field. The angle between the velocity(v)
of the charged particle and magnetic field(B) is . The trajectory of the charged particle varies with angle (a) Sources of light should be coherent.
(b) Sources of light should be narrow.
. Match the following table by choosing the appropriate trajectory traced by the charged particle for
(c) Sources of light should be very close.
different possible values of angle .
(A) only (a) (B) both (a) and (b) (C) both (b) and (c) (D) all (a), (b) and (c)
Angle Trajectory
Ans (D)
(i)  = 0 (a) Circle
(ii)  = 45 (b) Straight line 12. The de Broglie wavelength of a moving particle is independent of __________ of the particle.
(iii)  = 90 (c) helix (A) charge (B) mass (C) speed (D) momentum
(A) (i) – (a), (ii) – (b), (iii) – (c) (B) (i) – (b), (ii) – (c), (iii) – (a) Ans (A)
(C) (i) – (b), (ii) – (a), (iii) – (c) (D) (i) – (c), (ii) – (b), (iii) – (a)
Ans (B)

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13. For an electron revolving around the nucleus, 23. Define “magnetisation of a sample”. How is it related to magnetic intensity?
(A) kinetic energy and potential energy are positive, total energy is negative. Solution
(B) kinetic energy is positive, potential energy and total energy are negative. The net magnetic moment per unit volume developed in a material is called magnetisation.
(C) potential energy is negative, kinetic energy and total energy are positive. Magnetisation is directly proportional to magnetic intensity.
(D) kinetic energy and potential energy are negative, total energy is positive. 24. A boy peddles a stationary bicycle. The pedals of the bicycle are attached to a coil of 100 turns, each turn
Ans (B)
of area 0.20 m2. The coil rotates at 6 rotations per second and it is placed in a uniform magnetic field of
14. The ratio of nuclear densities of 13Al27 and 29Cu64 is 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum value of emf generated in
(A) 1 : 1 (B) 3 : 4 (C) 13 : 29 (D) 27 : 64 the coil.
Ans (A) Solution
15. The energy band gap in conductor, insulator and semiconductor are respectively E1, E2 and E3.  o  NBA
The relation between them is  o  100  0.01 0.20  2  3.14  6
(A) E1 = E2 = E3 (B) E1 < E2 < E3 (C) E1 > E2 > E3 (D) E1 < E3 < E2  o  7.536 V
Ans (D)
25. What is displacement current? Give the expression for it.
II. Fill in the blanks by choosing appropriate answer given in the bracket for ALL of Solution
the following questions: [5 × 1 = 5] The current due to a time varying electric field is called displacement current.
(mutual induction, inductance, diffraction, magnification, quantisation, interference)  d 
The displacement is given by I d   0  E 
16. One of the basic properties of electric charge is ______________.  dt 
Ans quantisation 26. Mention two uses of polaroids.
17. The ratio of the magnetic flux-linkage to the current in a coil is called ______________. Solution
Ans inductance  in sun glasses, window panes of aeroplanes, cameras, microscope objectives to eliminate glare of
18. The principle of working of a transformer is ______________. reflected light.
Ans mutual inductance  to view 3-D pictures and movies.

19. A microscope is used to produce large _____________ of small objects. 27. Write two limitations of Bohr’s atom model.
Ans magnification Solution
20. The phenomenon of bending of light around the edges of an obstacle is called ____________. 1. The theory can be applied only to hydrogen and hydrogen-like ions.
Ans diffraction 2. The theory is inadequate to explain fine structure of spectral lines.
3. The intensities of different lines and selection rules governing transitions cannot be explained from
PART – B this theory.
III. Answer any FIVE of the following questions: [5 × 2 = 10] 4. The model does not provide theoretical basis for the existence of stationary orbits.
21. Name the two factors on which the resistance of a metallic wire depends. 28. How can a semiconductor diode be forward biased? What happens to the width of the depletion region
Solution when forward bias voltage is increased?
1. Length of the wire Solution
2. Area of cross section The diode is said to be forward biased, when p-side is connected to the positive terminal and n-side to
3. Temperature of the wire the negative terminal of the battery.
4. Material of the wire The width of the depletion region decreases when forward bias voltage is increased.
22. When does a current carrying conductor placed in a uniform magnetic field experience (i) maximum
force and (ii) minimum force?
Solution
Magnetic force Fm = Bqv sin 
 When  = 0 or 180, Fm = Fmin = 0
 When  = 90, sin 90 = 1. Then, Fm  Fmax  qvB

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PART - C Explanation 

IV. Answer any FIVE of the following questions: [5 × 3 = 15]  Consider a current element MN  I dl of a conductor carrying current I.

29. Derive the expression for the torque on an electric dipole placed in a uniform electric field. Let r be the position vector of the point P from the current element and  be the angle between
 
dl and r .
Solution
 
Explanation: Consider a permanent electric dipole of dipole moment p held   According to Biot-Savart law, the magnitude of the magnetic field dB at P is
F
   I dl sin 
in a uniform electric field E such that p makes an angle  with E as shown  +q dB 
E B r2
in figure. 2a  
p I dl sin 
The positive charge (q) and the negative charge (q) experience forces of same dB  K
–q
A C
r2
magnitude but in opposite directions. 
F K  constant of proportionality.
These forces are written as
    
F  qE (along the direction of E ) and In SI system and for free space, K   0   107 henry (meter)–1
    4 
F   qE (opposite to the direction of E )
0 = 4   10–7 H m–1 is the absolute permeability of free space
The net force on the dipole system is zero.    I dl sin 
dB   0 
Hence there is no translation of the dipole.
   4  r2
The force F and F constitute a couple. This couple tends to rotate the dipole until its axis is along the
 32. Write the three differences between diamagnetic and ferromagnetic materials.
direction of E .
Moment of the couple also called torque is given by Solution
 = magnitude of either force  arm of the couple Diamagnetic substances Ferromagnetic substances
= F  2a sin  (1) Magnetic susceptibility () is negative and Magnetic susceptibility () is positive and
 = qE  2a sin  small large
= (q  2a) E sin  (2) These cannot be magnetised when placed in These gets strongly magnetised when placed in
The product (q  2a) represents dipole moment (p) of the dipole. an external magnetic field an external magnetic field
  pE sin  ...(1)
   (3) These are repelled by a magnet These are attracted by a magnet
In vector form,   p  E ...(2) (4) The relative permeability is always less than 1
The relative permeability is very large and
  
 acts through the point O perpendicular to both p and E obeying the rules of cross product of vectors. more than 1
30. Give three results of electrostatics of conductors. (5) Magnetic susceptibility is independent of Magnetic susceptibility depends on
Solution temperature temperature
1. Electrostatic field is zero everywhere inside a conductor. (6) Magnetic field lines are expelled out when a Magnetic field lines flow inside when a
2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every diamagnetic substance placed in an external ferromagnetic substance placed in an external
point. magnetic field magnetic field.
3. The interior of a conductor cannot have excess charge in the static situation. 33. Derive the expression for motional emf induced in a straight conductor moving perpendicular to uniform
4. Electrostatic potential is constant throughout the volume of the conductor and has the same value magnetic field.
(as inside) on its surface.
Solution
31. State and explain Biot-Savart’s law with a suitable diagram.  M
B I
Q P
Solution
Statement 
 l v

The magnetic field ( dB ) at a point due to a current element  I dl  is
R S
 directly proportional to the current through the current element. N
 directly proportional to the length (dl) of the current element. x

 directly proportional to the sine of the angle () between dl and r , and Consider a rectangular frictionless metal frame PQRS placed in a region of uniform and steady magnetic

 inversely proportional to the square of the distance between the point and the field B , directed normally into the plane of diagram.
current element.

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 2PBDBEPS(Vijaya Patake) 2PBDBEPS(Vijaya Patake)


Let a thin straight conductor x of length l be moved on the frame with a constant velocity v as shown in EAC and BAC are congruent
the figure.  i = r
At a certain time instant t. Let the conductor be at the position MN, at a distance x from the end QR of This is the law of reflection.
the frame. As the conductor moves, the area of the loop MNRQ changes with time t. The magnetic flux 36. Write the three features of nuclear force.
B through this area changes continuously. Hence, an emf is induced in the conductor. This emf is called Solution
motional emf. 1. Nuclear force is the strongest force in nature.
The magnetic flux, B enclosed by the loop MNRQ is B = BA cos  2. uclear force is a short-range force.
Since area, A = lx, B = Blx (  = 0 and cos 0 = 1) …(1) 3. Nuclear force is saturated force.
d d
e   B   ( Blx) 4. The nuclear force is charge independent.
dt dt
5. The nuclear force is spin dependent.
  dx 
e  Bl    Blv …(2) 6. The nuclear forces are non-central forces.
 dt 
dx
Here, v   , velocity of the conductor. The negative sign indicates that x decreases with time t. PART - D
dt
V. Answer any THREE of the following questions: [3 × 5 = 15]
34. A small candle is placed at a distance of 20 cm in front of a concave mirror of radius of curvature 30 cm.
At what distance from the mirror should a screen be placed in order to obtain a sharp image? What is the 37. State Gauss’s law in electrostatics. Derive an expression for the electric field at a point due to an
nature of the image? infinitely long thin uniformly charged straight wire using Gauss's law.
Solution Solution
u  20 cm, R  30 cm, f  15 cm 1
The total electric flux through a closed surface in an electric field is equal to times the total charge
0
1 1 1
  enclosed by it.
v u f
1
1 1 1   qi
  0
v f u
1 1 1 Electric field due to an infinitely long straight uniformly charged wire:
  Consider an infinitely long thin straight wire charged uniformly with a linear charge density  > 0
v 15 (20)
( = charge per unit length).
v  60 cm
Consider a point P at a distance r from the wire. Imagine a closed cylindrical surface of radius r and
Since v is negative, image is real and inverted.
height l as the Gaussian surface enclosing a portion of charged wire as shown in the figure.
35. Using Huygen’s principle, show that the angle of reflection is equal to the angle of incidence when a By symmetry, the electric field near the wire should be radially outwards and perpendicular to the wire.
plane wavefront is reflected by a plane surface. 
Let E be the electric field at P.
Solution The electric flux across the curved surface of the Gaussian cylinder is
Consider a plane wave AB incident at an angle i on a 1  E (2r l ).
reflecting surface XY. If v represents the speed of the Flux through the circular end faces of the Gaussian cylinder is

wave in the medium and if t represents the time taken by 2  0 [since  = 90, E is parallel to the end faces].
the wavefront from the point B to C, then the distance Total flux across the closed surface is

BC = vt. In order to construct the reflected wavefront, we   1  2  E (2rl )  0  E (2rl ) …(1) dS
draw a sphere of radius vt from the point A as shown in The total charge enclosed by the cylindrical surface is
fig. Let CE represent the tangent plane drawn from the q = l
point C to this sphere. q
Gauss's law:   . …(2)
Here, 0
E  B  90 each
Comparing eq. (1) and eq. (2), we get
AC = AC (common side) l
E  2r  l 
AE = BC = vt 0

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
E (In magnitude) …(3)
20 r
  
In vector form E  n
2 0 r
where n is the radial unit vector in the plane normal to the wire passing through the point P.
38. Arrive at the balance condition of Wheatstone bridge using Kirchhoff’s rules.
Solution
Wheatstone’s network is a mathematical model designed by Wheatstone,
B
for comparison of resistances. (I1  Ig)
P
A Wheatstone’s network consists of four resistances P, Q, S and R Q
I1 Ig
connected in a cyclic order to form a loop. A G C The moving coil galvanometer. Its elements are described in the text. Depending
A cell and a plug key are connected between the junctions A and C and a I2
on the requirement, this device can be used as a current detector or for measuring
the value of the current (ammeter) or voltage (voltmeter)
galvanometer is connected between B and D. Let G be the galvanometer S
R (I2 + Ig) The deflection of the coil is opposed by the restoring torque set up in the springs which control the
resistance. D
movement of the coil. In equilibrium, the two torques balance each other. The coil undergoes a steady
angular deflection (). The deflection of the coil () is directly proportional to the current (I) passed in
Condition for balance of a Wheatstone network
the coil. i.e.,   I.
Let the currents be marked as shown in figure. V
Theory: When a current I is passed through the coil of the galvanometer, the torque experienced by the
Applying KVL to the loop ABDA and BCDB we get coil is given by   NBIA …(1)
 I1 P  I g G  I 2 R  0 ...(1)
The restoring torque provided by the springs brings it to equilibrium at an angle  with respect to the
  I1  I g  Q   I 2  I g  S  I g G  0 ...(2) initial position.
The network is said to be balanced when the current through the galvanometer is zero ( I g  0) . Therefore, restoring torque = k …(2)
Then, eq. (1) and eq. (2) take the form where k is the torsional constant of the spring called couple per unit twist of the spring.
I1 P  I 2 R ...(3) From eq., (1) and eq., (2), we get NBIA  k 
I1 Q  I 2 S ...(4)  k 
I  
P R  NBA 
Dividing eq.(3) by eq. (4), we get  ...(5)
Q S  NBA 
 I
 k 
39. What is the principle behind the working of a moving coil galvanometer? With the help of a neat labelled
 NBA 
diagram, obtain the expression for the angular deflection produced in moving coil galvanometer. For a given galvanometer,   is a constant    I
 k 
Solution 40. Derive the expression for refractive index of the material of the prism in terms of angle of minimum
A moving coil galvanometer is a device used for detection and measurement of small currents. It is based deviation and angle of the prism.
on the principle that a current carrying coil placed in a uniform magnetic field experiences a torque, such Solution
that current is directly proportional to the deflection of the coil. ABC is the principal section of a prism made of a material of
Description refractive index n2 placed in a medium of refractive index n1. A
A pointer galvanometer consists of a coil of large number of turns of insulated copper wire wound over a ray of monochromatic light IO incident on the refracting face AB,
metal frame and placed symmetrically between curved pole pieces of a horse-shoe magnet. refracts along OR and emerges along RE, as in the Fig. Let i1 and
Working r1 be the angles of incidence and refraction respectively at AB. Let
A current carrying coil, free to rotate about a fixed axis is placed in a region of a uniform and radial r2 and i2 be the angles of incidence and emergence respectively at
magnetic field. The magnetic field exerts torque on the coil. As a result the coil undergoes deflection. R on AC. OM and RM are the normals at O and R to AB and AC Fig. Refraction of monochromatic light
through a prism
respectively.
ˆ ˆ
In the cyclic quadrilateral AOMR, OAR  OMR  180 ...(1)
ˆ  180
In the triangle ORM, r1  r2  OMR ...(2)

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From Eqs., (1) and (2), A = r1 + r2 ...(3) n type semiconductor

The angle  between the direction of incidence IOX and the direction of emergence TRE, is the angle of
deviation. From Fig. we have
Deviation produced at O, on AB, 1  i1  r1
Deviation produced at R, on AC,  2  i2  r2
Total deviation produced,   1  2  i1  r1  i2  r2  i1  i2  (r1  r2 )  i1  i2  A [ A  r1  r2 ]
Therefore, A    i1  i2 …(4)
(If i1 = i, i2 = e, the deviation  = i + e  A)
Angle of minimum deviation
A graph of  versus i is as shown in the figure. It is found that as the angle of incidence i is increased
from a small value, the angle of deviation  first decreases to a minimum value and then increases. The p type semiconductor
smallest angle of deviation of a ray passing through a prism is called the angle of minimum deviation
(Dm).
From the graph, it is clear that, for any particular angle of deviation, two angles of incidence are
possible.
When the prism is in the minimum deviation position, a ray of light passes through the prism
symmetrically and the refracted ray is parallel to the base.
When,  = Dm,
i1 = i2 = I, and r1 = r2 = r (from graph).
From the Eqs., (3) and (4), we get
A
A = 2r  r  and VI. Answer any TWO of the following questions: [2 × 5 = 10]
2
A  Dm 42. Three capacitors of capacitances 2 F, 3 F and 6 F are connected in series.
A + Dm = 2i  i 
2 (a) Determine the effective capacitance of the combination.
n2 sin i (b) Find the potential difference across 6 F capacitor if the combination is connected to a 60 V supply.
By Snell’s law, n21   ,
n1 sin r [2 + 3]
Fig. Graph of  versus i
 A  Dm  Solution
sin  
 2  1 1 1 1
n21    
 A Cs C1 C 2 C3
sin  
2 1 1 1
=  
41. (a) Write three differences between intrinsic semiconductor and extrinsic semiconductor. [3] 2 3 6
3  2 1
(b) Draw the energy band diagrams of =
6
(i) n-type and
Cs = 1F
(ii) p-type semiconductors at temperature T > 0 K [2]
Q = CsV = 1  10–6  60 = 60 C
Solution Q 60  106
V 
Intrinsic Semiconductor Extrinsic Semiconductor C 6  106
1. These are the pure semiconductors 1. These are the impure semiconductors. V = 10V
2. Electron density is equal to the hole density. 2. Electron density is not equal to hole density. 43. For copper, the number density of free electrons is 8.5  1028 m–3 and resistivity is 1.7  10–8 Ω m.
3. Conductivity is low. 3. Conductivity is high. Calculate the conductivity of copper and relaxation time of free electrons in copper. Take the mass of
4. Conductivity depends on temperature. 4. Conductivity depends on temperature and also electron = 9.1 10–31 kg and e = 1.6  10–19 C.
Example: Germanium on doping level. Solution
Example: doped Germanium 1 1
   5.88  107 Sm 1
 1.7  108

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ne 2
hc  1 1 
 V1  V2    
m e  1  2 
m
 2 h  3  108  1 1 
ne 0.96 – 0.34 =   
1.6  1019  400  109 500  109 
5.88  107  9.1  1031
= 0.62  1.6  109
8.5  1028  (1.6  1019 )2 = h  107 (0.25 – 0.2)
3  108
 = 2.45  10–14 s 0.62  1.6  1019
h=
44. A resistor of 50 Ω, a pure inductor of 250 mH and a capacitor are in series in a circuit containing an AC 3  108  0.05  107
source of 220 V, 50 Hz. In the circuit, current leads the voltage by 60. Find the capacitance of the h = 6.613  10–34 Js
capacitor.
Solution PART - E
XL = 2fL (FOR VISUALLY CHALLENGED STUDENTS ONLY)
= 2  3.14  50  250  10–3 4. A wire has a non-uniform cross-sectional area in which end A of the wire has smaller area than that of
XL = 78.5 end B. A steady current I flows through it. Which one of the following statements is correct?
Since current leads voltage (A) The drift speed of electron is constant.
X  XL (B) The drift speed of electron increases while moving from A to B.
tan = C
R (C) The drift speed of electron decreases while moving from A to B.
XC – XL = R tan (D) The drift speed of electron varies randomly.
XC – 78.5 = 50  tan60
XC
XC
1
XC 
2fc
1
C
2f X C
1
=
2  3.14  50  121.8
C = 0.0000261
C = 26.1  10–6 F
45. When light of wavelength 400 nm is incident on a photosensitive surface, the stopping potential for the
photoelectrons emitted is found to be 0.96 V. When light of wavelength 500 nm is incident on the same
photosensitive surface, the stopping potential is found to be 0.34 V. Calculate the Planck’s constant.
Given: speed of light in vacuum is 3  108 m s–1 and e = 1.6  10–19 C.
Solution
hc
eVs = 

hc o
Vs  
e e
hc o
V1   … (1)
1e e
hc o
V2   … (2)
 2e e

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