Top Chapter of Physics

Chapter Qs in 2021-24 Per shift Qs

Modern Physics 275 3.3

Current Electricity 189 2.3
Electrostatics 122 1.5
Magnetic Effects of Current 120 1.4
Alternating Current 112 1.3
EMI 77 0.9
Capacitance 76 0.9
Magnetism and Matter 38 0.5
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 Modern Physics

Atomic Structure
Complete JEE in One Shot
 Bohr’s Model
nh Shell 5
mvr = Shell 4
2π Shell 3
Shell 2
kq1 q2 mv 2 Shell 1
2
=
r r Nucleus +
n2 Shell K
rn = 0.529 Å Shell L
Z
Shell M
Shell N
Z
v = 2.2 × 106 m/s Shell O
n
Q) The radius of third stationary orbit of electron for Bohr′s atom is R.
The radius of fourth stationary orbit will be:
𝟒 𝟏𝟔 𝟑 𝟗
𝐀 𝐑 𝐁 𝐑 𝐂 𝐑 (𝐃) 𝐑
𝟑 𝟗 𝟒 𝟏𝟔
JEE MAIN 2024 PYQ
Ans. (B)
𝐄𝐧𝐞𝐫𝐠𝐲 𝐨𝐟 𝐄𝐥𝐞𝐜𝐭𝐫𝐨𝐧

Z2
E = −Rhc 2
n
R is ′Rydberg Constant ′ ≈ 1.1 x 107 m–1

𝐙𝟐
𝐄 = −𝟏𝟑. 𝟔 × 𝟐 eV
𝐧
𝐄𝐧𝐞𝐫𝐠𝐲 𝐨𝐟 𝐄𝐥𝐞𝐜𝐭𝐫𝐨𝐧
𝟏
𝐊𝐄 = 𝐦𝐯 𝟐
𝟐
−𝐊𝐙𝐞𝟐
𝐏𝐄 =
𝐫
TE = KE + PE
𝐏𝐄
TE = − KE =
𝟐
Q) An electron rotates in a circle around a nucleus having positive charge Ze.
Correct relation between total energy ( E ) of electron to its potential
energy ( U ) is:
(A) E = 2U (B) 2E = 3U (C) E = U (D) 2E = U
JEE MAIN PYQ
Ans. (D)
 Z2 n = 4, 3rd excited state
E = −13.6 × 2 eV −0.85 eV
n
n = 3, 2nd excited state
−1.51 eV

n = 2, 1st excited state

−3.4 eV

n = 1, ground state
−13.6 eV
ΔE = Efinal state – Einitial state
n = 4, 3rd excited state
−0.85 eV

1 1 2 n = 3, 2nd excited state

ΔE = 13.6 2 − 2 Z (in eV) −1.51 eV
n1 n2
n = 2, 1st excited state
−3.4 eV

n = 1, ground state
−13.6 eV
 𝟏 𝟏 𝟏 𝟐
ΔE = hv =𝐑 𝟐− 𝟐 𝐙
𝛌 𝐧𝟏 𝐧𝟐

hc
ΔE =
λ
Hydrogen Line Spectrum Le Balm Pasta Bread Fund
7 Q
6 P
Far I.R. region
Humphery series
5 O
I.R. region
0.31 eV
Pfund series
4 N
I.R. region 0.66 eV
Brackett series
Energy 3 M
Levels Infra Red region
or 1.89 eV
Paschen series
L
2
Visible region
or 10.2 eV
Balmer series
1 K
Ultra violet region
or
Lyman series
Q) If the wavelength of the first member of Lyman series of hydrogen is 𝛌.
The wavelength of the second member will be
𝟐𝟕 𝟑𝟐 𝟐𝟕 𝟓
𝐀 𝛌 𝐁 𝛌 𝐂 𝛌 (𝐃) 𝛌
𝟑𝟐 𝟐𝟕 𝟓 𝟐𝟕
(JEE Main PYQ)
Ans. (A)
If n levels are involved then
number of lines emitted are

𝑛) (𝑛 − 1
2
Photo of Students Hard-Work (Chits of Chapter)

Samarth S Bhise
Photo of Students Hard-Work (Chits of Chapter)
Nidhi Chaudhary
Photoelectric Effect
(Dual Nature)
Properties of Photon
𝐡𝐜
Energy of a photon E = = h𝛎
𝛌
Planck's constant ‘h’ = 6.626 × 10–34 Js
𝟏𝟐𝟒𝟐
E= 𝐢𝐧 𝐞𝐕
𝛌 (𝐢𝐧 𝐧𝐦)

𝐡 𝐄
Momentum of photon P = =
𝛌 𝐜
Photoelectric Effect

h −
e
Metal surface
− −
e e
Photoelectric Effect
Minimum frequency for which e–
𝐡𝛎𝟎 = 𝛟 just comes out is called threshold
frequency (𝛎0 ).

Photoelectric effect will take

place for 𝛎 ≥ 𝛎𝟎

h −
e
Metal surface
− −
e e
Photoelectric Effect
𝐡𝐜 Maximum wavelength for which e–
=𝛟 just comes out is called threshold
𝛌𝟎
wavelength ( 0 ).

Photoelectric effect will take

place for 𝛌 ≤ 𝛌𝟎

h −
e
Metal surface
− −
e e
𝐊𝐄𝐦𝐚𝐱 = 𝐡𝛎 − 𝛟

𝐡𝐜
KEmax = −𝛟
𝛌

h −
e
Metal surface
− −
e e
Q) UV light of 4.13 eV is incident on a photosensitive metal surface having
work function 3.13 eV . The maximum kinetic energy of ejected
photoelectrons will be :
(JEE Main PYQ)
(A) 4.13 eV
(B) 1.00 eV
(C) 3.13 eV
(D) 7.26 eV
Ans. (B)
Q) The threshold frequency of metal is f0. When the light of frequency 2f0 is
incident on the metal plate, the maximum velocity of photoelectron is v1.
When the frequency of incident radiation is increased to 5f0. the maximum
velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is:
𝐯𝟏 𝟏 𝐯𝟏 𝟏 𝐯𝟏 𝟏 𝐯𝟏 𝟏
𝐀 = 𝐁 = 𝐂 = (𝐃) =
𝐯𝟐 𝟐 𝐯𝟐 𝟖 𝐯𝟐 𝟏𝟔 𝐯𝟐 𝟒

(JEE Main PYQ)

Ans. (A)
 Photoelectric Effect

Same metal
i I 3 > I 2 > I1
𝐞𝐕𝐬𝐩 = 𝐊𝐄𝐦𝐚𝐱 = 𝐡𝛎 − 𝛟 I3
I2
For a given frequency of incident
radiation, the stopping potential I1
is independent of its intensity.

Vsp VA−VC
Q) The stopping potential in the context of photoelectric effect depends on the
following property of incident electromagnetic radiation :
(1) Phase (2) Intensity
(3) Amplitude (4) Frequency
(JEE Main PYQ)
Ans. (4)
De Broglie Wavelength

𝐡
𝛌= 𝐩 → Momentum
𝐩

𝐩𝟐
KE =
𝟐𝐦

𝐩= 𝟐𝐦(𝐊𝐄)
𝐡
𝛌=
𝟐𝐦(𝐊𝐄)
Q) A proton moving with one tenth of velocity of light has a certain de
Broglie wavelength of  . An alpha particle having certain kinetic energy
has the same de-Brogle wavelength  . The ratio of kinetic energy of
proton and that of alpha particle is:
(A) 2: 1 (B) 4:1 (C) 1: 2 (D) 1: 4
(JEE Main PYQ)
Ans. (B)
Q) The de Broglie wavelength of a proton and 𝛂-particle are equal.
The ratio of their velocity is :
(1) 4 : 2 (2) 4 : 1 (3) 1 : 4 (4) 4 : 3
(JEE Main PYQ)
Ans. (2)
Atomic Structure
Dual Nature
Nuclear Physics
Size of Nucleus

Radius of a nucleus:
R = R0 A1/3
R0 = 1.1 x 10-15 m
Density of nucleus (r) is independent of A
Q) Given below are two statements : one is labelled as Assertion A and the
other is labelled as Reason R. Assertion A : The nuclear density of nuclides
𝟏𝟎 𝟔 𝟓𝟔 𝟐𝟎 𝟐𝟎𝟗
𝟓 𝐁, 𝟑 𝐋𝐢, 𝟐𝟔 𝐅𝐞, 𝟏𝟎 𝐍𝐞 and 𝟖𝟑 𝐁𝐢 can be arranged as
𝛒𝐍𝐁𝐢 > 𝛒𝐍
𝐅𝐞 > 𝛒𝐍
𝐍𝐞 > 𝛒 𝐍
𝐁 > 𝛒 𝐍
𝐋𝐢 . (JEE Main PYQ)
Reason R : The radius R of nucleus is related to its mass number A as
R = R0A1/3, where R0 is a constant.
In the light of the above statement, choose the correct answer from the
options given below :
(A) Both A and R are true and R is the correct explanation of A
(B) A is false but R is true
(C) A is true but R is false
(D) Both A and R are true but R is NOT the correct explanation of A
Ans. (B)
Q) The mass number of nucleus having radius equal to half of the radius of
nucleus with mass number 192 is:
(A) 24 (B) 32 (C) 40 (D) –20

(JEE Main PYQ)

Ans. (A)
𝛂-particle: Doubly charged helium nucleus
with 2 protons and 2 neutrons
charge of + 2e.
𝟒
𝟐𝐇𝐞
𝟎 −
– (electron): Mass me and charge –e −𝟏𝐞

+ (positron): Mass me and charge +e 𝟎 +

+𝟏𝐞

-rays: Energetic photons having

rest mass zero.
 𝟗𝟐 𝐗 emits two -particles, one electron and two
Q) A radioactive element 𝟐𝟒𝟐
positrons. The product nucleus is represented by 𝟐𝟑𝟒
𝐏 𝐘. The value of P
is________
(JEE Main PYQ)
Ans. 87
 Mass Defect And Binding Energy of Nucleus

m+n
mp + nn X + Energy
BE = (m)c2
m1 m2 m3 BE of nucleus
Mass defect = m (m)c2
BE per nucleon =
𝐀
= (m1 + m2) – m3
Q-value of a Reaction

𝐀 + 𝐁 𝐂 + 𝐃 + 𝐄𝐧𝐞𝐫𝐠𝐲
𝐦𝐀 𝐦𝐁 𝐦𝐂 𝐦𝐃
Q-value of reaction

Q-value = [(mA + mB) – (mC + mD)] c2

= (m)c2

Q-value = Difference of rest mass energy or

Binding Energy of reactants and products
Q) Nucleus a having Z = 17 and equal number of protons and neutrons has 1.2
MeV binding energy per nucleon.
Another nucleus B of Z = 12 has total 26 nucleons and 1.8 MeV binding
energy per nucleons.
The difference of binding energy of B and A will be______ MeV.
(JEE Main PYQ)
Ans. 6
 Binding Energy/Nucleon Graph
9
8
BE per nucleon (MeV)

7
6
5
4
3
2
1
130 150 180 220
0 Mass number
0 20 40 60 80 100120 140 160 200 240
 Nuclear Fission

𝟐𝟑𝟓 𝐔 + 𝟎𝟏 𝐧→𝐗 + 𝐘 + (𝟐 𝐨𝐫 𝟑) 𝐧𝐞𝐮𝐭𝐫𝐨𝐧𝐬

𝐚𝐯𝐠=𝟐.𝟓 𝐧𝐞𝐮𝐭𝐫𝐨𝐧𝐬

𝟎
𝟏𝐧 = slow moving neutron i.e. thermal neutron

93% KE

7% -rays
Atomic Structure
Dual Nature
Nuclear Physics
Electrostatics
Coulomb’s Law
𝟏 𝐪𝟏 𝐪𝟐
𝐅= × 𝟐
𝟒𝛑 ∈𝟎 𝐫

𝐂𝟐
∈𝟎 = 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐
𝐍 − 𝐦𝟐

If charges kept in a medium

𝟏 𝐪𝟏 𝐪𝟐
𝐅= × 𝟐
𝟒𝛑 ∈ 𝐫
Q) Two point charges Q1 and Q2 exert a force F on each other when kept
certain distance apart. If the charge on each particle is halved and the
distance between the two particles is doubled, then the new force between
the two particles would be
𝐅 𝐅 𝐅 𝐅
𝐀 𝐁 𝐂 (𝐃)
𝟐 𝟒 𝟖 𝟏𝟔

(JEE Main PYQ)

Ans. (D)
Electric Field due to point charge
𝐤𝐐
𝐄= 𝟐
𝐫
EFI on the axis of uniformly charged ring
Q) Two equal positive point charges are separated by a distance 2a. The
distance of a point from the centre of the line joining two charges on the
equatorial line (perpendicular bisector) at which force experienced by a
𝐚
test charge q0 becomes maximum is . The value of x is _______.
𝐱

JEE Main PYQ

Ans (2)
EFI due to uniformly charged semi infinite long wire

𝐤𝛌 𝐤𝛌
𝐄⊥ = 𝐄|| =
𝐫 𝐫

EFI due to uniformly charged infinite long wire

𝟐𝐤𝛌
𝐄⊥ = 𝐄|| = 𝟎
𝐫
EFI due to uniformly charged arc at centre of curvature


𝟐𝐤𝛌
𝐄=
𝐫
R
O
EFI on axis of uniformly charged disc

𝛔
𝐄= 𝟏 − 𝐜𝐨𝐬 𝛉
𝟐𝝐𝟎
Q) Let 𝜎 be the uniform surface charge density of two infinite thin plane
sheets shown in figure. Then the electric fields in three different region EI ,
EII and JEE Main PYQ
𝟐𝜎 𝟐𝜎
𝐀 𝐄𝐈 = 𝐧ෝ, 𝐄𝐈𝐈 = 𝟎, 𝐄𝐈𝐈𝐈 = 𝑛ො
𝜖𝟎 𝜖𝟎
𝜎
𝐁 𝐄𝐈 = 𝟎, 𝐄𝐈𝐈 = 𝐧 ෝ, 𝐄𝐈𝐈𝐈 = 𝟎
𝜖𝟎
𝜎 𝜎
𝐂 𝐄𝐈 = 𝐧ෝ, 𝐄𝐈𝐈 = 𝟎, 𝐄𝐈𝐈𝐈 = ෝ
𝐧
𝟐𝜖𝟎 𝟐𝜖𝟎
𝜎 𝜎
(𝐃)𝐄𝐈 = − 𝐧 ෝ, 𝐄𝐈𝐈 = 𝟎, 𝐄𝐈𝐈𝐈 = 𝐧 ෝ
𝜖𝟎 𝜖𝟎
Ans (D)
 Electric Flux

𝛟=𝐄⋅𝐀
 𝟐𝐢Ƹ + 𝟔𝐣Ƹ + 𝟖𝐤ሶ
Q) An electric field, 𝐄 = passes through the surface of 4m2 area
𝟔
መ
𝟐𝐢ሶ + 𝐣Ƹ + 𝐤
having unit vector 𝐧
ෝ= . The electric flux for that surface
𝟔
is .....Vm (JEE Main PYQ)
Ans. (12)
Q) An electric field 𝐄 = (𝟐𝐱𝐢Ƹ )NC–1 exists in space. A cube of side 2 m is placed
in the space as per figure given below. The electric flux through the cube is
_____ .
(JEE Main PYQ)
Ans. (16)
 Gauss Law

σ 𝐪inside
ර𝐄. 𝐝𝐀 =
∈𝟎

𝐄 = Net EFI due to all charges

(both inside and outside
the closed surface)
Q) If a charge q is placed at the centre of a closed hemispherical non-
conducting surface, the total flux passing through the curved surface
would be :
𝐪 𝐪 (JEE Main PYQ)
(A) (B)
𝛆𝟎 𝟐𝛆𝟎
𝐪 𝐪
(C) (D)
𝟒𝛆𝟎 𝟐𝛑𝛆𝟎
Ans. (B)
Q) A charge q is placed at the center of one of the surface of a cube. The flux
linked with the cube is:- (JEE Main PYQ)
𝐪 𝐪 𝐪
𝐀 𝐁 𝐂 𝐃 𝐙𝐞𝐫𝐨
𝛆𝟎 𝟐𝛆𝟎 𝟔𝛆𝟎
Ans. (B)
EFI due to uniformly charged thin spherical shell

r < R, E =0

𝐤𝐐
r > R, 𝐄 = 𝟐
𝐫
EFI due to uniformly charged solid sphere

𝛒𝐫
r<R, 𝐄=
𝟑 ∈𝟎

𝛒𝐑
r =R, 𝐄 =
𝟑𝛜𝐨

𝐤𝐐
r >R, 𝐄 = 𝟐
𝐫
Potential energy of a system of two point charges

q1 r q2

𝐤𝐪𝟏𝐪𝟐
𝐔=
𝐫
 𝐔𝐢𝐧𝐭
Electric Potential 𝐕 =
𝐪𝟎

Due to point charge 𝐤𝐐

VA =
𝐫
Due to system of charges 𝐤𝐐𝟏 𝐤𝐐𝟐 𝐤𝐐𝟑
𝐕𝐀 = + +
𝐫𝟏 𝐫𝟐 𝐫𝟑
At the center of thin charged 𝐕𝐎 = 𝐤𝐐
𝐑
spherical shell
At the center of uniformly 𝐤𝐐
𝐕𝐎 =
charged ring 𝐑
On the axis of uniformly 𝐤𝐐
𝐕𝐀 =
charged ring 𝐱 𝟐 + 𝐑𝟐
Electric potential due to

a. Uniformly charged thin spherical shell

𝐤𝐐
r<R 𝐕=
𝐑
𝐤𝐐
r>R 𝐕=
𝐫
b. Uniformly charged solid sphere
𝛒
r< R 𝐕= 𝟑𝐑𝟐 − 𝐫 𝟐
𝟔𝛜𝐨
𝛒𝐑𝟐
r=R 𝐕=
𝟑𝝐𝐨
𝐤𝐐
r>R 𝐕=
𝐫
Q) Which of the following correctly represents the variation of electric
potential (V) of a charged spherical conductor of radius (R) with radial
distance (r) from the centre? JEE Main PYQ
(A) (B)

(C) (D)
Ans (C)
Electric potential from EFI EFI from EP
𝐝𝐕
𝐄𝐫 = −
𝐝𝐫
න𝐝𝐕 = − න𝐄. 𝐝𝐫
𝛛𝐕 𝛛𝐕 𝛛𝐕
𝐄 =− 𝐢Ƹ − 𝐉መ − መ
𝐤
𝛛𝐱 𝛛𝐲 𝛛𝐳
Electric Dipole moment

𝐩 = | 𝐩 |= 𝐪 𝐝
Electric Dipole moment
 Electric Dipole moment

𝐤𝐩
𝐕𝐀 = 𝟐
𝐫
𝐕=𝟎
Torque of uniform 𝐄 tries to align 𝐩 in
𝛕 = 𝐩 ×𝐄
direction of 𝐄 𝐭𝐡𝐫𝐨𝐮𝐠𝐡 𝐬𝐦𝐚𝐥𝐥𝐞𝐫 𝐚𝐧𝐠𝐥𝐞.
𝐄𝐥𝐞𝐜𝐭𝐫𝐢𝐜 𝐩𝐨𝐭𝐞𝐧𝐭𝐢𝐚𝐥 𝐞𝐧𝐞𝐫𝐠𝐲 𝐨𝐟 𝐞𝐥𝐞𝐜𝐭𝐫𝐢𝐜 𝐝𝐢𝐚𝐩𝐨𝐥𝐞 𝐨𝐟 Electric field.
Electric Conductors
Q) Two charged conducting spheres of radii a and b are connected to each
other by a conducting wire. The ratio of charges of the two spheres
respectively is: (JEE Main PYQ)
𝐚 𝐛
𝐀 𝐚𝐛 𝐁 𝐚𝐛 𝐂 𝐃
𝐛 𝐚
Ans. (C)
Modern Physics
Electrostatics
LET’S BOUNCE BACK
Current Electricity
 Electric Current Rate of flow of charge

Instantaneous Average
𝐝𝐐 ∆𝐐
i= <i>=
𝐝𝐭 ∆𝐭
Drift velocity 𝐯 = − 𝐞𝐄 𝛕
𝐝
𝐦

where  is avg. time between collisions

A
e- e-
e-
e- 𝐯𝐝
e- i = n e A vd
e- e- e-
e-
OHM's law(Microscopic) Macroscopic

𝐄=𝛒𝐣 𝐕=𝐢𝐑

I
Resistance of the body

𝛒𝓵
𝐑=
𝐀

r is a property of material
R is a property of body.
Q) A wire of resistance R and length L is cut into 5 equal parts. If these
parts are joined parallelly, then resultant resistance will be :
𝟏 𝟏
(A) R (B) R (C) 25 R (D) 5R
𝟐𝟓 𝟓

(JEE Main PYQ)

Ans. (A)
Q) The resistance of a wire is 𝟓𝛀. It's new resistance in ohm if stretched to
5 times of it's original length will be :
(A) 625 (B) 5 (C) 125 (D) 25

(JEE Main PYQ)

Ans. (C)
 Kirchhoff ’s Current Law (KCL)

i1 i2 At Junction σ Current Entering = 0

i3 Current Entering = Current Leaving

i5
i4 Based on conservation of charge.
 Kirchhoff ’s Voltage Law (KVL)

R2
The algebraic sum of changes in
E1 potential around any closed loop is zero.
R1
R3 Based on conservation of energy.
i
A
E2
Q) Equivalent resistance of the following network is .

(JEE Main PYQ)

Ans. (1)
 Wheatstone Bridge
B
𝐑𝟏 𝐑𝟑
= R1 R3
𝐑𝟐 𝐑𝟒
A C
G

R2 R4
D

V
Q) The equivalent resistance between A and B of the network shown in figure:
JEE Main PYQ

𝟐𝐑 𝟖
𝐀 𝟏𝟏 𝐁 𝟏𝟒𝐑 𝐂 𝟏𝟏𝐑 𝐃 𝐑
𝟑 𝟑
Ans (D)
 Meter Bridge

X R

It is used to find the unknown resistance.

Internal Resistance and Terminal Voltage of a Battery

Real Battery
𝐕𝐁 − 𝐕𝐀 = 𝐄 − 𝐢𝐑
A B
r
E
Internal Resistance
Q) In the given circuit, the terminal potential difference of the cell is:
(A) 2 V (B) 4 V (C) 1.5 V (D) 3 V

(JEE Main PYQ)

Ans. (A)
Series Combination of Batteries Parallel Combination of Batteries
𝟏 𝟏 𝟏
𝐫𝐞𝐪 = 𝐫𝟏 + 𝐫𝟐 + 𝐫𝟑 = +
𝐫𝐞𝐪 𝐫𝟏 𝐫𝟐
𝐄𝐞𝐪 𝐄𝟏 𝐄𝟐
𝐄𝐞𝐪 = 𝐄𝟏 + 𝐄𝟐 + 𝐄𝟑 = +
𝐫𝐞𝐪 𝐫𝟏 𝐫𝟐
Power Consumed by a Resistor
R

i V

V=iR
𝐕𝟐
Ploss = Vi = i2 R =
𝐑
𝐕𝟐 𝐭
Heat loss by resistor in time ′t′= i2 R t =
𝐑
Q) Ratio of thermal energy released in two resistor R and 3R connected in
parallel in an electric circuit is :
(A) 3: 1 (B) 1: 1 (C) 1: 3 (D) 1: 27

(JEE Main PYQ)

Ans. (A)
Maximum Power Transmission Theorem
Load Resistance
R

r E

Power consumed by load

resistance will be maximum
when its value is equal to
internal resistance of battery.
 Dependency of R on Temperature (T)

𝛒𝐓 = 𝛒𝟎 (𝟏 + 𝛂 𝐓𝟐 − 𝐓𝟏 )

Temperature coefficient of resistivity

R2 = R1 (1 + 𝛂 𝐓𝟐 − 𝐓𝟏 )

 is Temperature coefficient of Resistance

Q) Two conductors have the same resistances at 0°C but their temperature
coefficients of resistance are 1 and 2. The respective temperature
coefficients for their series and parallel combinations are :
𝛂𝟏 +𝛂𝟐 𝛂𝟏 +𝛂𝟐 𝛂𝟏 +𝛂𝟐
(A) 𝛂𝟏 + 𝛂𝟐 , (B) ,
𝟐 𝟐 𝟐

𝛂𝟏 𝛂𝟐 𝛂𝟏 +𝛂𝟐
(C) 𝛂𝟏 + 𝛂𝟐 , (D) , 𝛂𝟏 + 𝛂𝟐
𝛂𝟏 +𝛂𝟐 𝟐
(JEE Main PYQ)
Ans. (B)
Conversion of Galvanometer into Ammeter

𝐢𝐜 𝐑 𝐜 = (𝐈 − 𝐢𝐜 )𝐑 𝐬
Conversion of Galvanometer into Voltmeter

𝐕 = 𝐢𝐜 𝐑 + 𝐑 𝐜

𝐕𝐦𝐚𝐱 = 𝐢𝐦𝐚𝐱 𝐑 + 𝐑 𝐜
Q) Given below are two statements : One is labelled as Assertion A and the
other is labelled as Reason R. JEE Main PYQ
Assertion A : For measuring the potential difference across a resistance of
600 𝛀, the voltmeter with resistance 1000 𝛀 will be preferred over
voltmeter with resistance 4000 𝛀.
Reason R : Voltmeter with higher resistance will draw smaller current
than voltmeter with lower resistance.

(A) A is not correct but R is correct

(B) Both A and R are correct and R is the correct explanation of A
(C) Both A and R are correct but R is not the correct explanation of A
(D) A is correct but R is not correct
Ans (A)
Capacitor
Capacity (Capacitance) of Capacitor

+Q −Q

𝐐 = 𝐂𝐕
Capacity of a Capacitor depends on
1) Shape and size of the plates
2) Distance between plates
3) Medium between the plates

Parallel Plate Capacitor Spherical Capacitor

𝐀𝛆𝟎
𝐂= 𝟒𝛑𝛆𝟎 𝐚𝐛
𝐝 𝐂=
𝐛−𝐚
Series Combination (charge is same)

𝟏 𝟏 𝟏
= + +⋯
𝐂𝐞𝐪 𝐂𝟏 𝐂𝟐
Parallel Combination (V is same)

𝐂𝐞𝐪 = 𝐂𝟏 + 𝐂𝟐 + ⋯
Q) Two equal capacitors are first connected in series and then in parallel.
The ratio of the equivalent capacities capacities in the two cases will be –
(A) 2: 1 (B) 1: 4 (C) 4: 1 (D) 1: 2
(JEE Main PYQ)
Ans. (B)
Q) The total charge on the system of capacitance
C1 = 1 𝜇𝐅, C2 = 2 𝜇𝐅, C3 = 4 𝜇𝐅 and C4 = 3 𝜇𝐅 connected in parallel is
(Assume a battery of 20 V is connected to the combination)
𝐀 𝟐𝟎𝟎𝜇𝐂 𝐁 𝟐𝟎𝟎𝐂 𝐂 𝟏𝟎𝜇𝐂 𝐃 𝟏𝟎𝐂
JEE Main 2023 PYQ
Ans (A)
 𝟏 𝟐
Energy Stored in a Capacitor = 𝐂𝐕
𝟐

𝐐𝟐
=
𝟐𝐂

𝟏
= 𝐐𝐕
𝟐
Heat Generated in Circuit 𝐇 = ෍ 𝐖𝐛 − ෍ 𝚫𝐄
 = (Ef − Ei)
σ 𝐖𝐛 = Work done by all batteries
σ  = Energy change in all capacitors
 +𝐐𝟎 −𝐐𝟎
Dielectric Inside Capacitor
𝐄𝟎 𝐄𝟎
Enet = C = 𝐊𝐂𝟎 𝐄𝐧𝐞𝐭 =
𝐊 𝐊

𝐊
d
Bound Charges (Induced
Charges) on Dielectric

𝟏
𝐪=𝐐 𝟏−
𝐊
Q) A capacitor has capacitance 𝟓𝜇𝐅 when it's parallel plates are separated by
air medium of thickness d. A slab of material of dielectric constant 1.5
𝐝
having area equal to that of plates but thickness is inserted between the
𝟐
plates. Capacitance of the capacitor in the presence of slab will be____ 𝜇𝐅.
JEE Main PYQ
Ans (6)
Calculation of current at t = 0 and t = ∞
Q) In an electrical circuit drawn below the amount of charge stored in the
capacitor is …...C. (JEE Main PYQ)
Ans. (60)
Charging of Capacitor
 Discharging of Capacitor
Q0 q
+ − + −
C C
⇒

t=0 R at time t R
Modern Physics
Electrostatics
Current Electricity
Capacitor
Magnetic Effect
of Current
 MFI Due To Semi ∞ Wire

μ0 i
B= µ0 = 4π × 10–7 Tm/A
4πR

i
P
R
 MFI Due To ∞ Wire

μ0 i
B=
P 2πR
R
i
Q) The magnitude of magnetic induction at mid-point O due to current
arrangement as shown in Fig will be : JEE Main PYQ
𝜇𝟎 𝐈
𝐀 𝐁 𝟎
𝟐𝜋𝐚
𝜇𝟎 𝐈 𝜇𝟎 𝐈
𝐂 𝐃
𝟒𝜋𝐚 𝜋𝐚
Ans (D)
Magnetic Field at Centre of Current Carrying Circular Loop

R μ0 i
O B0 =
× 2R
Magnetic Field at the Centre of Current Carrying Circular Arc

ℓ
i μ0 i α
B= ×
2R 2π

α
Q) As shown in the figure, a long straight conductor with semicircular arc of
𝜋
radius 𝐦 is carrying current I = 3A. The magnitude of the magnetic
𝟏𝟎
field. at the center O of the arc is:
(The permeability of the vacuum = 𝟒𝜋 × 𝟏𝟎−𝟕 𝐍𝐀−𝟐 ) JEE Main PYQ

𝐀 𝟔𝜇𝐓 𝐁 𝟏𝜇𝐓 𝐂 𝟒𝜇𝐓 𝐃 𝟑𝜇𝐓

Ans (D)
Q) Find the magnetic field at the point P in figure. The curved portion is a
semicircle connected to two long straight wires.

𝛍𝟎 𝐢 𝟐 𝛍𝟎 𝐢 𝟏
𝐀 𝟏+ 𝐁 𝟏+
𝟐𝐫 𝛑 𝟐𝐫 𝛑

𝛍𝟎 𝐢 𝟏 𝟏 𝛍𝟎 𝐢 𝟏 𝟏
𝑪 + (𝑫) +
𝟐𝐫 𝟐 𝟐𝛑 𝟐𝐫 𝟐 𝛑

(JEE Main PYQ)

Ans. (C)
 MFI on the Axis of Current Carrying Circular Loop

i
μ0 i 3 θ) θ
B= (sin
2R P
R
 Ampere’s Circuital Law

i4
i5 i3

i2
ය B . dℓ = μ0 σ iin
i1
 MFI due to Current Carrying  Long Hollow Cylinder

r <R B=0
R
μ0 i
r>R B=
2πr
B

axis
r
R
 MFI due to Current Carrying  Long Solid Cylinder

μ0 j
r<R→B= r
2
R
μ0 i
r>R→B=
2πr
B

axis
r
R
Q) A long straight wire of circular cross-section (radius a) is carrying steady
current I. The current I is uniformly distributed across this cross-section.
The magnetic field is
(A) Zero in the region r < a and inversely proportional to r in the region
r>a (JEE Main PYQ)
(B) Inversely proportional to r in the region r < a and uniform throughout
in the region r > a
(C) Directly proportional to r in the region r <a and inversely
proportional to r in the region r > a
(D) Uniform in the region r < a and inversely proportional to distance r
from the axis, in the region r > a
Ans. (C)
 Long Solenoid

MFI inside is uniform

'n' turns per unit length.
and axial B = µ0 ni

MFI outside solenoid is

zero.
µ0 ni
MFI at ends is
2
Q) A long solenoid is formed by winding 70 turns cm–1. If 2.0 A current
flows, then the magnetic field produced inside the solenoid is________
𝛍𝟎 = 𝟒𝛑 × 𝟏𝟎−𝟕 𝐓𝐦𝐀−𝟏
(A) 1232 × 10–4 T (B) 176 × 10–4 T
(C) 352 × 10–4 T (D) 88 × 10–4 T

(JEE Main PYQ)

Ans. (B)
 Toroid

‘N' total turns

μ0 Ni
Binside =
2πr
Motion of a Charged Particle in Uniform Magnetic Field

Case 1: Motion of particle in plane ⊥ to B

B
× × × × × mv
R=
× × × v× × qB

× × × × ×
O (q , m)
× × × × ×
× × × × ×
× × × × ×
Motion of a Charged Particle in Uniform Magnetic Field

Case 1: Motion of particle in plane ⊥ to B

B 2πR 2πm 1 qB
× × × × × T= = f= =
v qB T 2πm
× × × v× ×
× × × × × 2π qB
(q , m) w= =
O T m
× × × × ×
× × × × × Time period does not depend on velocity
× × × × ×
Magnetic Force on a Current Carrying Wire Kept in Magnetic Field

B is Uniform and Wire is Straight

B

i

F = i ℓ ×B
B Uniform and Arbitrary Shaped Wire

× × Q ×
B × × F=iℓ×B
×
Vector joining end points
× i × ×
i ℓ
× × ×

× P × ×
 Magnetic Force Between Two Current Carrying Wires
Case 1 When wires are parallel to each other

F µ0 i1 i2
=
1 2 ℓ 2πd 1 2
i1 i2
i1 i2
➢Attraction if i1 and i2 are in
same direction

➢Repulsion if i1 and i2 are in

d opposite direction d
 Magnetic Moment or magnetic dipole moment of loop

N
μ=NiA Area vector of the loop

i
Number of turn A

Magnetic Moment of the loop A (inside the plane )

Torque on a Current Carrying Loop Kept in a Uniform Magnetic Field

μ p

B E

τ = μ ×B τ = p ×E

U = − μ .B
Q) Work done in rotating magnetic dipole from most unstable to stable
position is

(JEE Main PYQ)

Magnetic Moment due to Rotation of Charge
Q,M
R μ Q
O = Holds for all uniformly
w L 2m
charged and mass bodies

L=Iw
Q, m
angular momentum

R
O
w
Magnetism & Matter
Magnetic Field Lines
The magnetic field lines do not intersect.

More dense

S N Less
dense

Magnetic field lines always form closed loops.

Dipole in Uniform Magnetic Field

I
Time Period T = 2π
mB
 Magnetisation
mnet Net Magnetic Moment
M= =
V Volume

Magnetisation
 Classification of Magnetic Materials

Properties Diamagnetic Paramagnetic Ferromagnetic

χ −1 ≤ χ < 0 0 < χ< k χ >> 1
µr 0 ≤ µr < 1 1 < µr < 1 + k µr >> 1
µ µ < µ0 µ > µ0 µ >> µ0
Magnetisation Weak Weak Strong
Magnetisation Magnetisation Magnetisation
in opposite in Same in Same
direction direction direction
 Properties Diamagnetic Paramagnetic Ferromagnetic
Movement in (Weak tendency) (Weak tendency) (Strong tendency)
magnetic field From strong to From weak to From weak to
weak magnetic strong magnetic strong magnetic
field field field
Magnet Weak Repulsion Weak Attraction Strong Attraction
Properties Diamagnetic Paramagnetic Ferromagnetic
E.g. Bi, Au, Pb, Si, Al, Na, Ca, Fe, Co, Ni,Gd
H2O, NaCl, N2 O2(STP)
(STP)

Magnetic
Field Lines
Q) Paramagnetic substances:
A. align themselves along the directions of external magnetic field.
B. attract strongly towards external magnetic field.
C. has susceptibility little more than zero.
D. move from a region of strong magnetic field to weak (JEE Mainfield.
magnetic PYQ)
Choose the most appropriate answer from the options given below:
(A) A, B, C, D
(B) B, D Only
(C) A, B, C Only
(D) A, C Only
Ans. (D)
 Paramagnetic

µ0 1
χ=C χ∝
T T

Curie Law
C = Curie’s constant
Ferromagnetic

At high enough temperature, a

ferromagnet becomes a
paramagnet.

This temperature of transition

from ferromagnetic to
paramagnetic is called Curie
Temperature TC.
Alternating Current
Pure Resistive Circuit

R
i = i0 sin(wt)
𝐕𝟎 𝐕𝐫𝐦𝐬
i 𝐢0 = 𝐢rms =
𝐑 𝐑
~
V = V0 sin (wt) 𝐢𝟎 𝐕𝟎
𝐢𝐫𝐦𝐬 = 𝐕𝐫𝐦𝐬 =
𝟐 𝟐
𝐕 𝐕𝟎 𝐬𝐢𝐧(w𝐭)
𝐢= =
𝐑 𝐑 In pure resistive circuit V and i
are in same phase.
 Pure Capacitive Circuit
C
𝐕𝟎 𝟏
i 𝐢𝟎 = 𝐗𝐂 =
𝐗𝐂 w𝐂
~ (𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐢𝐯𝐞 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐜𝐞)
V = V0 sin(wt)
In pure Capacitive circuit Current
𝛑 𝛑
𝐢 = 𝐢𝟎 𝐬𝐢𝐧 w𝐭 + leads Voltage by phase
𝟐 𝟐
Pure Inductive Circuit
𝛑
L 𝐢 = 𝐢𝟎 𝐬𝐢𝐧 𝛚𝐭 −
𝟐
𝐕𝟎 𝐕𝐫𝐦𝐬
𝐢𝟎 = 𝐢𝐫𝐦𝐬 =
𝐗𝐋 𝐗𝐋
i
𝐗 𝐋 = w𝐋 inductive reactance

~
In pure Inductive circuit Voltage
𝛑
V = V0 sin(wt) leads Current by phase
𝟐
Q) An alternating voltage source V = 260 sin (628 t) is connected across a
pure inductor of 5mH. Inductive reactance in the circuit is:
(A) 3.14 𝛀 (B) 6.28 𝛀 (C) 0.5 𝛀 (D) 0.318 𝛀

JEE Main PYQ

Ans. (A)
Impedance
Z
R R 𝟐
𝐙= 𝐗𝐋 − 𝐗𝐂 + 𝐑𝟐
C XC
𝐕𝟎
L XL 𝐢𝟎 =
𝐙
RC 𝐗 𝟐𝐜 + 𝐑𝟐 𝐕𝐫𝐦𝐬
𝐢𝐫𝐦𝐬 =
𝐙
LR 𝐗 𝟐𝐋 + 𝐑𝟐

LCR 𝐗𝐋 − 𝐗𝐂 𝟐 + 𝐑𝟐
L−C−R Circuit

𝐕𝐋𝟎 − 𝐕𝐂𝟎
𝐕𝟎

R C L 
𝐕𝐑 𝟎

i = i0sin(wt)
𝐕𝟎 𝐕𝐫𝐦𝐬
𝐢0 = 𝐢rms =
𝐙 𝐙
V = V0sin(wt + )
𝐗 𝐋 −𝐗 𝐂
𝐙= 𝐗𝐋 − 𝐗𝐂 𝟐 + 𝐑𝟐 𝐭𝐚𝐧  =
𝐑
Q) A series LCR circuit consists of R= 𝟖𝟎𝛀 𝐗 𝐋 = 𝟏𝟎𝟎𝛀, and 𝐗 𝐂 = 𝟒𝟎𝛀
The input voltage is 𝟐𝟓𝟎𝟎 𝐜𝐨𝐬 𝟏𝟎𝟎𝛑𝐭 V. The amplitude of current, in
the circuit, is___________ A. JEE Main PYQ
Ans. (25)
Q) In an ac circuit, the instantaneous current is zero when the instantaneous
voltage is maximum. In this case, the source may be connected to:
A. pure inductor.
B. pure capacitor.
C. pure resistor. JEE Main PYQ
D. combination of an inductor and capacitor.
Choose the correct answer from the options given below:
(A) A, B and C only (B) B, C and D only
(C) A and B only (D) A, B and D only
Ans. (D)
Q) If wattless current flows in the AC circuit, then the circuit is
(A) Purely Resistive circuit (JEE Main PYQ)
(B) Purely Inductive circuit
(C) LCR series circuit
(D) RC series circuit only
Ans. (B)
Power Delivered by Source

R C L
Power Factor

< 𝐏 > = 𝐢𝐫𝐦𝐬 𝐕𝐫𝐦𝐬 𝐜𝐨𝐬

i = i0sin(wt)
𝐑
𝐜𝐨𝐬𝛟 =
V = V0sin(wt + ) 𝐙
Power Analysis

< 𝐏𝐑 > = 𝐢𝟐𝐫𝐦𝐬 R < 𝐏𝐂 > = 𝟎 < 𝐏𝐋 > = 𝟎

Q) An AC voltage 𝐕 = 𝟐𝟎𝐬𝐢𝐧(𝟐𝟎𝟎𝛑𝐭) is applied to a series LCR circuit which
𝛑
drives a current 𝐈 = 𝟏𝟎𝐬𝐢𝐧 𝟐𝟎𝟎𝛑𝐭 + . The average power dissipated is:
𝟑
(A) 21.6 W (B) 200 W (C) 173.2 W (D) 50 W
JEE Main PYQ
Ans. (D)
Resonance in LCR Circuit
i

capacitive inductive

At Resonance
1) i0 will be maximum
2) XL = XC w
w𝟏 w𝟎 w𝟐
3) Z=R
𝚫w
4) cos =1 Resonant Frequency
𝟏 𝟏
𝟓) w𝟎 = f𝟎 =
𝐋𝐂 𝟐𝝅 𝐋𝐂
Q) In series LCR circuit, the capacitance is changed from C to 4 C . To keep the
resonance frequency unchanged, the new inductance should be :
𝟏 JEE Main PYQ
(A) reduced by L
𝟒
(B) increased by 2 L

𝟑
(C) reduced by L
𝟒
(D) increased to 4 L
Ans. (C)
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G.O.C
HC ≡ CH sp hybrid

CH2 = CH2 sp2 hybrid

CH3 ⎼ CH3 sp3 hybrid

EN Order: sp > sp2 > sp3

 Inductive Effect
Partial displacement of -bond electrons
toward more electronegative atom within
same orbital is known as inductive effect.
 Inductive Effect
— — —
+4 + 3 + 2 + 1 –
C4 — C3 — C2— C1— F
x3  x2 x 1 x

(i) x > x1 > x2 > x3

(ii)  = (1 + 2 + 3 + 4)
(iii)  > 1 > 2 > 3 > 4
(iv) 4 ~ 0 (experimentally)
 –I effect
(i) If the electron withdrawing power of
group is higher than that of hydrogen then
these groups are known as –I groups.

–I-Series
+ + +
- NF3 ˃- NR3 ˃- NH3 ˃ -NO2 ˃ -CN ˃ -SO3H
˃ -CHO ˃ -COOH ˃ -F ˃ -CI ˃ -Br ˃ -I ˃ -OH
˃ - C ≡ CH ˃ - NH2 ˃ -PH ˃ -CH = CH2 ˃ H
“+I effect”
If the electron donating power of group is
higher than that of hydrogen then these
groups are known as +I groups.

+I series
CH3
CH3
CH2 ‾ > NH ‾ > O ‾ > COO ‾ > – C–CH3 > – CH > CH2-CH3
CH3
CH3

> CT3 > CD3 > CH3 > T > D > H

Note
(i) For I-effect C–H bond is taken as reference.
(ii) It is additive effect.
Application of I-effect
Stability of Carbocation ∝ + I effect

𝟏
Stability of carbocation ∝
− 𝐈 𝐞𝐟𝐟𝐞𝐜𝐭

𝟏
Stability of carbanion ∝
+ 𝐈 𝐞𝐟𝐟𝐞𝐜𝐭

Stability of Carbanion ∝ - I effect

 F
NO2 CN
(1)
NO2 F
I I I I

a>b>c>d
BAAP Rule
[D N P]

Distance Power
No. of groups
Resonance
Hypothetical Hypothetical Real
1 1
6 6

(I) (II) (III)

3C=C 3C-C
1.39Å

1.34Å 1.54Å
Important Points
1. Hypothetical.

2. R.H. is formed with contribution of various

resonating structure according to their stability.

3. Distance independent effect.

4. Intra molecular phenomena.

5. Due to resonance potential energy of the system is

decreased so stability is increased.
 Conditions for Resonance
1. System must be planar.

2. System must be in conjugation.

(i.e. parallel p orbitals are required)
1.  → vacant orbital conjugation

2. Lone pair →  conjugation

3.  → free radical e– conjugation

4.  →  conjugation

5. Lone pair → +ve charge conjugation

Q) (JEE Main PYQ)

Among the given species the Resonance stabilised carbocations are:

(1) (C) and (D) only (2) (A), (B) and (D) only
(3) (A) and (B) only (4) (A),(B) and (C) only

Ans. (3)
Q) Which of the following compounds does not exhibit resonance?
𝟏 𝐂𝐇𝟑 𝐂𝐇𝟐 𝐎𝐂𝐇 = 𝐂𝐇𝟐
(JEE Main PYQ)
𝐂𝐇𝟐𝐎𝐇
𝟐

𝟑 𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐎𝐍𝐇𝟐

𝟒 𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇 = 𝐂𝐇𝐂𝐇𝟐 𝐍𝐇𝟐

Ans. (4)
Stability Of Resonating Structure
Resonating structures contribute to resonance
hybrid in proportion to their stability.
1. More Pi bond more will be the stability of RS
2. Resonating structure with complete octet of all
atom is more stable than incomplete octet R.S.

+
CH3–CH–O–CH3 CH3–CH = O –CH3
+
3. Neutral or non-polar resonating
structure (R.S.) is more stable
than polar-resonating structure.

_
+
(I) (II)
4. Negative charge on more E.N. atom is more
stable, positive charge on less E.N. atom is
more stable.
According to Coulomb's law like charges
should be away from each other & unlike
charges should be closer.
Fries rule : Resonating structure having more
number of benzenoid ring is more stable.
Q)
(JEE Main PYQ)

The correct order of stability of given carbocation is:

𝟏 𝐀>𝐂>𝐁>𝐃 𝟐 𝐃>𝐁>𝐂>𝐀
𝟑 𝐃>𝐁>𝐀>𝐂 𝟒 𝐂>𝐀>𝐃>𝐁

Ans. (1)
Equivalent Resonating Structure
 CH3 – C – O CH3 – C = O
1.
O O
x y
x=y
2.

3.
HO O

HO O
(Squaric acid)
 Mesomeric Effect or Resonance Effect
The atoms or groups which donate or withdraw
electrons from conjugated system via resonance
show resonance or mesomeric effect.

Permanent Effect
 +M effect
For +M effect, group should have either a lone
pair of electron or should have negative
charge.

Due to + M effect negative charge comes over

conjugate system or electron density increase
on conjugate system.
Group which shows + M effect are -
O‾, NH‾, –NR2 , –NHR, –NH2 , – OH, –OR, –
SH – SR, – F, – Cl, – NHCOR , –O–COR etc.
 -M effect
If transfer of pi-bond electron takes place from
conjugated system (withdrawal) to group then
it is known as negative mesomeric (–M) effect.

O

CH2 =CH — N
O

 O

CH2 —CH = N
O
For –M effect, group should have either
positive charge or should have vacant orbital.
Due to –M effect positive charge comes over
conjugate system or due to –M effect electron
density decreases in conjugate system.
 O
CH2 =CH — N
O
Group which show –M effect are -
–NO2 , –CN, –SO3H , – CHO, –COR, –
COOH, –COOR , –COX, –CONH2 etc.
Q)
JEE Main PYQ

Among the given species the Resonance stabilised carbocations are:

(1) (C) and (D) only (2) (A), (B) and (D) only
(3) (A) and (B) only (4) (A),(B) and (C) only

Ans. (3)
Q) Which of the following carbocations is most stable
(JEE Main PYQ)
(A) (B)

(C) (D)

Ans. (D)
Q) The order of stability of the following carbocations
(A) III > I > II (B) I > II > III (JEE Main PYQ)
(C) II > III > I (D) III > II > I

Ans. (A)
Q) Arrange the carbanions, 𝐂𝐇𝟑 𝟑 𝐂, ത 𝐂𝐂𝐥
ത 𝟑 , 𝐂𝐇𝟑 𝟐 𝐂𝐇, 𝐂𝟔 𝐇𝟓 𝐂𝐇
ത 𝟐 , in order
of their decreasing stability
(A) 𝐂𝐇𝟑 𝟐 𝐂𝐇ത > 𝐂𝐂𝐥ത 𝟑 > 𝐂𝟔 𝐇𝟓 𝐂𝐇𝟐 > 𝐂𝐇𝟑 𝟑 𝐂ത (JEE Main PYQ)
ത 𝟐 > 𝐂𝐂𝐥
(B) 𝐂𝟔 𝐇𝟓 𝐂𝐇 ത 𝟑 > 𝐂𝐇𝟑 𝟑 𝐂ത > 𝐂𝐇𝟑 𝟐 𝐂𝐇ത
(C) 𝐂𝐇𝟑 𝟑 𝐂ത > 𝐂𝐇𝟑 𝟐 𝐂ത + 𝐂𝟔 𝐇𝟓 𝐂ത𝟐 > 𝐂𝐂𝐥
ത 𝟑
(D) 𝐂𝐂𝐥𝟑 > 𝐂𝟔 𝐇𝟓 𝐂𝐇ത 𝟐 > 𝐂𝐇𝟑 𝟐 𝐂𝐇
ത > 𝐂𝐇𝟑 𝟑 𝐂ത

Ans. (D)
Q) Among the following, total number of meta directing functional groups is
(Integer based)
−𝐎𝐂𝐇𝟑 , −𝐍𝐎𝟐 , −𝐂𝐍, −𝐂𝐇𝟑 − 𝐍𝐇𝐂𝐎𝐂𝐇𝟑 , JEE Main 2024 PYQ
−𝐂𝐎𝐑, −𝐎𝐇, −𝐂𝐎𝐎𝐇, −𝐂𝐥

Ans. (4)
Q) The set of meta directing functional groups from the following sets is:
(1) −𝐂𝐍, −𝐍𝐇𝟐 , −𝐍𝐇𝐑, −𝐎𝐂𝐇𝟑 (2) −𝐍𝐎𝟐 , −𝐍𝐇𝟐 , −𝐂𝐎𝐎𝐇, −𝐂𝐎𝐎𝐑
(3) −𝐍𝐎𝟐 , −𝐂𝐇𝐎, −𝐒𝐎𝟑 𝐇, −𝐂𝐎𝐑 (4) −𝐂𝐍, −𝐂𝐇𝐎, −𝐍𝐇𝐂𝐎𝐂𝐇𝟑 , −𝐂𝐎𝐎𝐑

JEE Main 2024 PYQ

Ans. (3)
 HYPERCONJUGATION
Complete transfer of e‾ of C–H  bond towards
 bond or positive charge or free electron is
called as H-effect (permanent effect). It is also
called as non-bonded resonance or baker
Nathan effect.
Condition: at least one alpha hydrogen should be
present at sp3 C in conjugation with carbocation, alkene,
alkyne, free radical.
1. Hyperconjugation In Alkene
2. Hyperconjugation In Carbocation
3. Hyperconjugation In Free Radical
 CH2 = CH2

CH3 CH = CH2
4. Hyperconjugation on Benzene Ring
Alkyl groups activates the benzene
ring towards electrophile.
Because it generates negative charge
on ortho and para position .
So they are called activating groups
towards aromatic electrophilic
Substitution.
 Applications of Hyperconjugation
(1) Stability of Alkenes
"More alkylated alkenes are more stable".

αH ∝ 𝐒𝐭𝐚𝐛𝐢𝐥𝐢𝐭𝐲
Q) Stability order of alkenes is
(2) Stability of Carbocation
 + + +
CH3 C CH3 CD3 C CD3 CT3 C CT3

CH3 CD3 CT3

Inductive 3 > 2 > 1

Hyperconjugation 1 > 2 > 3

H bond weak hence H dominates over I

(3) Stability of Free Radical
(4) Heat of Hydrogenation

CH2 = CH2 CH3 –CH =CH2 CH3 – CH = CH – CH3

(5) Bond Length
Bond length is also affected
by hyperconjugation
(6) Heat of Combustion (HOC)
Enthalpy change when 1 mole of
compound is oxidized.
𝟏
HOC  No. of C-atom ∝ ∝ 𝐬𝐭𝐫𝐚𝐢𝐧
𝐬𝐭𝐚𝐛𝐢𝐥𝐢𝐭𝐲
 Arrange In Order Of HOC
1.
(I)

(II)

(III)

(IV)

IV > III > II > I , ↑ no. of C-atom

 3+ 2+
 Ni ( H 2 O )6  , Cu ( H 2O )6 

Q) Number of carbocation from the following that are not stabilized by

hyperconjugation is............
(JEE Main PYQ)

Ans. (5)
Q) Match List -I with List II: (JEE Main PYQ)
Choose the correct answer from the options
given below :
(A) (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(B) (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
(C) (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
(D) (A) - (I), (B) - (II), (C) - (IV), (D) - (III)

Ans. (A)
Aromaticity
S. No. Aromatic Anti-aromatic Non-aromatic
Non-planarity
1. Cyclic Cyclic (presence of sp3
atom)
Each atom of ring Each atom of ring
2. is sp2 hybridised is sp2 hybridised

Compound should Compound should

3. follow Huckel’s Rule. not follow Huckel’s
(4n+2)𝛑 𝐝𝐞𝐥𝐨𝐜𝐚𝐥𝐢𝐳𝐞𝐝 e‾ Rule. 4n𝛑 delocalized
n = 0 to ∞ electron. n ≠ 𝟎
 Stability Order
Aromatic > Non-Aromatic > Anti
Aromatic
Q) Which of the following compounds is not aromatic ? (JEE Main PYQ)

Ans. (3)
Q) Which compound(s) out of the following is/are not aromatic ?

(JEE Main PYQ)

+ +
(A) (B) (C) (D)
(1) C and D (2) B, C and D
(3) A and C (4) B

Ans. (2)
Q) Which of the following compounds will produce a precipitate with AgNO3 ?

(A) (B) (C) (D)

(JEE Main PYQ)

Ans. (D)
Q) Which of the following is an aromatic compound?
(JEE Main PYQ)

Ans. (1)
Stability of carbocation/free radical  EDG (+M / +H / +I / _ / -I / -M)

Stability of carbonion  EWG (–M / –I / _ / +I / +H / +M)

AERHI Concept
CH3 OCH 3 NO 2
Bredt's Rule

According to this rule, planarity

(sp2 hybrid) can't be achieved at
bridge head centre of bicyclic
compound having less than 8C.
Q) Which of the following molecule/species is most stable? JEE Main PYQ
(1) (2) (3) (4)
Condition :- Same period elements

CH4 NH3 H2O HF

2p 2p 2p 2p
-H+ -H+ -H+ -H+
_ _ _ _
CH3 NH2 OH F
SiH4 PH4 H2S HCl
I II III IV
2. Size

H2O H2S H2Se H2Te

3. Hybridisation
4. Electronic Effect

Ph–O CH3–C–O
O
 CH2 – C – CH– CH2
Ha O Hb Hc

Ans. a > b > c

 Hb
Ha

CF3 – SO3H CH3 – SO3H CH3 – C – OH

O
Q) The increasing order of the pKa values of the following compounds is :

(JEE Main PYQ)

(1) D < A < C < B (2) B < C < D < A

(3) C < B < A < D (4) B < C < A < D

Ans. (4)
Q) The correct order for acid strength of compounds
𝐂𝐇 ≡ 𝐂𝐇, 𝐂𝐇𝟑 − 𝐂 ≡ 𝐂𝐇 𝐚𝐧𝐝 𝐂𝐇𝟐 = 𝐂𝐇𝟐 is as follows:
𝟏 𝐂𝐇 ≡ 𝐂𝐇 > 𝐂𝐇 = 𝐂𝐇 > 𝐂𝐇 − 𝐂 ≡ 𝐂𝐇 (JEE Main PYQ)
𝟐 𝟐 𝟑
𝟐 𝐇𝐂 ≡ 𝐂𝐇 > 𝐂𝐇𝟑 − 𝐂 ≡ 𝐂𝐇 > 𝐂𝐇𝟐 = 𝐂𝐇𝟐
𝟑 𝐂𝐇𝟑 − 𝐂 ≡ 𝐂𝐇 > 𝐂𝐇𝟐 = 𝐂𝐇𝟐 > 𝐇𝐂 ≡ 𝐂𝐇
𝟒 𝐂𝐇𝟑 − 𝐂 ≡ 𝐂𝐇 > 𝐂𝐇 ≡ 𝐂𝐇 > 𝐂𝐇𝟐 = 𝐂𝐇𝟐

Ans. (2)
Q) Which among the following is the strongest acid?
(JEE Main PYQ)
𝟏 𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
(2)

(3) (4)

Ans. (4)
Q) The correct decreasing order for acid strength is :-
𝟏 𝐍𝐎𝟐 𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐍𝐂𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐅𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐂𝐥𝐂𝐇𝟐 𝐂𝐎𝐎𝐇
𝟐 𝐅𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐍𝐂𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐍𝐎𝟐 𝐂𝐇𝐂𝐎𝐎𝐇 > 𝐂𝐥𝐂𝐇𝟐 𝐂𝐎𝐎𝐇
𝟑 𝐍𝐎𝟐 𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐅𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐂𝐍𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐂𝐥𝐂𝐇𝟐 𝐂𝐎𝐎𝐇
𝟒 𝐂𝐍𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐎𝟐 𝐍𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐅𝐂𝐇𝟐 𝐂𝐎𝐎𝐇 > 𝐂𝐥𝐂𝐇𝟐 𝐂𝐎𝐎𝐇

(JEE Main PYQ)

Ans. (1)
Q) Arrange the following in decreasing acidic strength.

(A) (B) (C) (D)

(A) A > B > C > D (JEE Main PYQ)

(B) B > A > C > D
(C) D > C > A > B
(D) D > C > B > A

Ans. (A)
Q) Which will undergo deprotonation most readily in basic medium?

(A) a only (B) c only

(C) a and c only (D) b only JEE Main PYQ

Ans. (A)
Q) What is the correct order of acidity of the protons marked A-D in the
given compounds?
(A) 𝐇𝐂 > 𝐇𝐃 > 𝐇𝐁 > 𝐇𝐀
(B) 𝐇𝐂 > 𝐇𝐃 > 𝐇𝐀 > 𝐇𝐁
(C) 𝐇𝐃 > 𝐇𝐂 > 𝐇𝐁 > 𝐇𝐀
(D) 𝐇𝐂 > 𝐇𝐀 > 𝐇𝐃 > 𝐇𝐁

(JEE Main PYQ)

Ans. (B)
 S.I.R. Effect

Some small groups and linear

group which do not show SIR
are –OH, –NH2, –CN, F etc.
Steric crowding of some group is
not considered -H, -T, -D, -F, – OR.
 Some Large Groups Which Show S.I.R.

O O O O O
O
–NR2 , –C–OH , –N , –C–OR , –C–NH2 , –C–Cl , –C –O –C–
O
O
 Give Acidic Strength Order
(1)
COOH COOH

<

NO2 CN
(2)
OH OH

N N
Me Me Me Me
Ortho Effect
Ortho derivative of benzoic acid is
generally more acidic than benzoic
acid itself. It is called ortho effect

CO2 H CO2 H
G
>
Example

Ortho effect a>c>b>d

Solublity in NaHCO3
Q) Compound(s) which will liberate carbon dioxide with sodium bicarbonate
solution is/are:
(A) A and B only (JEE Main PYQ)
(B) B and C only
(C) C only
(D) B only

Ans. (B)
 Basic Strength

Ability to donate Lp is known as basic strength .

 Basic Strength in Aqueous Medium

By considering + I effect of methyl we may

expect basic nature as
NH3 < MeNH2 < Me2NH < Me3N
(Which is not true always)
Thus, due to two opposite effects i.e.
solvation of cation and + I effect, the
jumbled order comes to be

Me2NH > Me – NH2 > Me3N > NH3

Similarly on the same reasons ethyl amines
and other amine follow the following order
for basic strength

Et3N > Et2NH > EtNH2 > NH3

(gas phase)

Et2NH > Et3N > EtNH2 > NH3

(water as solvent)
 SIP Effect(Steric Inhibition of Protonation)
Ortho derivative of aniline is generally less basic
than aniline itself and it’s p,m derivatives.

NH2 NH2
G
<
Arrange the following in order of their basic strength

–
Solution :Aniline > m > p > o
NH2 NH2 NH2

I II III

III < I < II

Q) In the following compound, the favorable site/s for protonation is/are :-

(JEE Main PYQ)

(1) (b), (c) and (d) (2) (a)

(3) (a) and (e) (4) (a) and (d)

Ans. (1)
Q) Given below are two statements :
Statement I : Aniline is less basic than acetamide.
Statement II : In aniline, the lone pair of electrons on nitrogen atom is
delocalised over benzene ring due to resonance and hence less available to a
proton.
Choose the most appropriate option : (JEE Main PYQ)
(1) Statement I is true but statement II is false.
(2) Statement I is false but statement II is true.
(3) Both statement I and statement II are true.
(4) Both statement I and statement II are false.

Ans. (2)
Q) Which of the following is least basic ?
.. ..
𝟏 𝐂𝐇𝟑 𝐂𝐎 𝐍𝐇𝐂𝟐 𝐇𝟓 𝟐 𝐂𝟐 𝐇𝟓 𝟑 𝐍 (JEE Main PYQ)
.. ..
𝟑 𝐂𝐇𝟑 𝐂𝐎 𝟐 𝐍𝐇 𝟒 𝐂𝟐 𝐇𝟓 𝟐 𝐍𝐇

Ans. (3)
 3+ 2+
 Ni ( H 2 O )6  , Cu ( H 2O )6 

Q) What will be the decreasing order of basic strength of the following

conjugate bases? (JEE Main PYQ)
−
𝐎𝐇, 𝐑𝐎 ഥ , 𝐂𝒍ҧ
ഥ , 𝐂𝐇𝟑 𝐂𝐎𝐎
− −
ҧ ഥ
(A) 𝐂𝐥 > 𝐎𝐇 > 𝐑𝐎 > 𝐂𝐇𝟑 𝐂𝐎𝐎 ഥ ഥ
(B) 𝐑𝐎 > 𝐎𝐇 > 𝐂𝐇𝟑 𝐂𝐎𝐎 ഥ > 𝐂𝒍ҧ
−
− ഥ ഥ ҧ ҧ ഥ
(C) 𝐎𝐇 > 𝐑𝐎 > 𝐂𝐇𝟑 𝐂𝐎𝐎 > 𝐂𝒍 (D) 𝐂𝒍 > 𝐑𝐎 > 𝐎𝐇 > 𝐂𝐇𝟑 𝐂𝐎𝐎 ഥ

Ans. (B)
Atomic Structure
Radius of Bohr Orbits

𝐧𝟐
𝐫𝐧 = 𝟎. 𝟓𝟐𝟗 Å
𝐙
Q) The radius of the second Bohr orbit for hydrogen atom is : (Plank's
const. h = 𝟔. 𝟔𝟐𝟔𝟐 × 𝟏𝟎−𝟑𝟒 Js ; mass of electron = 𝟗. 𝟏𝟎𝟗𝟏 × 𝟏𝟎−𝟑𝟏 kg ;
charge of electron e = 𝟏. 𝟔𝟎𝟐𝟏𝟎 × 𝟏𝟎−𝟏𝟗 C; permittivity of vacuum
𝛜𝟎 = 𝟖. 𝟖𝟓𝟒𝟏𝟖𝟓 × 𝟏𝟎−𝟏𝟐 kg−𝟏 m−𝟑 A𝟐 ൯ JEE Main PYQ
(1) 𝟏. 𝟔𝟓Å (2) 𝟒. 𝟕𝟔Å (3) 𝟎. 𝟓𝟐𝟗Å (4) 𝟐. 𝟏𝟐Å
Q) If the radius of the 3rd Bohr's orbit of hydrogen atom is r3 and the
radius of 4th Bohr's orbit is r4. Then :
𝟗 𝟏𝟔 𝟑 𝟒
𝐀 𝐫𝟒 = 𝐫𝟑 𝐁 𝐫𝟒 = 𝐫𝟑 𝐂 𝐫𝟒 = 𝐫𝟑 (𝐃)𝐫𝟒 = 𝐫𝟑
𝟏𝟔 𝟗 𝟒 𝟑

JEE Main PYQ

Ans. (B)
Energy of Electrons
Energy Negative????
E = KE + PE
E electron = 0 at ꚙ
Total Energy = − KE = PE/2

𝐙𝟐
𝐄 = −𝟏𝟑. 𝟔 × 𝟐 eV/atom
𝐧
Velocity of Electron

𝐙
𝐯 = 𝟐. 𝟏𝟖𝟖 × 𝟏𝟎𝟔 m sec−1
𝐧
Q) According to Bohr’s atomic theory :-
𝐙𝟐 JEE Main PYQ
(A) Kinetic energy of electron is ∝ 𝟐 .
𝐧
(B) The product of velocity (v) of electron and principal quantum
number 𝐧 , ′′𝐯𝐧′′ ∝ 𝐙𝟐
𝐙𝟑
(C) Frequency of revolution of electron in an orbit is ∝ 𝟑 .
𝐧
𝐙𝟑
(D) Coulombic force of attraction on the electron is ∝ .
𝐧𝟒
Choose the most appropriate answer from the options given below :
(A) C Only (B) A Only
(C) A, C and (D) only (D) A and D only

Ans. (D)
Ground State
Excited State
Excitation Energy Shell 5
Shell 4
Ionisation Energy (IE) Shell 3
Shell 2
Shell 1
Binding Energy
Nucleus +
Shell K
Shell L
Shell M
Shell N
Shell O
Energy states of atom other than
the ground state are called excited
states

n=2 first excited state

n=3 second excited state
n=4 third excited state
n=n+1 nth excited state
Hydrogen Line Spectrum

400 to 700 nm
Visible region
 5 lines 4 lines 3 lines 2 lines 1 line
6 lines
7 Q
6 P
Far I.R. region
Humphery series
5 O
I.R. region
0.31 eV
Pfund series
4 N
I.R. region 0.66 eV
Brackett series
Energy 3 M
Levels Infra Red region
or 1.89 eV
Paschen series
L
2
Visible region
or 10.2 eV
Balmer series
1 K
Ultra violet region
or
Lyman series
First line of any series =  line

Second line of any series =  line

Third line of any series =  line

Total no. of emission lines between

𝐧𝟐 − 𝐧𝟏 )(𝐧𝟐 − 𝐧𝟏 + 𝟏
n2 & n1 =
𝟐
(n2 > n1)
Q) Number of spectral lines obtained in H spectra, when an electron makes
transition from fifth excited state to first excited state will be
JEE Main PYQ

Ans. (10)
Q) Match List I with List II
List I List II (Spectral Region/Higher
(Spectral Series for Hydrogen) Energy State)
A. Lyman I. Infrared region JEE Main PYQ
B. Balmer II. UV region
C. Paschen III. Infrared region
D. Pfund IV. Visible region
Choose the correct answer from the options given below :-
(1) A-II, B-III, C-I, D-IV (2) A-I, B-III, C-II, D-IV
(3) A-II, B-IV, C-III, D-I (4) A-I, B-II, C-III, D-IV

Ans. (3)
Q) In a hydrogen spectrum if electron moves from 7 to 1 orbit by
transition in multi steps then find out the total number of lines
in the spectrum. JEE Main PYQ

Ans. (21)
Energy During Transitions
𝟏 𝟐
𝟏 𝟏
𝐯= = 𝐑𝐙 𝟐
− 𝟐
𝛌 𝐧𝟏 𝐧𝟐

value of R = 109737.31 cm–1

= 10973731m–1
≈ 10973700 m–1
Q) The wave number of the first emission line in the Balmer series
of H-Spectrum is : R = Rydberg constant : JEE Main PYQ
𝟑 𝟗 𝟓 𝟕
(1) R (2) R (3) R (4) R
𝟒 𝟒𝟎𝟎 𝟑𝟔 𝟔

Ans. (3)
Planck’s Theory

The radiant energy emitted or absorbed 𝐄 ∝ 𝛎

by a body is not radiated continuously
but discontinuously in the form of small 𝐄 = 𝐧𝐡𝛎
discrete packets of energy. These packets 𝐡𝐜
are called quantum. 𝐄 =
𝛌
𝐡 = 𝟔. 𝟔 × 𝟏𝟎−𝟑𝟒 𝐉𝐨𝐮𝐥𝐞 sec
 𝐡𝐜
𝐄=
𝛌
hc = 𝟔. 𝟔 × 𝟏𝟎−𝟑𝟒 x 3 x 108 Jm
eV = 1.6 x 10-19 J
Å = 10-19 m
hc = 12400 eV Å
Q) Calculate the energy of a photon of wave length 4000 Å
JEE Main PYQ
Photoelectric Effect

It is the phenomenon of emission

of electrons from the metal
surface when light falls on it.
Q) A source of monochromatic radiation of wavelength 400 nm provides
1000J of energy in 10 seconds. When this radiation falls on the surface of
sodium, x × 1020 electrons are ejected per second. Assume that wavelength
400 nm is sufficient for ejection of electron from the surface of sodium
metal. The value of x is______ (Nearest integer) JEE Main PYQ
–34
(h= 6.626 × 10 Js)

Ans. (2)
Albert Einstein
 Electrons which are knocked out of
the metal are called photoelectrons.

.. .... .. .. .... .... .... .... .. .. ..

.. .... .. .. .... .... .... .... .. .. ..
.
On increasing the intensity of light, photocurrent
increases but the kinetic energy of photo electrons
does not change.

Energy of photoelectrons will

depend on energy of photons!

𝐄𝐩𝐡𝐨𝐭𝐨𝐧 = 𝐡𝛎
 Ephoton Work Function

KE 𝐦𝐚𝐱 = h𝛎 − 𝛟
Work Function (ϕ)

𝐊𝐄𝐦𝐚𝐱 𝐞𝐕
The minimum energy per electron
given to the free electrons of the metal
which enables them to cross the energy
barrier present at the surface of the
metal, is called work function. 𝛎 𝐇𝐞𝐫𝐭𝐳
De Broglie’s Hypothesis

Louis de Broglie suggested that, like light,

matter also has dual character. It exhibits
Louis De Broglie
wave as well as particle nature. According to
de Broglie, the wavelength λ of a particle is
inversely proportional to its momentum p.
 𝐡
𝛌=
𝐩
Here h = Planck's constant
p = momentum of particle
 Momentum (p) = Mass (m) × Velocity (v)
 𝐡
𝛌=
𝟐𝐦 𝐊. 𝐄.

𝐡
𝛌= 𝟏𝟓𝟎
𝟐𝐦 𝐐𝐕 𝐅𝐨𝐫 𝐞𝐥𝐞𝐜𝐭𝐫𝐨𝐧 𝛌= Å
𝐕
Q) According to the wave-particle duality of matter by de-Broglie, which of
the following graph plot presents most appropriate relationship between
wavelength of electron (  ) and momentum of electron (p) ?
JEE Main PYQ
(1) (2)

(3) (4)

Ans. (1)
Q) A proton and a 𝐋𝐢𝟑+ nucleus are accelerated by the same potential. If 𝛌𝐋𝐢
and 𝛌𝐏 denote the de Broglie wavelengths of Li3+ and proton respectively,
𝛌𝐋
then the value of is × 10–1. The value of x is
𝛌𝐩
JEE Main PYQ
(Rounded off to the nearest integer)
(Mass of Li3+ = 8.3 mass of proton)

Ans. (2)
Q) A stream of electrons from a heated filament was passed between two
charged plates kept at a potential difference V esu. If e and m are charge
and mass of an electron respectively, then the value of hΤ𝛌 (𝐰𝐡𝐞𝐫𝐞 𝛌 is
wavelength associated with electron wave) is given by:
(1) 𝟐meV (2) meV (3) 𝟐meV (4) meV
JEE Main PYQ
𝐡
𝛌=
𝟐𝐦 𝐐𝐕

Ans. (1)
 𝐧𝐡
𝐦𝐯𝐫 =
𝟐𝛑

n=5
∴ Waves made =
5
A whole number of wavelength must fit
into the circumference of the circle (2).
Heisenberg’s Uncertainty Principle

It is impossible to measure
simultaneously the exact position
and exact momentum of a body.
 𝐡
𝛌=
𝐩
 𝐡
△ 𝐱.△ 𝐩 ≥
𝟒𝛑

𝐡
△ 𝐱. 𝐦 △ 𝐯 ≥
𝟒𝛑

𝐡
𝚫𝐱. 𝚫𝐯 ≥
𝟒𝛑𝐦
Q) In an atom, an electron is moving with a speed of 𝟔𝟎𝟎 mΤs with an
accuracy of 𝟎. 𝟎𝟎𝟓%. Certainty with which the position of the electron
can be located is ቀh = 𝟔. 𝟔 × 𝟏𝟎−𝟑𝟒 kg m𝟐 s−𝟏 , mass of electron,
𝐞𝐦 = 𝟗. 𝟏 × 𝟏𝟎−𝟑𝟏 kg൯: − JEE Main PYQ
(1) 𝟏. 𝟗𝟐 × 𝟏𝟎−𝟑 m (2) 𝟑. 𝟖𝟒 × 𝟏𝟎−𝟑 m
(3) 𝟏. 𝟓𝟐 × 𝟏𝟎−𝟒 m (4) 𝟓. 𝟏𝟎 × 𝟏𝟎−𝟑 m

Ans. (1)
Quantum-Mechanical Model
These regions of space around the nucleus
where probability of finding the electron is
maximum are called orbitals.

Orbital Orbit
Electron-cloud Bohr’s orbit
representation of an orbital
Quantum Numbers
 It represents shell number/energy level.
Quantum Number Maximum electrons that a shell can
Principal (n) accommodate = 2n2
The values of l depends upon the
Azimuthal (l) value of ‘n’ and possible values are
‘0’ to (n –1) .
The value of m depends upon the
Magnetic (m) value of l and it may have integral
value –l to +l including zero.
𝟏 𝟏
Spin (s) Two spins + &−
𝟐 𝟐
Q) The correct set of four quantum numbers for the valence electron of
rubidium atom (Z = 37) is:
𝟏 𝟏 𝟏 𝟏
(1) 𝟓, 𝟎, 𝟎, + (2)𝟓, 𝟎, 𝟏, + (3) 𝟓, 𝟏, 𝟎, + (4) 𝟓, 𝟏, 𝟏, +
𝟐 𝟐 𝟐 𝟐

JEE Main PYQ

Ans. (1)
Q) The number of electrons present in all the completely filled subshells
𝟏
having 𝐧 = 𝟒 and 𝐬 = + is________.
𝟐

JEE Main PYQ

Ans. (16)
Q) The four quantum numbers for the electron in the outer most orbital of
potassium (atomic no. 19) are
𝟏 𝟏
(1) 𝐧 = 𝟒, 𝐥 = 𝟐, 𝐦 = −𝟏, 𝐬 = + (2) 𝐧 = 𝟒, 𝐥 = 𝟎, 𝐦 = 𝟎, 𝐬 = +
𝟐 𝟐
𝟏 𝟏
(3) 𝐧 = 𝟑, 𝐥 = 𝟎, 𝐦 = 𝟏, 𝐬 = + (4) 𝐧 = 𝟐, 𝐥 = 𝟎, 𝐦 = 𝟎, 𝐬 = +
𝟐 𝟐
JEE Main PYQ

Ans. (2)
The orbitals having same value of
n and l but different value of m,
have same energy in absence of
external electric & magnetic field.

These orbitals having same energy

of a particular subshell are known
as Degenerate orbitals.
Radial Probability Distribution Function (RPDF)

Radial probability distribution = [R(r)]2 4r2dr

H Atom
 1s 1s

4πr2ψ2(r)
ψ2(r)

r rmax r
4πr2ψ2(r) 2s

rmax r Node
Q) Which of the following is the correct plot for the probability density 𝛙𝟐 𝐫
as a function of distance ‘ r ' of the electron form the nucleus for 2 s orbital?
JEE Main PYQ

(A) (B) (C) (D)

Ans. (B)
 2px

ψ(r)

r
 4πr2ψ2(r) 2p

rmax
r
4πr2ψ2(r)
3p

rmax
r

4πr2ψ2(r)
4p

rmax
r
Node

The place in three dimensional space

in which electron finding probability
is zero.
Number of radial nodes = (n – l – 1).

Number of angular nodes = l

Q) The number of radial nodes for 3p orbital is:
(1) 1 (2) 4 (3) 2 (4) 3

JEE Main PYQ

Ans. (1)
Q) The orbital having two radial as well as two angular nodes is –
(A) 3p (B) 4f (C) 4d (D) 5d
JEE Main PYQ

Ans. (D)
Pauli’s exclusion principle
No two electrons in an atom can
have same values of all the four
quantum numbers.

Aufbau Rule
Electrons are filled from lower
energy level to higher energy level.

If two orbitals have the same

(n + l) value, then orbital with
lower value of n has the lower
energy.
Hund’s rule of maximum multiplicity

Electrons are distributed among the

orbitals of a subshell in such a way as
to give the maximum number of
unpaired electron with parallel spins.
Exceptional Configuration

Chromium and Copper

Magnetic Moment
𝛍𝐞𝐟𝐟𝐞𝐜𝐭𝐢𝐯𝐞 = 𝐧(𝐧 + 𝟐)
Q) The spin only magnetic moments (in BM) for free Ti3+ , V2+ and Sc3+ ions
respectively are
(At . No. Sc: 21, Ti: 22, V: 23) JEE Main PYQ
(A) 3.87,1.73,0 (B) 1.73,3.87,0
(C) 1.73,0,3.87 (D) 0,3.87,1.73

Ans. (B)
Periodic Table
This is also known as
'Modern Periodic Law'.
Q) It is observed that characteristic X-ray spectra of elements show
regularity. When frequency to the power 'n' i.e. vn of X-rays emitted
is plotted against atomic number ' Z ', following graph is obtained.

JEE MAIN 2023 PYQ

The value of 'n' is

𝟏
(A) 1 (B) 2 (C) (D) 3
𝟐

Ans. (C)
Q) Element "E" belongs to the period 4 and group 16 of the periodic table.
The valence shell electron configuration of the element, which is just above
'E' in the group is
𝟐 𝟒 𝟏𝟎 𝟐 𝟒 JEE Main PYQ
(A) 𝟑𝐬 𝟑𝐩 (B) 𝟑𝐝 𝟒𝐬 , 𝟒𝐩
(C) 𝟒𝐝𝟏𝟎 𝟓𝐬𝟐 𝟓𝐩𝟒 (D) 𝟐 𝐬𝟐 , 𝐩𝟒

Ans. (A)
 Nomenclature of elements
0 Nil
1 Un
2 Bi
3 Tri
4 Quad
5 Pent 101 unnilunium Unu
6 Hex 102 unnilbium Unb
7 Sept 103 unniltrium Unt
8 Oct
9 Enn
Q) The IUPAC symbol for the element with atomic number 119 would be :
(1) uue (2) une (3) unh (4) uun
JEE Main PYQ

Ans. (1)
Size
Slaters rule
e‾- e‾ repulsion

Zeff = Z – σ
(a)Zeff increases, atomic radius (b) Number of shell (n) increases,
decreases atomic radius increases
Li < Na < K < Rb < Cs

Li > Be > B > C > N > O > F

Ionic Size

O < O– < O–2 Mn > Mn+2 > Mn+3 > Mn+4

Isoelectronic Species

Na+ < F– < O–2

Q) The ionic radii of 𝐎𝟐− , 𝐅 − , 𝐍𝐚+ 𝐚𝐧𝐝 𝐌𝐠 𝟐+ 𝐚𝐫𝐞 𝐢𝐧 𝐭𝐡𝐞 𝐨𝐫𝐝𝐞𝐫:
𝟏 𝐅 − > 𝐎𝟐− > 𝐍𝐚+ > 𝐌𝐠 𝟐+ 𝟐 𝐎𝟐− > 𝐅 − > 𝐍𝐚+ > 𝐌𝐠 𝟐+
𝟑 𝐌𝐠 𝟐+ > 𝐍𝐚+ > 𝐅 − > 𝐎𝟐− 𝟒 𝐎𝟐− > 𝐅 − > 𝐌𝐠 𝟐+ > 𝐍𝐚+

Ans. (2)
Q) The correct order of increasing ionic radii is
(A) Mg2+ < Na+ < F− < O2− < N3− JEE Main PYQ
(B) N3− < O2− < F− < Na+ < Mg2+
(C) F− < Na+ < O2− < Mg2+ < N3−
(D) Na+ < F− < Mg2+ < O2− < N3−

Ans. (B)
Q) The correct order of the atomic radii of C, Cs , Al, and S is :
(1) C < S < Al < Cs
JEE Main PYQ
(2) S < C < Cs < Al
(3) S < C < Al < Cs
(4) C < S < Cs < Al

Ans. (1)
 190
180
170

Radius/pm
160
150 Radius of Ga < Radius of Al
140
130
120
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
La Hf Ta W Re Os Ir Pt Au Hg
Trends in atomic radii of
transition elements
Effect of Lanthanides cancelling out the effect of
last shell added in the sixth period and therefore
the transition series 4d and 5d elements having
the same size is called as Lanthanide contraction
Q) The atomic radius of Ag is closest to:
(1) Au (2) Ni (3) Cu (4) Hg
JEE Main PYQ

Ans. (1)
 𝟏
𝐈𝐨𝐧𝐢𝐬𝐚𝐭𝐢𝐨𝐧 𝐄𝐧𝐞𝐫𝐠𝐲 ∝
𝐚𝐭𝐨𝐦𝐢𝐜 𝐬𝐢𝐳𝐞

𝐈𝐨𝐧𝐢𝐬𝐚𝐭𝐢𝐨𝐧 𝐄𝐧𝐞𝐫𝐠𝐲 ∝ 𝐙𝐞𝐟𝐟

s>p>d>f

𝟏
𝐑𝐞𝐚𝐜𝐭𝐢𝐯𝐢𝐭𝐲 𝐨𝐟 𝐦𝐞𝐭𝐚𝐥𝐬 ∝
𝐈𝐄
Charge Stable EC
Fe+3 > Fe+2 > Fe IE1 N > O
1s2, 2s2 2p3 1s2, 2s2 2p4
Ionisation Energy Ga > Al (due to poor shielding of 3d)

Ionisation Energy of 5d > 4d (due to lanthanide contraction)

Ex. Hf > Zr
Q) Identify the elements X and Y using the ionisation energy values given
below: Ionization energy (kJ/mol) JEE Main PYQ
1st 2nd
X 495 4563
Y 731 1450

(1) X = Na ; Y = Mg (2) X = Mg ; Y = F
(3) X = Mg ; Y = N (4) X = F ; Y = Mg

Ans (1)
Q) Match the column.
Column-I Column-II
Valence electronic configuration Successive ionisation energies
(a) ns1 (p) 19, 27, 36, 48, 270
(b) ns2 (q) 16, 28, 34, 260
(c) ns2 np1 (r) 18, 26, 230, 250
(d) ns2 np2 (s) 14, 200, 220, 240
Q) Consider the elements Mg, Al, S, P and Si, the correct increasing order of
their first ionization enthalpy is:
(1) Al < Mg < Si < S < P
(2) Al < Mg < S < Si < P
(3) Mg < Al < Si < S < P
(4) Mg < Al < Si < P < S

Ans (1)
Q) The correct order of first ionization enthalpy values of the following
elements is: JEE Main PYQ
(1) O (2) N (3) Be (4) F
(5) B
Choose the correct answer from the options given below:
(A) 2 < 4 < 3 < 5 < 1 (B) 5 < 3 < 1 < 2 < 4
(C) 3 < 5 < 1 < 2 < 4 (D) 1 < 2 < 4 < 3 < 5

Ans. (B)
Electron Affinity = -EGE
 𝟏
𝐄𝐀 ∝
𝐒𝐢𝐳𝐞

(a) In period (b) In Group

EGE becomes more negative In a group, the electron affinity
generally as we go from left to decreases on moving from top to
right in a period. bottom.

EGE of N, Be, Ne are positive

Electron affinity of 3rd period element is greater than electron
affinity of 2nd period elements of the respective group.
F Cl
[He] 2s22p5 [Ne] 3s23p5
Q) The absolute value of the electron gain enthalpy of halogens satisfies:
(𝐀) I > Br > Cl > F (𝐁) Cl > Br > F > I JEE Main PYQ
(𝐂) Cl > F > Br > I (𝐃) F > Cl > Br > I

Ans. (C)
Q) The correct order of electron gain enthalpies of Cl, F, Te and Po is
(A) F < Cl < Te < Po (B) Po < Te < F < Cl
(C) Te < Po < Cl < F (D) Cl < F < Te < Po
JEE Main PYQ

Ans. (B)
Q) The correct order of electron gain enthalpy is
(A) S > O > Se > Te (B) O > S > Se > Te
JEE Main PYQ
(C) S > Se > Te > O (D) Te > Se > S > O

Ans. (C)
Q) The set of elements that differ in mutual relationship from those of
the other sets is :
(1) Li – Mg (2) B – Si (3) Be – Al (4) Li – Na
JEE Main PYQ

Ans. (4)
 𝟏
𝐄𝐥𝐞𝐜𝐭𝐫𝐨𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐢𝐭𝐲 ∝
𝐀𝐭𝐨𝐦𝐢𝐜 𝐬𝐢𝐳𝐞
𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐜𝐡𝐚𝐫𝐠𝐞
𝐄𝐥𝐞𝐜𝐭𝐫𝐨𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐢𝐭𝐲 ∝ 𝐙𝐞𝐟𝐟 ∝
𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞 𝐜𝐡𝐚𝐫𝐠𝐞

(a)In Group:
Electronegativity decreases down
the group.

(b) In period:
On moving from left to right
electronegativity increases.
E.N. Fe+3 > Fe+2 > Fe N3– < N‾ < N
Nature of oxides

a). Along a period: Acidic nature increases.

b). Down the group: Basic nature increases

Li Be B C N O F
Na Mg Al Si P S Cl

Basic Amphoteric Acidic

 𝐀𝐜𝐢𝐝𝐢𝐜 𝐍𝐚𝐭𝐮𝐫𝐞 ∝ 𝐎𝐱𝐢𝐝𝐚𝐭𝐢𝐨𝐧 𝐬𝐭𝐚𝐭𝐞

SO3 > SO2

N2O5 > N2O3

Q) Match List-I with List-II.

JEE Main PYQ

Choose the correct answer from the options given below:

(A) (A) – (IV), (B) – (III), (C) – (I), (D) – (II)
(B) (A) – (IV), (B) – (II), (C) – (I), (D) – (III)
(C) (A) – (II), (B) – (IV), (C) – (III), (D) – (I)
(D) (A) – (I), (B) – (II), (C) – (IIII), (D) – (IV)

Ans. (B)
Q) Among the following basic oxide is :
(A) SO3 (B) SiO2 (C) CaO (D) Al2O3
JEE Main PYQ

Ans. (C)
Q) Given below are two statements:
Statement I : The metallic radius of Na is 1.86 A° and the ionic radius of
Na+ is lesser than Na.
Statement II : Ions are always smaller in size than the corresponding
elements.
In the light of the above statements, choose the correct answer from the
options given below:
JEE Main PYQ
(A) Statement I is correct but Statement II is false
(B) Both Statement I and Statement II are true
(C) Both Statement I and Statement II are false
(D) Statement I is incorrect but Statement II is true

Ans. (A)
 Sequence and Series
A.P. G.P. H.P.

AM, GM, HM

Special
A.G.P.
Sequences
 A.P. G.P.
a + (a+d) + (a+2d) …. – a + (ar) + (ar2) …
𝐓𝐓𝟐𝟐 𝐓𝐓𝟑𝟑 𝐓𝐓𝐧𝐧
d = T2 – T1 = T3 – T2 = ..... = Tn – Tn–1 𝐫𝐫 = = = ......=
𝐓𝐓𝟏𝟏 𝐓𝐓𝟐𝟐 𝐓𝐓𝐧𝐧−𝟏𝟏

If d > or < or = 0 If r > 1 or ∈ (0,1) or < 0

⇒ Incg, Decg, Const AP ⇒ Incg, Decg, Alter + - GP
 A.P. G.P.
a + (a+d) + (a+2d) …. – a + (ar) + (ar2) …

Tn = Sn – Sn–1
 A.P. G.P.
𝐧𝐧
Sn = [𝟐𝟐𝐚𝐚 + 𝐧𝐧 − 𝟏𝟏 𝐝𝐝] 𝐚𝐚(𝟏𝟏 − 𝐫𝐫 𝐧𝐧 ) 𝐚𝐚(𝐫𝐫 𝐧𝐧 −𝟏𝟏)
𝟐𝟐 Sn = = , 𝐫𝐫 ≠ 𝟏𝟏
𝟏𝟏 − 𝐫𝐫 𝐫𝐫−𝟏𝟏
𝐧𝐧
Sn = [𝐚𝐚 + 𝐥𝐥]
𝟐𝟐 If −1 < r < 1 and n → ∞, then rn → 0
Sn = Pn2 + Qn 𝐚𝐚
𝐒𝐒∞ = ; 𝐫𝐫 < 𝟏𝟏
𝟏𝟏−𝐫𝐫
d = 2[Coefficient of ‘n2’] = 2P
 Properties of an A.P. Properties of G.P.
P-1 If a, b, c are in AP, then If a, b, c are in G.P., then
𝐛𝐛 𝐜𝐜
⇒ b – a = c – b ⇒ 2b = a + c = ⇒ b2 = ac
𝐚𝐚 𝐛𝐛

P-2 If each term of an A.P. is increased or If each term of an G.P. is increased or

decreased, multiplied, divided by the decreased, multiplied, divided by the
same non-zero (k), then the resulting same non-zero (k), then the resulting
sequence is also an A.P. sequence is also an G.P.
P-3

1 3 5 7 9 11 13 1 3 9 27 81 243
243
14
14 243
14 243
In an A.P., summation of kth term In an G.P., Product of kth term
from beginning and kth term from from beginning and kth term from
the last is always constant which the last is always constant which is
is equal to summation of first and equal to Product of first and last
last terms. terms.
Tk + 𝐓𝐓𝐧𝐧−𝐤𝐤+𝟏𝟏 = constant = 𝐚𝐚 + 𝓵𝓵 Tk . 𝐓𝐓𝐧𝐧−𝐤𝐤+𝟏𝟏 = constant = 𝐚𝐚 . 𝓵𝓵
P-4

1 3 5 7 9 11 13 a , b , c , ….. are in G.P , then

ak , bk , ck , ….. are also in G.P
14
14
14
Any term of an AP (except the If a , b, c .. is a G.P. of positive
first & last) is equal to half the terms, then
sum of terms which are log(a) , log(b) , log(c) ..
equidistant from it. will be in A.P.
 Supposition of terms
A.P. G.P.
𝐚𝐚
(A) ODD terms , a , ar
𝐫𝐫
a – d, a, a + d
𝐚𝐚 𝐚𝐚
a – 2d, a – d, a, a + d, a + 2d , , a, ar , ar2
𝐫𝐫𝟐𝟐 𝐫𝐫
(B) EVEN
𝐚𝐚 𝐚𝐚 , ar, ar3
a – 3d, a – d, a + d, a + 3d , 𝐫𝐫
𝐫𝐫 𝟑𝟑

NOTE : Here, 'a' is not the first term

Arithmetic Mean (AM) Geometric Mean (GM)
𝐚𝐚 + 𝐛𝐛 𝐆𝐆 = 𝐚𝐚𝐚𝐚
𝐀𝐀 =
𝟐𝟐 a, b Positive real numbers
𝟏𝟏
a1 +a2 +a3 +......+an 𝐆𝐆𝐧𝐧 = 𝐚𝐚𝟏𝟏 . 𝐚𝐚𝟐𝟐 . 𝐚𝐚𝟑𝟑 . . . 𝐚𝐚𝐧𝐧 𝐧𝐧
An =
n
 Insertion of 'n' A.M.'s between two numbers ‘a’ and ‘b’
a , A1 , A2 , A3 , ....., An, b are in A.P. a, G1 , G2 , ...., Gn , b are in G.P.,

⇒ b = a + [(n+2) − 1]d ∴ b = ar(n + 2)- 1

𝟏𝟏
𝐛𝐛−𝐚𝐚 𝐛𝐛
⇒d= 𝐫𝐫 =
𝐧𝐧+𝟏𝟏
𝐧𝐧+𝟏𝟏 𝐚𝐚
A1 = a + d G1 = ar
A2 = a + 2d G2 = ar2
An = a + nd Gn = arn
n 𝐧𝐧
𝐧𝐧
� Ai = nA � 𝐆𝐆i = 𝐚𝐚𝐚𝐚 = Gn
i=1 i=𝟏𝟏
Harmonic Progression (H.P.)
a1, a2, a3, ..... are in H.P., if and only if

𝟏𝟏 𝟏𝟏 𝟏𝟏
, , , ..... are in A.P.
𝐚𝐚𝟏𝟏 𝐚𝐚𝟐𝟐 𝐚𝐚𝟑𝟑
Harmonical Mean (H.M.)
Single H.M. between two numbers a and b

𝟐𝟐𝟐𝟐𝟐𝟐
Η=
𝐚𝐚 + 𝐛𝐛

Harmonic Mean of 'n' positive real numbers a1, a2, a3, ........ an
𝐧𝐧
𝐇𝐇𝐧𝐧 =
𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
+ + .....
𝐚𝐚𝟏𝟏 𝐚𝐚𝟐𝟐 𝐚𝐚𝟑𝟑 𝐚𝐚𝐧𝐧
 Relation Between A.M., G.M. & H.M.
If a1, a2, a3 ..... an are
'n’ positive numbers, then

𝐀𝐀. 𝐌𝐌 ≥ 𝐆𝐆. 𝐌𝐌 ≥ 𝐇𝐇. 𝐌𝐌

Note: Moreover equality holds only
when a1 = a2 = a3 = ........ = an .
Q) Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is
(A) 30 (B) 32
[JEE Main PYQ]
(C) 36 (D) 40
 𝟏𝟏𝟏𝟏
Q) Find sum of 10 + 100 + 1000 + 10000 ...... upto n terms . = [𝟏𝟏𝟏𝟏𝐧𝐧 − 𝟏𝟏]
(Chupa Rustam)
𝟗𝟗

𝟏𝟏𝟏𝟏
Find sum of 9 + 99 + 999 + ...... upto n terms . = [𝟏𝟏𝟏𝟏𝐧𝐧 − 𝟏𝟏] − 𝐧𝐧
𝟗𝟗

Q) Find sum of 5 + 55 + 555 + ...... upto n terms .

 𝟏𝟏𝟏𝟏
Q) If S = 11 + 103 + 1005 + ⋯ 𝐧𝐧 terms Sn = 𝟏𝟏𝟏𝟏𝐧𝐧 − 𝟏𝟏 + 𝐧𝐧𝟐𝟐 .
𝟗𝟗

𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
Q) + 𝟐𝟐 + 𝟑𝟑 + 𝟒𝟒 + 𝟓𝟓 + 𝟔𝟔 +. . . . . ∞ = ?
𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑

Q) 1002 – 992 + 982 – 972 + 962 – 952 + ....... + 22 – 12 = ?

 Special Sequences
𝐧𝐧 𝐧𝐧
(a) � 𝐤𝐤 = 𝐤𝐤 � 𝟏𝟏 = 1 + 1 + 1 + ……. + 1 = 𝐤𝐤 . 𝐧𝐧
𝐫𝐫 = 𝟏𝟏 𝐫𝐫 = 𝟏𝟏
𝐧𝐧
𝐧𝐧(𝐧𝐧 + 𝟏𝟏)
(b) 𝟏𝟏 + 𝟐𝟐 + 𝟑𝟑 + 𝟒𝟒 + . . . . . . . . . . . . + 𝐧𝐧 = � 𝐫𝐫 =
𝟐𝟐
𝐫𝐫 =𝟏𝟏

𝐧𝐧

(c) 𝟏𝟏𝟐𝟐 + 𝟐𝟐𝟐𝟐 + 𝟑𝟑𝟐𝟐 +. . . . . . . . . . . . + 𝐧𝐧𝟐𝟐 = � 𝐫𝐫 𝟐𝟐 = 𝐧𝐧 𝐧𝐧+𝟏𝟏 𝟐𝟐𝟐𝟐+𝟏𝟏

𝟔𝟔
𝐫𝐫 = 𝟏𝟏
 𝐧𝐧 𝐧𝐧 𝟐𝟐
𝐫𝐫 𝟑𝟑 𝐧𝐧(𝐧𝐧+𝟏𝟏) 𝟐𝟐
(d) 𝟏𝟏𝟑𝟑 + 𝟐𝟐𝟑𝟑 + 𝟑𝟑𝟑𝟑 + . . . . . . . . . + 𝐧𝐧𝟑𝟑 = � = � 𝐧𝐧 =
𝟐𝟐
𝐫𝐫 = 𝟏𝟏 𝐫𝐫 = 𝟏𝟏

(e) 1 + 3 + 5 + .….. + (2n – 1) = 𝒏𝒏𝟐𝟐

(f) 2 + 4 + 6 + .….. + 2n = n(n + 1)

 Special Sequences
𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
+ + + . . . . . 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝐧𝐧 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 = ? = −
𝟏𝟏. 𝟐𝟐 𝟐𝟐. 𝟑𝟑 𝟑𝟑. 𝟒𝟒 𝟏𝟏 𝐧𝐧 + 𝟏𝟏

𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
+ + + ........ ? =
𝟐𝟐 𝟏𝟏.𝟐𝟐
−
𝐧𝐧+𝟏𝟏)(𝐧𝐧+𝟐𝟐
𝟏𝟏⋅𝟐𝟐⋅𝟑𝟑 𝟐𝟐⋅𝟑𝟑⋅𝟒𝟒 𝟑𝟑⋅𝟒𝟒⋅𝟓𝟓
 Special Sequences
𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
+ + + . . . . . 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝐧𝐧 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 = ? = −
𝟏𝟏. 𝟐𝟐 𝟐𝟐. 𝟑𝟑 𝟑𝟑. 𝟒𝟒 𝟏𝟏 𝐧𝐧 + 𝟏𝟏

𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏
+ + + ........ ? =
𝟐𝟐 𝟏𝟏.𝟐𝟐
−
𝐧𝐧+𝟏𝟏)(𝐧𝐧+𝟐𝟐
𝟏𝟏⋅𝟐𝟐⋅𝟑𝟑 𝟐𝟐⋅𝟑𝟑⋅𝟒𝟒 𝟑𝟑⋅𝟒𝟒⋅𝟓𝟓

𝟏𝟏 𝟏𝟏 𝟏𝟏
Short Trick =
𝐋𝐋𝐋𝐋−𝐅𝐅𝐅𝐅 𝐅𝐅𝐅𝐅.𝐒𝐒𝐒𝐒….
−
….. 𝐒𝐒𝐒𝐒𝐒𝐒)(𝐋𝐋𝐋𝐋
 𝟏𝟏 𝟏𝟏 𝟏𝟏
Q) + + + ....=?
𝟏𝟏⋅𝟐𝟐⋅𝟑𝟑⋅𝟒𝟒 𝟐𝟐⋅𝟑𝟑⋅𝟒𝟒⋅𝟓𝟓 𝟑𝟑⋅𝟒𝟒⋅𝟓𝟓⋅𝟔𝟔

𝟏𝟏 𝟏𝟏 𝟏𝟏
= −
𝟑𝟑 𝟏𝟏.𝟐𝟐.𝟑𝟑 𝐧𝐧+𝟏𝟏)(𝐧𝐧+𝟐𝟐)(𝐧𝐧+𝟑𝟑
 Special Sequences
Q) The sum of n term of 𝟏𝟏. 𝟐𝟐. 𝟑𝟑 + 𝟐𝟐. 𝟑𝟑. 𝟒𝟒 + 𝟑𝟑. 𝟒𝟒. 𝟓𝟓 + . . . . .
𝐧𝐧

is equal to ? OR � 𝐫𝐫) (𝐫𝐫 + 𝟏𝟏) (𝐫𝐫 + 𝟐𝟐 = ?

𝐫𝐫=𝟏𝟏

Sol.
Tn = n(n + 1)(n + 2)= n[n2 + 3n + 2]
= n3 + 3n2 + 2n
𝐒𝐒𝐧𝐧 = ∑ 𝐓𝐓𝐧𝐧 = � 𝐧𝐧𝟑𝟑 + 𝟑𝟑𝐧𝐧𝟐𝟐 + 𝟐𝟐𝟐𝟐
𝐧𝐧(𝐧𝐧+𝟏𝟏) 𝟐𝟐 𝟑𝟑𝟑𝟑(𝐧𝐧+𝟏𝟏)(𝟐𝟐𝟐𝟐+𝟏𝟏) 𝟐𝟐𝟐𝟐(𝐧𝐧+𝟏𝟏)
= + + 𝟐𝟐
𝟐𝟐 𝟔𝟔
𝟏𝟏
= 𝐧𝐧(𝐧𝐧 + 𝟏𝟏)(𝐧𝐧 + 𝟐𝟐)(𝐧𝐧 + 𝟑𝟑)
𝟒𝟒
Q) 𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ 𝟒𝟒 + 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ 𝟒𝟒 ⋅ 𝟓𝟓 + 𝟑𝟑 ⋅ 𝟒𝟒 ⋅ 𝟓𝟓 ⋅ 𝟔𝟔 + . . . . . . . . . upto n terms ?

𝟏𝟏
Sn = [n(n + 1)(n + 2)(n + 3)(n + 4)]
𝟓𝟓
 Arithmetico - Geometric Progression (AGP)
General Form of an A.G.P.

Let a, (a + d), (a + 2d), ...., [a + (n – 1)d] , ....... are in A.P.

b, br, br2, ...., brn–1 , ....... are in G.P.
a.b , (a + d).br , (a + 2d).br2, ...., [a + (n – 1)d].brn–1 , .......
are in A.G.P.
General Term of A.G.P.
 𝟏𝟏 𝟏𝟏 𝟏𝟏
Q) If 𝟖𝟖 = 𝟑𝟑 + 𝟑𝟑 + 𝐩𝐩 + 𝟑𝟑 + 𝟐𝟐𝟐𝟐 + 𝟑𝟑 𝟑𝟑 + 𝟑𝟑𝟑𝟑 + ⋯ ∞, then the value of
𝟒𝟒 𝟒𝟒𝟐𝟐 𝟒𝟒
p is [JEE Main PYQ]
Ans. (9)
 Determinants
Value, Minor, Cofactors
Properties (7)
Multiplication
Differentiation
Cramer Rule
Value of the Determinant
𝐱𝐱 = 𝐱𝐱
𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏
= a1b2 − a2b1
𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐

𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏 𝐜𝐜𝟏𝟏 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 𝐚𝐚𝟐𝟐 𝐜𝐜𝟐𝟐 𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐

𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 = 𝐚𝐚𝟏𝟏 𝐛𝐛 𝐜𝐜 −𝐛𝐛𝟏𝟏 𝐚𝐚 𝐜𝐜𝟑𝟑
+𝐜𝐜𝟏𝟏
𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑
𝟑𝟑 𝟑𝟑 𝟑𝟑
𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑 𝐜𝐜𝟑𝟑
Minor (𝐌𝐌𝐢𝐢𝐢𝐢 )
 Cofactors
𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟐𝟐 𝐚𝐚𝟏𝟏𝟑𝟑
D = 𝐚𝐚𝟐𝟐𝟏𝟏 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟑𝟑 𝐂𝐂𝐢𝐢𝐢𝐢 = −𝟏𝟏 𝐢𝐢+𝐣𝐣 . 𝐌𝐌
𝐢𝐢𝐢𝐢
𝐚𝐚𝟑𝟑𝟏𝟏 𝐚𝐚𝟑𝟑𝟐𝟐 𝐚𝐚𝟑𝟑𝟑𝟑

∴ Cofactor is nothing but minor with a proper sign.

𝟏𝟏+𝟐𝟐 . 𝐌𝐌 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐

𝐂𝐂𝟏𝟏𝟏𝟏 = −𝟏𝟏 𝟏𝟏𝟏𝟏 = (−𝟏𝟏)
𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑
 3 Magics of determinant

01 Sum of product of elements of any row (column) with their

corresponding cofactors is equal to the value of DETERMINANT.

𝐃𝐃 = 𝐚𝐚𝟏𝟏𝟏𝟏 𝐂𝐂𝟏𝟏𝟏𝟏 + 𝐚𝐚𝟏𝟏𝟐𝟐 𝐂𝐂𝟏𝟏𝟏𝟏 + 𝐚𝐚𝟏𝟏𝟑𝟑 𝐂𝐂𝟏𝟏𝟏𝟏

02 Sum of product of elements of any row (column) with cofactors of

corresponding elements of any other row (column) is ZERO.

⇒ 𝐚𝐚𝟏𝟏𝟏𝟏 𝐂𝐂𝟐𝟐𝟏𝟏 + 𝐚𝐚𝟏𝟏𝟏𝟏 𝐂𝐂𝟐𝟐𝟐𝟐 + 𝐚𝐚𝟏𝟏𝟏𝟏 𝐂𝐂𝟐𝟐𝟑𝟑 = 𝟎𝟎

 𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏
03 𝐃𝐃 = 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐
𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑

𝐂𝐂𝟏𝟏𝟏𝟏 𝐂𝐂𝟏𝟏𝟏𝟏 𝐂𝐂𝟏𝟏𝟏𝟏

𝐃𝐃𝐂𝐂 = 𝐂𝐂𝟐𝟐𝟐𝟐 𝐂𝐂𝟐𝟐𝟐𝟐 𝐂𝐂𝟐𝟐𝟐𝟐
𝐂𝐂𝟑𝟑𝟑𝟑 𝐂𝐂𝟑𝟑𝟑𝟑 𝐂𝐂𝟑𝟑𝟑𝟑

Then, 𝐃𝐃𝐂𝐂 = 𝐃𝐃𝟐𝟐

∴ In general,
𝐃𝐃𝐂𝐂 = 𝐃𝐃𝐧𝐧−𝟏𝟏
Where ‘n’ is the order of determinant.
Properties of Determinants

1 The value of a determinant remains unchanged, if the

rows & columns are interchanged.
𝐃𝐃 = 𝐃𝐃T

2 If any two rows (or columns) of a determinant are interchanged, the

value of determinant is changed in sign only.

3 If a determinant has any two rows (or columns)

identical, then its value is ZERO.
4 If all the elements of any row (or column) are multiplied by the same
number, then value of the determinant is multiplied by that number.
Kachra
 𝐚𝐚 𝐛𝐛 𝐜𝐜 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐
Q) If 𝐝𝐝 𝐞𝐞 𝐟𝐟 = 𝐱𝐱 then find 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑 in terms of x.
𝐠𝐠 𝐡𝐡 𝐢𝐢 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒 𝟒𝟒𝟒𝟒

= 𝟒𝟒. 𝟑𝟑. 𝟐𝟐. 𝐱𝐱 = 𝟐𝟐𝟐𝟐𝐱𝐱

 If each element of any row (or column) can be expressed as a sum
5
of two or more terms, then the determinant can be expressed as
the sum of two or more determinants.

= +
 𝟎𝟎 𝟏𝟏 𝟎𝟎 𝟏𝟏⁄𝟐𝟐 𝟎𝟎 𝟏𝟏⁄𝟒𝟒
Q) Find 𝐃𝐃 = + + + ........∞
𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑 𝟐𝟐 𝟑𝟑

𝐀𝐀𝐀𝐀𝐀𝐀. −𝟒𝟒
 𝐧𝐧(𝐧𝐧 + 𝟏𝟏)
𝐫𝐫 𝐱𝐱 𝐧𝐧
𝟐𝟐
Q) Let 𝐃𝐃𝐫𝐫 = 𝟐𝟐𝟐𝟐 − 𝟏𝟏 𝐲𝐲 𝐧𝐧𝟐𝟐 , then find the value of � 𝐃𝐃𝐫𝐫
𝟐𝟐
𝐧𝐧(𝐧𝐧 + 𝟏𝟏)(𝟐𝟐𝟐𝟐 + 𝟏𝟏) 𝐫𝐫=𝟏𝟏
𝐫𝐫 𝐳𝐳
𝟔𝟔 [JEE Main PYQ]
Ans. (0)
6 If a constant multiple of any row or column is added to or
subtracted from any other row or column, then the value
of determinant remains unchanged.

Note :

While applying this property ATLEAST ONE

ROW (OR COLUMN) must remain unchanged.
 If all the rows (or columns) of a determinant are in A.P., then
7
value of the determinant is ZERO.
 𝟏𝟏 𝐱𝐱 𝐱𝐱 + 𝟏𝟏
Q) 𝐈𝐈𝐟𝐟 𝐟𝐟 𝐱𝐱 = 𝟐𝟐𝟐𝟐 𝐱𝐱(𝐱𝐱 − 𝟏𝟏) 𝐱𝐱(𝐱𝐱 + 𝟏𝟏) 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 𝐟𝐟 𝟏𝟏𝟏𝟏𝟏𝟏 = ?
𝟑𝟑𝟑𝟑(𝐱𝐱 − 𝟏𝟏) 𝐱𝐱(𝐱𝐱 − 𝟏𝟏)(𝐱𝐱 − 𝟐𝟐) 𝐱𝐱(𝐱𝐱 + 𝟏𝟏)(𝐱𝐱 − 𝟏𝟏)

(A) 0 (B) 1 (C) 100 (D) 1003

⇒ 𝐟𝐟(𝟏𝟏𝟏𝟏𝟏𝟏) = 𝟎𝟎
Special Determinant (We may use Factor Theorem)

𝟏𝟏 𝐱𝐱 𝐱𝐱 𝟐𝟐
𝟏𝟏 𝐲𝐲 𝐲𝐲 𝟐𝟐 = (𝐱𝐱 − 𝐲𝐲)(𝐲𝐲 − 𝐳𝐳)(𝐳𝐳 − 𝐱𝐱)
𝟏𝟏 𝐳𝐳 𝐳𝐳 𝟐𝟐

𝟏𝟏 𝐱𝐱 𝐱𝐱 𝟑𝟑
𝟏𝟏 𝐲𝐲 𝐲𝐲 𝟑𝟑 = 𝐱𝐱 − 𝐲𝐲 𝐲𝐲 − 𝐳𝐳 𝐳𝐳 − 𝐱𝐱 (x + y + z)
𝟏𝟏 𝐳𝐳 𝐳𝐳 𝟑𝟑

𝐚𝐚 𝐛𝐛 𝐜𝐜
𝐛𝐛 𝐜𝐜 𝐚𝐚 = (3abc – a3 – b3 – c3)
𝐜𝐜 𝐚𝐚 𝐛𝐛
Multiplication of Determinants
𝐚𝐚 𝐛𝐛 𝐩𝐩 𝐪𝐪
𝐃𝐃𝟏𝟏 = 𝐃𝐃𝟐𝟐 =
𝐜𝐜 𝐝𝐝 𝐫𝐫 𝐬𝐬
𝐃𝐃 = 𝐃𝐃𝟏𝟏 × 𝐃𝐃𝟐𝟐

𝐃𝐃 =

(Row by column multiplication)

Multiplication can also be done
1) Row by Row
2) Column by Row
3) Column by Column
Differentiation of Determinant :
 𝐱𝐱𝟐𝟐 + 𝟑𝟑𝟑𝟑 𝐱𝐱𝟐𝟐 + 𝟏𝟏 𝐱𝐱 − 𝟏𝟏
Q) If 𝐚𝐚𝐚𝐚 𝟒𝟒 + 𝐛𝐛𝐱𝐱 𝟑𝟑 + 𝐜𝐜𝐱𝐱 𝟐𝟐 + 𝐝𝐝𝐝𝐝 + 𝐞𝐞 = 𝐱𝐱 − 𝟐𝟐 𝟑𝟑𝟑𝟑 𝐱𝐱 + 𝟐𝟐 then :
𝐱𝐱 + 𝟏𝟏 𝟐𝟐𝟐𝟐 𝐱𝐱 − 𝟑𝟑
𝐞𝐞 = ? And d = ?
 𝐱𝐱𝟐𝟐 + 𝟑𝟑𝟑𝟑 𝐱𝐱𝟐𝟐 + 𝟏𝟏 𝐱𝐱 − 𝟏𝟏
Q) If 𝐚𝐚𝐚𝐚 𝟒𝟒 + 𝐛𝐛𝐱𝐱 𝟑𝟑 + 𝐜𝐜𝐱𝐱 𝟐𝟐 + 𝐝𝐝𝐝𝐝 + 𝐞𝐞 = 𝐱𝐱 − 𝟐𝟐 𝟑𝟑𝟑𝟑 𝐱𝐱 + 𝟐𝟐 then :
𝐱𝐱 + 𝟏𝟏 𝟐𝟐𝟐𝟐 𝐱𝐱 − 𝟑𝟑

𝟎𝟎 𝟏𝟏 −𝟏𝟏
e = −𝟐𝟐 𝟎𝟎 𝟐𝟐 =−4
𝟏𝟏 𝟎𝟎 −𝟑𝟑

𝟑𝟑 𝟎𝟎 𝟏𝟏 𝟎𝟎 𝟏𝟏 −𝟏𝟏 𝟎𝟎 𝟏𝟏 −𝟏𝟏
⇒ 𝐝𝐝 = −𝟐𝟐 𝟎𝟎 𝟐𝟐 + 𝟏𝟏 𝟑𝟑 𝟏𝟏 + −𝟐𝟐 𝟎𝟎 𝟐𝟐 = 0+7+8 = 15
𝟏𝟏 𝟎𝟎 −𝟑𝟑 𝟏𝟏 𝟎𝟎 −𝟑𝟑 𝟏𝟏 𝟐𝟐 𝟏𝟏
 Non – Homogeneous System Cramer’s Rule
a1x + b1y + c1z = d1 .... (i)
a2x + b2y + c2z = d2 .... (ii)
a3x + b3y + c3z = d3 .... (iii)

𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏 𝐜𝐜𝟏𝟏 𝐝𝐝𝟏𝟏 𝐛𝐛𝟏𝟏 𝐜𝐜𝟏𝟏 𝐚𝐚𝟏𝟏 𝐝𝐝𝟏𝟏 𝐜𝐜𝟏𝟏 𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏 𝐝𝐝𝟏𝟏
𝐃𝐃 = 𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 𝐃𝐃𝐱𝐱 = 𝐝𝐝𝟐𝟐 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 𝐃𝐃𝐲𝐲 = 𝐚𝐚𝟐𝟐 𝐝𝐝𝟐𝟐 𝐜𝐜𝟐𝟐 𝐃𝐃𝐳𝐳 = 𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐 𝐝𝐝𝟐𝟐
𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑 𝐜𝐜𝟑𝟑 𝐝𝐝𝟑𝟑 𝐛𝐛𝟑𝟑 𝐜𝐜𝟑𝟑 𝐚𝐚𝟑𝟑 𝐝𝐝𝟑𝟑 𝐜𝐜𝟑𝟑 𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑 𝐝𝐝𝟑𝟑

𝐃𝐃𝐱𝐱 𝐃𝐃𝐲𝐲 𝐃𝐃𝐳𝐳

Then, 𝐱𝐱 = 𝐲𝐲 = 𝐳𝐳 =
𝐃𝐃 𝐃𝐃 𝐃𝐃
Types of solutions and consistency of the system
 Types of solutions

Solution exist (At least one solution) No solution

(Consistent) (Inconsistent)

Unique/ Exactly one sol. Infinite solutions

All variables zero is the At least one non – zero

only solution variable satisfy the system
𝐱𝐱 = 𝟎𝟎, 𝐲𝐲 = 𝟎𝟎, 𝐳𝐳 = 𝟎𝟎
Non - Trivial solution
Trivial solution Non - Zero solution
 𝐃𝐃 𝐃𝐃𝐱𝐱 ′ 𝐃𝐃𝐲𝐲 ′ 𝐃𝐃𝐳𝐳 Type of Solution Type of System
Unique
≠ 𝟎𝟎 Atleast any one of 𝐃𝐃𝐱𝐱 , 𝐃𝐃𝐲𝐲 , 𝐃𝐃𝐳𝐳 ≠ 𝟎𝟎 Consistent
Non – Trivial

≠ 𝟎𝟎 𝐃𝐃𝐱𝐱 = 𝐃𝐃𝐲𝐲 = 𝐃𝐃𝐳𝐳 = 𝟎𝟎 Unique / Trivial Consistent

= 𝟎𝟎 𝐃𝐃𝐱𝐱 = 𝐃𝐃𝐲𝐲 = 𝐃𝐃𝐳𝐳 = 𝟎𝟎 Infinite Sols. Consistent

= 𝟎𝟎 𝐃𝐃𝐱𝐱 ≠ 𝟎𝟎, 𝐨𝐨𝐨𝐨 𝐃𝐃𝐲𝐲 ≠ 𝟎𝟎, 𝐨𝐨𝐨𝐨 𝐃𝐃𝐳𝐳 ≠ 𝟎𝟎 No Solution Inconsistent

Q. If the system of equations 𝟐𝟐𝐱𝐱 − 𝐲𝐲 + 𝟐𝟐𝐳𝐳 = 𝟐𝟐; 𝐱𝐱 − 𝟐𝟐𝟐𝟐 + 𝐳𝐳 = −𝟒𝟒 & 𝐱𝐱 + 𝐲𝐲 + 𝝀𝝀𝐳𝐳 = 𝟒𝟒
Has no solution then the value of λ is :-
(A) 3 (B) 1 (C) -3 (D) 4

OR Has Unique solution then the value of λ can not be equal to :-

No solution means 𝐢𝐢) 𝐃𝐃 = 𝟎𝟎 𝐢𝐢𝐢𝐢) Atleast one of 𝐃𝐃 𝐱𝐱 , 𝐃𝐃 𝐲𝐲 , 𝐃𝐃𝐳𝐳 ≠ 𝟎𝟎

Q) Find x, y & z if 𝐱𝐱 + 𝟐𝟐𝟐𝟐 + 𝟑𝟑𝟑𝟑 = 𝟔𝟔 ; 𝟒𝟒𝟒𝟒 + 𝟓𝟓𝟓𝟓 + 𝟔𝟔𝐳𝐳 = 𝟏𝟏𝟏𝟏 & 𝟕𝟕𝟕𝟕 + 𝟖𝟖𝟖𝟖 + 𝟗𝟗𝟗𝟗 = 𝟐𝟐𝟐𝟐

(All the rows are in A.P.) 𝐃𝐃 = 𝟎𝟎 𝐃𝐃𝐱𝐱 = 𝟎𝟎 𝐃𝐃𝐲𝐲 = 𝟎𝟎 𝐃𝐃𝒛𝒛 = 𝟎𝟎

⇒ Infinite Solutions and system is consistent
 Homogeneous System
a1x + b1y + c1z = 0 .... (i)
a2x + b2y + c2z = 0 .... (ii)
a3x + b3y + c3z = 0 .... (iii)

𝟏𝟏. 𝐃𝐃𝐱𝐱 = 𝐃𝐃𝐲𝐲 = 𝐃𝐃𝐳𝐳 = 𝟎𝟎

𝟐𝟐. Atleast one solution : x = 0, y = 0, z = 0 Trivial solution

𝟑𝟑. System is always Consistent

𝟒𝟒. For Non – Zero or Non – Trivial Sol D = 0.

 𝐃𝐃
𝐃𝐃𝐱𝐱 = 𝐃𝐃𝐲𝐲 = 𝐃𝐃𝐳𝐳 = 𝟎𝟎
𝐃𝐃 ≠ 𝟎𝟎 𝐃𝐃 = 𝟎𝟎

𝟏𝟏. Unique Solution 𝟏𝟏. ∞ No. of Solutions

(x = 0, y = 0, z = 0)
𝟐𝟐. Trivial Solution 𝟐𝟐. Non – Trivial Solution
Zero Solution Non – Zero Solution
Q) If the system of linear equations
𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝐚𝐚𝐚𝐚 = 𝟎𝟎
𝟐𝟐𝟐𝟐 + 𝟑𝟑𝟑𝟑𝟑𝟑 + 𝐛𝐛𝐛𝐛 = 𝟎𝟎 [JEE Main PYQ]
𝟐𝟐𝟐𝟐 + 𝟒𝟒𝟒𝟒𝟒𝟒 + 𝐜𝐜𝐜𝐜 = 𝟎𝟎
where a, b, c ∈ 𝐑𝐑 are non-zero and distinct; has a non-zero solution, then:
𝟏𝟏 𝟏𝟏 𝟏𝟏
𝐀𝐀 , , 𝐚𝐚𝐚𝐚𝐚𝐚 𝐢𝐢𝐢𝐢 𝐀𝐀. 𝐏𝐏. (B) a, b, c are in G.P.
𝐚𝐚 𝐛𝐛 𝐜𝐜
(C) a + b + c = 0 (D) a, b, c are in A.P.

For Non – Zero or Non – Trivial Sol D = 0.

Q) If the system of equations
𝐱𝐱 +( 𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬 𝛂𝛂) 𝐲𝐲 + ( 𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜 𝛂𝛂) 𝐳𝐳 = 𝟎𝟎,
[JEE Main PYQ]
𝐱𝐱 + 𝐜𝐜𝐜𝐜𝐜𝐜 𝛂𝛂 𝐲𝐲 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝛂𝛂 𝐳𝐳 = 𝟎𝟎
and 𝐱𝐱 + 𝐬𝐬𝐬𝐬𝐬𝐬 𝛂𝛂 𝐲𝐲 − 𝐜𝐜𝐜𝐜𝐜𝐜 𝛂𝛂 𝐳𝐳 = 𝟎𝟎
𝛑𝛑
has a non−trivial solution, then 𝛂𝛂 ∈ 𝟎𝟎, is equal to
𝟐𝟐
𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏𝟏𝟏 𝟕𝟕𝟕𝟕 𝟓𝟓𝟓𝟓
𝐀𝐀 𝐁𝐁 𝐂𝐂 𝐃𝐃
𝟒𝟒 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐𝟐𝟐

For Non – Zero or Non – Trivial Sol D = 0.

 Matrices
Types
Multiplication
Transpose
Adjoint
Inverse
Application
Matrix
Arrangement of m × n elements into m − rows
and n − columns is called matrix

𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏 ...... 𝐚𝐚𝟏𝟏𝟏𝟏

𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐 ...... 𝐚𝐚𝟐𝟐𝟐𝟐

....
....

....
....

𝐚𝐚𝐦𝐦𝐦𝐦 𝐚𝐚𝐦𝐦𝐦𝐦 ...... 𝐚𝐚𝐦𝐦𝐦𝐦

Unlike determinants it has no value.

m × n called order
Types of Matrices
Row Matrix

Column Matrix

Zero or Null Matrix (O)

Singleton Matrix
Square Matrix (Order n)
𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏
A= 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐
𝐚𝐚𝟑𝟑𝟏𝟏 𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑

Diagonal Matrix
𝐝𝐝𝟏𝟏 𝟎𝟎 𝟎𝟎
A= 𝟎𝟎 𝐝𝐝𝟐𝟐 𝟎𝟎 𝐚𝐚𝐢𝐢𝐢𝐢 = 𝟎𝟎 𝐢𝐢𝐢𝐢 𝐢𝐢 ≠ 𝐣𝐣
𝟎𝟎 𝟎𝟎 𝐝𝐝𝟑𝟑
 Scalar Matrix
𝐚𝐚 𝟎𝟎 𝟎𝟎
A = 𝟎𝟎 𝐚𝐚 𝟎𝟎
𝟎𝟎 𝟎𝟎 𝐚𝐚

Unit or Identity Matrix

𝟏𝟏 𝟎𝟎 𝟎𝟎
I = 𝟎𝟎 𝟏𝟏 𝟎𝟎
𝟎𝟎 𝟎𝟎 𝟏𝟏

Det (I) = 𝐈𝐈 = 𝟏𝟏
 𝐀𝐀

𝐀𝐀 = 𝟎𝟎 𝐀𝐀 ≠ 𝟎𝟎
Singular Matrix Non − Singular Matrix
 Trace of a Matrix
𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏 𝐚𝐚𝟏𝟏𝟏𝟏
A= 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐 𝐚𝐚𝟐𝟐𝟐𝟐
𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑 𝐚𝐚𝟑𝟑𝟑𝟑

tr(A) = � 𝐚𝐚𝐢𝐢𝐢𝐢 = 𝐚𝐚𝟏𝟏𝟏𝟏 + 𝐚𝐚𝟐𝟐𝟐𝟐 + 𝐚𝐚𝟑𝟑𝟑𝟑

1 tr. (λA) = λ tr (A)

2 tr (A + B) = tr (A) + tr(B)

3 tr(AB) = tr(BA)
 Multiplication of Matrix by a Scalar
𝐚𝐚 𝐛𝐛 𝐜𝐜 𝐤𝐤𝐚𝐚 𝐤𝐤𝐛𝐛 𝐤𝐤𝐜𝐜
If 𝐀𝐀 = 𝐛𝐛 𝐜𝐜 𝐚𝐚 𝐓𝐓𝐓𝐓𝐓𝐓𝐓𝐓, 𝐤𝐤𝐀𝐀 = 𝐤𝐤𝐛𝐛 𝐤𝐤𝐜𝐜 𝐤𝐤𝐚𝐚
𝐜𝐜 𝐚𝐚 𝐛𝐛 𝐤𝐤𝐜𝐜 𝐤𝐤𝐚𝐚 𝐤𝐤𝐛𝐛

Note
If A is a square matrix then,

1 𝐭𝐭𝐭𝐭 𝐤𝐤𝐤𝐤 = 𝐤𝐤[𝐭𝐭𝐭𝐭(𝐀𝐀)]

2 𝐤𝐤𝐤𝐤 = 𝐤𝐤 𝐧𝐧 𝐀𝐀 Kachra
Where, n is order of matrix A.
 Multiplication of Matrices
AB = Am×n . Bn×p

Then, order of AB : m × p

AB exists only if,

No. of columns in pre multiplier (A)
= No. of rows in post multiplier (B)
It is row x column wise
If AB exists, but BA may or may not exist.
 a b c
1 2 3
If A = B= d e f Find AB
4 5 6 2×3 g h i 3×3
Properties of Matrix Multiplication
1 Not Commutative (In General) ⇒ AB ≠ BA

2 Matrix Multiplication is Associative

If A, B & C are conformable for the product AB & BC,
Then, (AB)C = A(BC)

3 Distributivity A(B + C) = AB + AC
(A + B)C = AC + BC
4 Positive Integral Powers of a Square Matrix

For a square matrix A,

A.A = A2
𝐀𝐀𝐦𝐦 𝐀𝐀𝐧𝐧 = 𝐀𝐀𝐦𝐦+𝐧𝐧
(𝐀𝐀𝐦𝐦 )𝐧𝐧 = 𝐀𝐀𝐦𝐦𝐦𝐦 (for m, n ∈ N)
Note

 (A + B)2 = (A + B) (A + B)
= A2 + AB + BA + B2

𝐈𝐈𝐈𝐈 𝐀𝐀 & B commute 𝐈𝐈𝐈𝐈 𝐀𝐀 & B anticommute

(ie. AB = BA) (ie. AB = −BA)
𝐀𝐀𝟐𝟐 + 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝐁𝐁𝟐𝟐 𝐀𝐀𝟐𝟐 + 𝐁𝐁𝟐𝟐

In general (A + B)2 ≠ A2 + B2 + 2AB

5

 If A = dia (d1,d2 ,d3 )

Then, An = dia (d1n ,d2n , d3n)

 for a unit matrix I of any order,

𝐈𝐈𝐦𝐦 = I for all m ∈ N

5 For square matrices A & B,

i) AB = A B

Ak = A k, k∈ N

kA = k n A

Where, where k is
scaler and n is order
of A
Q) If A = dia [2, −1, 3], B = dia [− 1, 3, 2], then A2B = [JEE Main PYQ]
(A) dia [5, 4, 11] (B) dia [− 4, 3, 18]
(C) dia [3, 1, 8] (D) dia [3, 2, 6]
 1 1 1 2 1 3 1 n 1 210
Q If …………. = then n equals to.
0 1 0 1 0 1 0 1 0 1
 1 1 1 2 1 3 1 n 1 210
Q If …………. = then n equals to.
0 1 0 1 0 1 0 1 0 1
Sol.
1 1 1 2
0 1 0 1
1 1+2 1 3
0 1 0 1

1 1 + 2 + 3 …………. 1 n = 1 210
0 1 0 1 0 1

1 1 + 2 + 3 + ⋯ n = 1 210 n(n + 1)
⇒ = 210 n = – 1, 20
0 1 0 1 2
Q If A, B, C are square matrices of order 2 × 2 and A = 4, B = 1 and
C = −1 then 2A2 B1 C7 is equal to :-
(A) −8 (B) −16 (C) −32 (D) −64

= −64
Transpose of A Matrix
Transpose of a Matrix is obtained by changing its rows
into columns and columns into rows.

𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏 𝐜𝐜𝟏𝟏 𝐚𝐚𝟏𝟏 𝐚𝐚𝟐𝟐 𝐚𝐚𝟑𝟑

𝐀𝐀 = 𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 𝐀𝐀𝐓𝐓 = 𝐛𝐛𝟏𝟏 𝐛𝐛𝟐𝟐 𝐛𝐛𝟑𝟑
𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑 𝐜𝐜𝟑𝟑 𝐜𝐜𝟏𝟏 𝐜𝐜𝟐𝟐 𝐜𝐜𝟑𝟑

A = [aij] m × n
Then,

AT = [aji] n × m
Properties of Transpose
Matrices : A and B Transpose : AT & BT

1 (AT)T = A
2 (A + B)T = AT + BT (A & B have the same order.)

3 (KA)T = KAT K be any scalar Kachra

4 (A B)T = BT AT Provided, A & B are conformable for matrix product AB

(A1.A2 ................. .An)T = 𝐀𝐀𝐓𝐓𝐧𝐧 . .............. 𝐀𝐀𝐓𝐓𝟐𝟐 .𝐀𝐀𝐓𝐓𝟏𝟏

(reversal law for transpose)
5 (An)T = AT 𝐧𝐧

6 tr(AT) = tr(A)

7 |AT|=|A|
 A = [aij]
Square Matrix

Symmetric Skew Symmetric

AT = A AT = −A
A − AT = 0 A + AT = 0
𝐝𝐝 𝐚𝐚 𝐛𝐛 𝟎𝟎 𝐚𝐚 𝐛𝐛
𝐚𝐚 𝐞𝐞 𝐜𝐜 −𝐚𝐚 𝟎𝟎 𝐜𝐜
𝐛𝐛 𝐜𝐜 𝐟𝐟 −𝐛𝐛 −𝐜𝐜 𝟎𝟎
For a skew symmetric matrix A of odd order |A| = 0.
Properties of Symmetric & Skew Symmetric Matrix

1 If A be a square matrix then,

a) A + AT is a symmetric matrix
b) A − AT is a skew symmetric matrix
c) A AT and AT A are symmetric matrices.

2 Every square matrix can be uniquely expressed as the sum of a

symmetric and a skew symmetric matrix.
𝟏𝟏 𝐓𝐓
𝟏𝟏
𝐀𝐀 = 𝐀𝐀 + 𝐀𝐀 + 𝐀𝐀 − 𝐀𝐀𝐓𝐓
𝟐𝟐 𝟐𝟐

Symmetric Skew Symmetric

 𝟐𝟐 𝟏𝟏 𝟏𝟏 𝟎𝟎
Q) If 𝐀𝐀 = , 𝐁𝐁 = , 𝐂𝐂 = 𝐀𝐀𝐀𝐀𝐀𝐀𝐓𝐓 𝐚𝐚𝐚𝐚𝐚𝐚 𝐗𝐗 = 𝐀𝐀𝐓𝐓 C2𝐀𝐀, then det X
−𝟏𝟏 𝟐𝟐 𝟏𝟏 𝟏𝟏
is equal to [JEE Main PYQ]

(A) 243 (B) 729 (C) 27 (D) 891

Q) Order of skew symmetric matrix A is 4 and |A|= 5. Then find |A – AT|.
Ans. (80)
Q) The number of symmetric matrices of order 3, with all the entries from the
set 0,1,2,3,4,5,6,7,8,9, is [JEE Main PYQ]
(A) 910 (B) 106 (C) 109 (D) 610
 Adjoint of a Square Matrix
Transpose of the matrix of cofactors of elements of A in det(A)
T
adjA = adj(A) = C
 Note
𝐩𝐩 𝐪𝐪
1 If 𝐀𝐀 = 𝐫𝐫 𝐬𝐬
𝐬𝐬 −𝐪𝐪
Then, adj 𝐀𝐀 =
−𝐫𝐫 𝐩𝐩
e.g.
𝟏𝟏 𝟐𝟐
𝐀𝐀 =
𝟑𝟑 𝟒𝟒
𝟒𝟒 −𝟐𝟐
adj 𝐀𝐀 =
−𝟑𝟑 𝟏𝟏
Properties of Adjoint A is Square Matrix

1 A (adj. A) = (adj. A).A = |A| In.

2 |adj A| = 𝐀𝐀 𝐧𝐧−𝟏𝟏

adj(KA) = Kn–1 adj(A) kachra

3

4 adj AB = (adj B) (adj A)

5 adj AT = (adj A)T

6 adj (adj A) =|A|n–2 A

𝐧𝐧−𝟏𝟏 𝟐𝟐
7 |adj (adj A)|= |𝐀𝐀|

Note
(adj I) = I
(adj O) = O
Q) If A is a 3 × 3 matrix such that |A| = 1, then value of |adj(adj 7A)| is
equal to - [JEE Main PYQ]
(A) 73 (B) 74 (C) 77 (D) 712
Inverse of Matrix
A square matrix ‘A’ is said to be invertible (non singular) if
there exist a matrix B such that :
AB = I = BA
B = A–1 (Inverse of A)
Thus,

A–1 = B ⇔ AB = I = BA
𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚
𝐀𝐀−𝟏𝟏 =
|𝐀𝐀|
Note

1 For a Square matrix A to be

invertible.
|𝐀𝐀| ≠ 𝟎𝟎

2 Let,
A = dia (k1 k2 k3 ......kn)
(Non Singular)
Then,
−𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏 −𝟏𝟏
𝐀𝐀 = dia 𝐤𝐤 𝟏𝟏 , 𝐤𝐤 𝟐𝟐 , 𝐤𝐤 𝟑𝟑 . . . . . . , 𝐤𝐤 𝐧𝐧
Some Important Theorems
1 Every invertible matrix posses a unique inverse.

2 If, A → Invertible & Square Then, AT → Invertible

& (AT)–1 = (A–1)T
3 If, A → Non – Singular
𝟏𝟏
Then, |A–1| =
|𝐀𝐀|
4 If A & B are Invertible Matrices
Then, (AB)–1 = B–1 A–1

5 If, A → Invertible
Then,
(a) (A–1)–1 = A
(b) (Ak)–1 = (A–1)k = A–k
Q) Let A & B are two invertible matrices of order 3 × 3 If 𝐝𝐝𝐝𝐝𝐝𝐝 𝐀𝐀𝐀𝐀𝐀𝐀⊤ = 𝟖𝟖
& 𝐝𝐝𝐝𝐝𝐝𝐝 𝐀𝐀𝐁𝐁−𝟏𝟏 = 𝟖𝟖 then 𝐝𝐝𝐝𝐝𝐝𝐝 𝐁𝐁𝐀𝐀−𝟏𝟏 𝐁𝐁⊤ =
 𝟏𝟏
Ans.
𝟏𝟏𝟏𝟏
 Applications of Matrices & Determinants
Solution of System of Linear Equations
a1x + b1y + c1z = d1 .... (i)
a2x + b2y + c2z = d2 .... (ii)
a3x + b3y + c3z = d3 .... (iii)

𝐚𝐚𝟏𝟏 𝐛𝐛𝟏𝟏 𝐜𝐜𝟏𝟏 𝐱𝐱 𝐝𝐝𝟏𝟏

A = 𝐚𝐚𝟐𝟐 𝐛𝐛𝟐𝟐 𝐜𝐜𝟐𝟐 𝐗𝐗 = 𝐲𝐲 𝐁𝐁 = 𝐝𝐝𝟐𝟐 Then, A X = B
𝐚𝐚𝟑𝟑 𝐛𝐛𝟑𝟑 𝐜𝐜𝟑𝟑 𝐳𝐳 𝐝𝐝𝟑𝟑 ⇒ X = A–1B
Note
𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚
X= . 𝐁𝐁
|𝐀𝐀|
Types of solutions

𝐀𝐀 ≠ 𝟎𝟎 𝐀𝐀 ≠ 𝟎𝟎 & (adj A).B ≠ 𝐎𝐎 𝐀𝐀 ≠ 𝟎𝟎 & (adj A).B = 𝐎𝐎

(Unique Solution) (Unique non−trivial sol. ) (Trivial sol. X = 0)
(Consistent) (Consistent) (Consistent)
Note
𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚
X= . 𝐁𝐁
|𝐀𝐀|
Types of solutions
If 𝐀𝐀 = 𝟎𝟎

(adj A).B = O (adj A).B ≠ 𝟎𝟎

Infinite sols. No solution
Consistent Inconsistent
All the Best !!