JEE-Main-29-01-2024 (Memory Based)

[MORNING SHIFT]

Physics
Question: A block of mass 100 kg is moved along a horizontal surface 10 m from the
starting point. If coefficient of friction between ground and the block is 0.4 find work done
against friction
Options:
(a) 3.9 kJ
(b) 4.2 kJ
(c) 3.7 kJ
(d) 4.1 kJ
Answer: (a)

Question: A particle is executing SHM with an amplitude A. If potential energy of the

system is zero about mean position x = 0, Find ratio of total energy to kinetic energy at x =
A/3
Options:
(a) 8/9
(b) 9/8
(c) 3/2√2
(d) 2√2/3
Answer: (b)

Question: i = 20 + 3/2 t Find charge flown in 20 S

Options:
(a) 1600 C
(b) 1200 C
(c) 1000 C
(d) 800 C
Answer: (c)

Question: Match the following

A ∮ B  dA = 0 P Faraday & Lenz’s law

B Qin Q Gauss law on magnetism

∮ E  dA =
0

C ∮ B  d l = .ienc R Ampere’s law

 D dB S Gauss law of electrostatics
∮ E dl = −
dt
Options:
(a) (A-Q),(B-S), (C-R), (D-P)
(b) (A-S),(B-Q), (C-R), (D-P)
(c) (A-Q),(B-R), (C-S), (D-P)
(d) (A-Q),(B-S), (C-P), (D-R)
Answer: (a)

Question: In the Following Circuit the resistance of square loop ABCD is 16 Ohm. Find the
Voltage Across Capacitor in steady State

Options:
(a) 4.5 V
(b) 4 V
(c) 3 V
(d) 1 V
Answer: (a)

Question: A Square loop of side 0.1 m is in East West Plane and magnetic field is along
North East of 0.2 T. If B is Removed in 10 s find EMF Induced?
Options:
(a) 14 mV
(b) 0.14 mV
(c) 1 mV
(d) 0.2 mV
Answer: (b)

Question: If debroglie wavelength of an electron is same as wavelength of a photon and

speed of the electron is 25% of speed of EM waves in vacuum. Find ratio of kinetic energy of
electron & energy of photon.
Options:
(a) 1/8
(b) ¼
(c) ½
(d) 1
Answer: (a)
Question: P-V graph of a gas is given. Find the work done by the gas.
Options:
(a) 400 J
(b) 600 J
(c) 800 J
(d) 100 J
Answer: (c)

Question: A convex lens made of glass (µ glass = 1.5) has focal length of 20 cm in air If this
lens is put inside a fluid of refractive index 1.6. The new focal length will be
Options:
(a) 160 cm
(b) -160 cm
(c) -180 cm
(d) 80 cm
Answer: (b)

Question: If R is the radius of Earth's and Particle has Equal weight at “d” distance below the
surface of Earth's and “d” distance above it, find “d”
Options:
(a) d = √5 R/2
(b) d = √3 R
(c) d = (√5-1) R/2
(d) d = R
Answer: (c)

Question: The flow speeds on upper & lower surfaces of the wings are 70 m/s & 64 m/s
respectively on an airplane in a wind tunnel. What is the lift force on the wing? Area of wing
is 0.2 m2 Given: density of air = 1.2 kg/m3
Options:
(a) 16
(b) 36
(c) 81
(d) 144
Answer: (c)

Question: In a concave mirror of radius of curvature R = 30 cm the size of inverted image is

half the size of object. Find the distance of the object from pole.
Options:
(a) 30
(b) 45
(c) 60
(d) 20
Answer: (b)
Question: A Galvanometer shows deflection corresponding to 25 division when a certain
current is passed. The deflection becomes 5 divisions when galvanometer is shunted with
24Ω. Find the resistance of galvanometer
Options:
(a) 24 Ω
(b) 48 Ω
(c) 96 Ω
(d) 120 Ω
Answer: (c)

Question: In the given nuclear reaction, which of the following expression correctly
represent the Q value
3 Li + 1 H → 2 2 He
6 2 4

Given masses:
6
3Li = 6.015122 amu, 42He = 4.00 2603 amu
2
1H - 2.014101 amu, 1 amu = 931.5 MeV
Options:
(a) 22.37 MeV
(b) 21.42 MeV
(c) 22.02 MeV
(d) 21.90 MeV
Answer: (a)

Question: S1: When a capillary tube is dipped in cold water and then hot water, the height of
water increases
S2: When a capillary tube is dipped in hot water and then cold water, the height of water
decreases
[Assume negligible change in density of water or radius of capillary]
Options:
(a) 1 true, 2 false
(b) 1 false, 2 true
(c) Both false
(d) Both True
Answer: (c)

Question: In YDSE experiment source is placed exactly in front of one slit.

The distance between slits & screen is 0.2m. Wavelength used is 400 nm. Find the minimum
distance between slits such that point O is dark

Options:
(a) 0.28 mm
(b) 0.36 mm
(c) 0.14 mm
(d) 0.49 mm
Answer: (a)

Question: A galvanometer with resistance Rg = 8Ω has a fall scale deflection current of

Ig = 3 mA. What is the shunt resistance required to create an ammeter of 8 ampere range?
Options:
(a) 0.001 Ω
(b) 0.003 Ω
(c) 0.009 Ω
(d) 0.01 Ω
Answer: (b)

Question: Calculate the flux passing through a sphere of radius 4a whose center is at the
origin, if two changes 5q and -2q are placed at
(2q, 0) and (-5q, 0) respectively
Options:
(a) 5q/ε0
(b) -2q/ε0
(c) 7q/ε0
(d) 3q/ε0
Answer: (a)

Question: If the magnetic potential due to a small magnetic dipole along the axis at a
distance of 20 cm is 1.5 × 10-5 J Am-1 find its magnetic dipole moment
Options:
(a) 4 Am2
(b) 6 Am2
(c) 8 Am2
(d) 2 A2
Answer: (b)
 JEE-Main-29-01-2024 (Memory Based)
[MORNING SHIFT]

Chemistry

Question: Which of the following pair will be formed by the decomposition of KMnO4 ?
Options:
(a) MnO4, MnO2
(b) K2 MnO4 , MnO2
(c) KMnO4 , MnO2
(d) MnO2 , H2O
Answer: (b) K2 MnO4 , MnO2
Solution:
Potassium permanganate forms dark purple (almost black) crystals which are isostructural
with those of KCLO4. The salt is not very soluble in water (6.4 g / 100 g of water at 293 K),
but when heated it decomposes at 513 K.
2KMnO4 → K2MnO4 + MnO2 + O2

Question: Interaction b/w 𝞹𝞹. Bond & lone pair l-s on adjacent atoms
Options:
(a) Resonance
(b) Hyper conjugation
(c) Inductive Effect
(d) Electronic Effect
Answer: (a) Resonance
Solution:

Question: Assertion. Electronegativity increase across a period

Reason. Effective increase in nuclear charge is more than effective shielding.
Options:
(a) Step 1: Electronegativity increase down the group 14 is to pb
(b) Step 2: Group 14 contains metals, non metals and also metalloids
Solution: Assertion true reason true
Step : 1 is incorrect but Step : 2 is correct

Question:
Column - I Column - II

Ziegler Natta Catalyst Rh

Blood Pigment CO

Wilkinson Catalyst Fe

Vitamin B12 Ti
Solution:
1 → Ti
2 → Fe
3 → Rh
4 → Co

Question: Appearance of Red colour on treatment with Na fusion extract of an organic

compound with FeSO4 in presence of conc. H2SO4 indicate element
Options:
(a) N
(b) Br
(c) S
(d) N & S
Answer: (d) N & S
Solution:(d) N & S

Question: Cl- shows disproportionation in alkaline meol :

a cl2 + b OH- → c cl O- + d cl- + H2O
Options:
(a) 1 1 1 3
(b) 3 6 2 4
(c) 1 2 1 1
(d) 2 4 1 3
Answer: (b)
Solution: 3Cl2 + 6OH- → 2ClO-3 + 4Cl- + 3H2O

Question: The correct set of 4 Quantum numbers of Valence e- of Rb(37)

Options:
(a) n = 5 ., l= 0 ., m = 1.,
(b) n = 5., l = 0., m = 0.,
(c) n = 5., l = 1., m = 0.,
(d) n = 5., l = 1., m = 1.,
Answer:
Solution:
Rb ⇒ 5 sl
↓
n=5
l=0
Ml = 0
Mg = +1/2 or - 1/2
The electronic configuration of rubidium atom (Z = 37) is given by
Rb = [Kr] 5 s1
Hence, the quantum numbers for 5 s1 electron is given by
n = 5, I = 0, m = 0, s=+1/2 or - 1/2

Question: Type of amino acids obtained on hydrolysis of proteins

Options:
(a) 𝝰𝝰
(b) 𝞫𝞫
(c) 𝜸𝜸
(d) δ
Answer: (a)
Solution: Alpha amino acid

Question: CO forms a bridge b/w M atoms

Options:
(a) Os3 (CO)12
(b) Co2 (CO)8
(c) Ru3 (CO)12
(d) Mn2 (CO)10
Solution:

Question: Calculate the Molarity of a Solution having density = 1.25 g/ml. % (w/w) of
Solute is 31.4% of H2SO4 solution
Options:
(a) 4
(b) 9
(c) 8
(d) 6
Answer: (a)
Solution:
M = 10 × w/w % × d
Msolute
M =10 × 31.4 × 125 x 100
98
=4

Question: Find all quantum numbers Z = 37

Options:
(a) n = 5 ., l= 0 ., m = 1.,
(b) n = 5., l = 0., m = 0.,
(c) n = 5., l = 1., m = 0.,
(d) n = 5., l = 1., m = 1.,
Answer: (a)

Question: Among the heterocyclic compound that contain Sulphur atom is :

Options:
(a) Pyradizine
(b) Furan
(c) Thiophene
(d) Pyrrole
Answer: (c)
Solution:

Question: Find weight of Zinc in Zinc sulphate electrolysis i = 0.015 A t = 15 minutes

Solution:
Zn+2 + 2e- → Zn
1 mol Zn = 65.3 gm = 2 F
Number of Faradays = 0.015 ✖ 15 ✖ 60
965
= 0.00013 g F
= .0046

Question: Number of compound in which B.O = 1 and is paramagnetic

He2+., O2+, O2-2, N2+
Answer: 0
Solution:
B.O Magnetic nature
He2+ 0.5 Paramagnetic
+
O2 1.5 Paramagnetic
O2-2 1 Diamagnetic
N+2 2.5 Paramagnetic

Question: Number of compounds that gives positive fehling test Benzaldehyde,

acetophenone, methanal
Answer: 1
Solution: Aliphatic aldehyde group. Aromatic aldehydes and ketones do not a give Fehling’s
test.
 JEE-Main-29-01-2024 (Memory Based)
[MORNING SHIFT]

Maths
Question: f(x) = 2x – x2 m = number of solution such that f(x) with x axis
N = number of solutions such that f’(x) with x axis m + n ?
Answer: 5
Solution:
f ( x=
) 2x − x2
′ ( x ) 2 x n 2 − 2 x
f=
m=3
n=2

Question: (1 + y2) (1 + ln x) dx + x dy = 0
Answer:

π
2  x 2 cos x 1 + sin 2 x 
Question:
= Find I ∫ 
 1 + ex
π
+
1+ e
sin ( x 2023 )



−
2
Solution:
π
2
x 2 sin x 1 + cos 2 x
=I ∫π 1 + ex
+
1+ e (
sin x 2023)
dx
−
2
π
2
x 2 sin x 1 + cos 2 x
=I ∫π 1 + ex
− ex +
1+ e (
sin x 2023)
dx
−
2
π
2

∫π x
2
=αI sin x + 1 + cos 2 x
−
2
π
2

∫π x
2
αI = α sin x + cos 2 xdx
−
2
π
π π
I = x 2 ( − cos x ) + ( 2 x )( + sin x ) + 2 ( cos x ) 02 + +
2 4
 π  3π
I 0 + 2.
=  − (0 + 0 + 2) +
 2  4
7π
I
= −2
4
 Question: If an AP with terms <ai>, a6 = 2 and a1, a4 a5 is maximum.
Find the common difference.
Solution:
=a6 2&
= a1a4 a5 max.(given)
⇒M = ( a6 − 5d )( a6 − 2d )( a6 − d )
( 2 − sd )( 2 − 2d )( 2 − d )
=

(
M =2 −5d 3 + 17 d 2 − 16d + 4 )
dM
dQ
(
=2 −15d 2 + 24d + 10d − 16 =0 )
= 2 ( −3d ( 5d − 8 ) + 2 ( sd − 8 ) ) = 0
=−2 ( 3d − 2 )( 5d − 8 ) =0
8
d=
5

11
C1 11 C2 11
C n
Question: + +−−−−−−+ =
2 3 9 m
gcd (M, n) = 1, find m + n.
Solution:
r =11

∑ h Cr x r
(1 + x ) =
11

r =0
l
r =11 11
1 Cr x r
∫ (1 + x ) ∑
11
dx =
r =0 r + 1
0

212 − 1 r =11 11 Cr
=∑
12 r =0 r + 1
11
11
C0 Cg 11 C10 11 C11
= + ( 5) + + +
1 10 11 12
12
2 −1 91 4096 − 91 4095 1365
=5+ ⇒ 5 = = =
12 12 12 12 4
m+n = 1369

(2 x
+ 2− x ) ( tan x ) tan −1 ( x 2 − x + 1)
Question: f ( x ) =
(x − x 2 + 1)
3 3

Find f′(a) =
Solution:
 (2 x
+ 2− x ) tan x tan −1 ( x 2 − x + 1)
f ( x) =
(x − x 2 + 1)
3 3

f ( 0) = 0
π
2 ⋅1 ⋅
f ( x) 4
f ' ( 0 ) lim
= = = π
x →0 x 1

Question:
x2 + y 2 =46, 

x2 y 2 
+ 2 = 1 
16 b 
POI lives on y2 = 3x2 find 3 3 times of areas of rectangle formed by POI of conics

Question:
3
π 
   1
1.∫ cos  x 3  dx
2
x3
lim  
2
x→
π
 π
2
x− 
 2
Solution:
3π 2
=
8

Question: GTWENTY, find rank of GTWENTY

Solution:
2 4 513 4 6
G T W N T Y
1 2
! !3 0 0 0 0
2 2
1 2
Rank = × 6!+ × 5!+ 3 × 4!+ 1
2! 2!
= 360 + 120 + 72 + 1
= 480 + 73 = 553

Question:
1 
A ⋅ AT I Value of  A   A + AT ( ) + ( A − A ) 
2 2
T
=
 2  
Options:
(a) A3 + AT
(b) (A3 + AT)2
(c) (A3+I)
(d) A3
Solution:
 A ⋅ AT =
I
1
(( ))
2
) (
2
A A + AT + A − AT
2
1
2 ( 2
( )
A A2 + AT + 2 I + A2 ( AT ) − 2 I
2
)
( ( ))
2
A A2 + AT A−1
AT =

= A3 + AAT AT
= A3 + AT

Question:
1 0 0 
=  det ( 2 A ) 221
3
0 β α  A=
0 α β 
Find one of value of
𝛼𝛼(or 𝛽𝛽) 𝛼𝛼, 𝛽𝛽 both integers.
Options:
(a) 3
(b) 17
(c) 9
(d) 6
Solution:
1 0 0 
0 β α  = A
 
0 α β 
Net ( 2 A ) = 2 A
3 3

(=
8 A)
3
= 221
3
=29 ⋅ A =221
3
212
⇒ A =
= β 2 − α 2 = 24 =16

Question:
f ( x ) = 4 2 x3 − 2 2 x − 1
1 
S − 1 f  ;1 → R ; f(x) intersection x axis at 1 point S-2 f(x) intersection x axis at
2 
π
x = cos
12
Solution:
 S −1
f ( x ) = 4 2 x3 − 2 2 x − 1
1 1 1
f   =4 2 × − 2 2 × − 1 = 2 − 2 − 1 =−14
2 8 2
f (1) = 4 2 − 2 2 − 1
1
f   ⋅ f (1) < 0
2
f ' ( x=
) 12 2 x 2 − 2 2= 0
2 1
= x=
6
1 1 
x=
± ∈
/ ,1
6  2 
= 2 2 −1 > 0
S − 1 is true

Question: Sum of all 64 terms is 7(sum of terms at odd), find common ratio.
Options:
(a) 6
(b) 7
(c)
(d)
Solution:
a ( r 04 − 1) ar ( r 64−1 ) 1
=7 ⇒r=
r −1 r −1 7

Question: Event of tossing a dice and setting 2 in even no of throws.

Options:
5
(a)
11
6
(b)
11

3
α tan −
Question: (1 +y2) (1 + ln x) dx + x dy = 0 Passes thrown (1, 1) find f ( e ) = 2 .
3
β + tan −1
2
Find 𝛼𝛼 + 2𝛽𝛽

1
Question: Z = + 2i, z + 1 = α z + β ( l + i )
2
Find 𝛼𝛼 + 2𝛽𝛽 or 2𝛼𝛼 + 𝛽𝛽
Solution:
 1
Z= + 2i, z + 1 = α z + β + β i
2
3 9 3 
Z + 1 = + 2i, + 4 = α  + 2i  + β + β i
2 4 2 
3α 5
+ β = , 2α + β =⇒ 0 β= −2d
2 2
⇒ 3α − 4α =⇒ 5 α= −5, β = −10

Question: (a, b) R(c, d) a, b, c, d ∈ Z ab - bd is divided by 5.

Options:
(a) S, R not T
(b) Not Transitive
Solution: Not Transitive

Question: x2 + y2 = 169, 5x - y = 13, find area inside circle lying below the line.

 −π π 
Question: 4cos𝜃𝜃 + 5 sin 𝜃𝜃 = 1, x is a solution Find tan x. ∈  ,
 2 2 

1 − f 2   2f  x
Question: 4  2
+5 2
= I when f = tan
1 + f  1 + f  2