JEE-Mains-06-04-2023 [Memory Based]

[Morning Shift]

Physics
Question: Current 2I in both rings, find resultant B?

Solution:

Current = 2 I

B= B=
µ0 ( 2I )
1 2
2a
So,
BN
= B12 + B22
µ0 µ I
= =
2a
( )
2I 2  0 
 a 

Question: If rate of heat supplied to the system is 1000 watt and the rate of work done by the
system is 200 watt. Find rate of change of internal energy.
Answer: 800.00
Solution:
dQ
= +1000 watt
dt
dw
= +200 watt
dt
dQ dw du
= +
dt dt dt
dw
+1000 =
+200 +
dt
dw
= +800 watt
dt

Question: Find the ratio of energy density of E and B in EM waves.

Options:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) None of these
Answer: (a)
Solution:
1
Average electric field energy density = ε 0 E 2
2
B2
Average magnetic field energy density =
2 µ0
As both are equal
1
ε0E2
2 =1
B2
2 µ0

Question: Percentage error in equivalent resistance if connected in parallel (10 ± 0.5) ohm
and (15 ± 0.5) ohm
Options:
(a) 13 %
(b) 3 %
(c) 13/5 %
(d) 13/3 %
Answer: (d)
Solution:
 ∆Req ∆R1 ∆R2
= +
Req2 ∆R12 ∆R22
∆Req  0.5 0.5  15 ×10
=  2 + 2 ×
Re q  10 15  15 + 10
 1 1  150
= 0.5  2 + 2  ×
 10 15  25
 1 1   225 + 100 
= 3 +  = 3 
 100 225   225 ×100 
∆R 3 × 325 13
×= 100 100
×= %
R 225 ×100 3

Question: Assertion: Earth has atmosphere while moon does not.

Reason: Escape velocity in moon is very small than earth.
Options:
(a) A correct, R correct & R is correct explanation
(b) A correct, R correct but not correct explanation
(c) A correct, R false
(d) A false, R false
Answer: (a)
Solution: A correct, R correct & R is correct explanation

Question: A mass of 100 g is rotated with a spring of natural length 20 cm, with angular
velocity 5 rads–1. Find tension in spring [R = spring constant 7.5 nm–1]
Answer: 0.75
Solution:

= mω 2 (l0 + x)
T= kx
1
7.5 x =× 25(0.2 + x)
10
3=x 0.2 + x
x = 0.1
so T = kx = 7.5 × 0.1 = 0.75 N
 Question: A solid infinite cylindrical wire with radius a is carrying current I find the graph of
magnetic field inside & outside the wire.
Answer:
Solution:

Dependency of magnetic fixed in solid current carrying wire

µi
Bout = 0
2π r
µi
Bsurface = o
2π R
Binside

Current through h loop

i
=i1 ×π r2
πR 2

∫ B ⋅ dl =
µi o

i ×π r2
B.2π r = µo
π R2
µo i
=B ×r
2π R 2
B∝r

Question: A: range is max at θ = 45°

R : range is max when sin 2θ = 1
Options:
(a) R : true A : False
(b) R : true A : True
(c) R : False A : False
(d) R : False A : True
Answer: (b)

Question: Graph of electric potential inside conducting solid sphere is

 Solution:
kQ
vinside
= vsurface
= = constant
R

Question: In a capacitor when liquid of dielectric constant ‘k’ is filled upto height d/3 then
capacitance is 2μF. Find capacitance when it is filled till x = 2d/3 Take k = 2

Answer: 2.5
Solution:

3 Akε 3 Aε 0
.
Ceq = d 2d
Aε  1
3 0 k + 
d  2
3K Aε 0 3 × 2 Aε 0 Aε 0 10
=2 = ⇒2 ⇒=
 1  2d 5 2d d 6
K +  2
 2
 3 AK ε 0 3 Aε 0
.
2 d d 3 Aε 0 k
=Ceq =
3 Aε 0  k  2d  k 
 + 1  2 + 1
d 2 
3  10   2 
=Ceq =   2.5 F
2  6   2 

Question: If retardation of a body of mass 10 gram is given as 2x, where x is the position of
−n
10 
the particle starting from origin at rest. If loss of kinetic energy is   find n .
x
Answer: 2.00
Solution:
x
∆KE =W =∫ mad x
0
x −2
10 1 2  x 2   10 
∆KE = ∫ ( −2 x ) dx =
− .x =
− = 
100 0 100 100   x 
So n = 2.

Question: Two spheres of mass 2 kg each placed on the ends of a light rod and r = 10 cm and
dist b/w the centres = 40 cm find MOI about centre of the rod perpendicular to the line
joining centres.
Answer: 0.17
Solution:
 I I cm + ml 2
=
2
= MR 2 + Ml 2
5
2
= × 2 × ( 0.1) + 2 ( 0.2 )
2 2

5
4 8
=I +
500 100
4 + 40 44
=I = kg m 2
500 500
For 2 spheres
44
I final = 2 × I = 2 × = 0.176 kg m2
500

Question: Resistivity of semiconductor changes with temp according to which graph

Options:
(a)

(b)

(c)

(d)

Answer: (d)
Question: Alpha, electron, proton has KE is such that Kα = 4K, Ke = 2K, Kp = K write order
of de broglie wave
Solution: λe > λP > λα

Question: For the oscillations exhibited by the spring block system on the smooth surface
along the spring, the time period is equal to

Options:
m ( K1 + K 2 )
(a) 2π
K1 K 2
m ( K1 + K 2 )
(b) 2π
2 K1 K 2
m
(c) 2π
K1 + K 2
m
(d) π
K1 + K 2
Answer: (c)
Solution:

Since the springs are in parallel connection

keq= k1 + k2
M
T = 2π
K1 + K 2

Question: A car is moving with speed of 15 m/s towards a stationary wall. A person in the
car press the horn and experience the change in frequency of 40 Hz due to reflection from the
stationary wall. Find the frequency of horn. (Use vsound = 330 m/s)
Answer: 420.00
Solution:
  V 
f ′ = f0  
V − VC 
V + VC 
f ′′ = f ′  
 V 

 V  V + VC 
f ′′ = f 0  
V − VC   V 
V + VC 
f ′′ = f 0  
V − VC 
f ′′ − f 0 =
40
 330 + 15 
f0  40
 − f0 =
 330 − 15 
 23 
f 0  − 1 = 40
 21 
f 0 = 420

Question: Communication system

Height of the tower increased 21% percentage increase in range.
Options:
(a) 10
(b) 12
(c) 14
(d) 15
Answer: (a)
Solution:
Range = 2 RE h
R1
= 2 RE ⋅=
h 2 RE ⋅ h
 21 
R2
= 2 RE  h + ⋅h
= 2 RE (1.21h)
 100 
R1 2 RE h 1 1
= = =
R2 2 RE (1.21h ) 1.21 1.1
1.1R1
⇒ R2 =
( R2 − R1 ) ×100
% change in R =
R1
 1.1R1 − R1
= ×100
R1
1.1 − 1
×100 =
10%
1

Question: If length of wire is increased 20% and area is increased 4% the % change in
resistance is
Answer: 15.00
Solution:
ρl
R=
A
ρ (1.2) 12
=R′ R′
⇒= =R 1.15 R
1.04 A 1.04
⇒↑ 15%

Question: A Body has mass m and moving with const vel in viscous fluid having coiff. of
viscosity 𝜂𝜂 density is ⍴b liquid density ⍴L find vel v
Solution:
 ρ
mg 1 − 
 S  =v
6πη r

Question: Which gate is this

Options:
(a) NOR
(b) OR
(c) AND
(d) NOT
Answer: (a)
 JEE-Mains-06-04-2023 [Memory Based]
[Morning Shift]

Chemistry
Question: Polymer which is named as orlon is?
Options:
(a) Polyacrylonitrile
(b) Polycarbonate
(c) Polyethene
(d) Polyamide
Answer: (a)
Solution: Orlon is also called Acrilan or Polyacrylonitrile

Question: The correct set of strong oxidising and reducing agent

Ce4+, Yb2+, Tb4+ and Eu2+
Options:
(a) Ce4+, Tb4+, Yb2+, Eu2+
(b) Tb4+, Yb2+, Ce4+, Eu2+
(c) Tb4+, Eu2+, Yb2+, Ce4+
(d) Yb2+, Eu2+, Tb4+, Ce4+
Answer: (a)
Solution: Ce4+, Tb4+ act as oxidising agent and Yb2+, Eu2+ act as reducing agent

Question: Match column I (Deficiency) with column II (Disease)

Vitamins Deficiency Disease
(P) Vitamin A (1) Scurvy
(Q) Vitamin C (2) Xeropthalmia
(R) Vitamin B1 (3) Cheilosis
(S) Vitamin B2 (4) Beri-Beri
Options:
(a) P-2, Q-1, R-4, S-3
(b) P-2, Q-4, R-3, S-1
(c) P-4, Q-2, R-4, S-1
(d) P-3, Q-2, R-4, S-1
Answer: (a)
Solution: Fact based

Question: Y form FCC lattice in which X occupies 1/3 of tetrahedral Voids. Then formula of
the compound will be
Options:
(a) X3Y2
(b) XY3
(c) X2Y3
(d) X3Y
Answer: (c)
Solution: tetrahedral voids are 8 in count in FCC thus X is 8/3 and Y = 4 hence the formula

Question: Which of the following have highest electron gain enthalpy difference?
Options:
(a) F, Ne
(b) Ar, F
(c) Ne, Cl
(d) Ar, Cl
Answer: (a)
Solution: Fact based
EA values are F = –333, Cl = -349, Ne = 116, Ar = 96

Question: Name reactions Matching

Name Reaction Reagents
(P) Etard Reaction (1) NaOI
(Q) Iodoform (2) CO/HCl, Anh. AlCl3
(R) Gatterman aldehyde (3) CrO2Cl2, CS2, H3O+
(S) HVZ (4) X2/red P, H2O
Options:
(a) P-3, Q-1, R-2, S-4
(b) P-3, Q-2, R-1, S-4
(c) P-3, Q-4, R-2, S-1
(d) P-1, Q-3, R-2, S-4
Answer: (a)
Solution: Fact based

Question: Match column I (Compound) with column II (Type of Bond)

Nitrogen oxides Type of Bonds
(P) N2O (1) N-N bond
(Q) N2O5 (2) N-O-N bond
(R) NO2 (3) N=N or N triple bond N
(S) N2O4 (4) N=O
Options:
(a) P-1, Q-4, R-2, S-3
(b) P-3, Q-2, R-4, S-1
(c) P-1, Q-2, R-4, S-3
(d) P-1, Q-3, R-2, S-4
Answer: (b)
Solution: structure-based question

Question: Photochemical smog is maximum in

Options:
(a) Himalayan Region
(b) Green Healthy vegetation
(c) Marshy Lands
(d) Industrial Region
Answer: (d)
Solution: Hydrocarbons and nitrogen oxides produced by automobiles and factories.

Question: Which of the reaction is correct among the following with appropriate enzyme?
Options:
(a) Sucrose → Glucose + fructose : Enzyme – Invertase
(b) Glucose → CO2 + Ethanol : Enzyme : Maltase
(c) Protein → Amino acid : Enzyme : Zymase
(d) Starch → Maltose : Enzyme : Pepsin
Answer: (a)
Solution: Sucrose → Glucose + fructose : Enzyme – Invertase

Question: Which of the following is used for settling of cement?

Options:
(a) Gypsum
(b) Limestone
(c) Clay
(d) Silica
Answer: (a)
Solution: Setting of cement: When mixed with water, the setting of cement takes place to
give a hard mass. This is due to the hydration of the molecules of the constituents and their
rearrangement.

Question: which of the following is having square Pyramidal shape

Options:
(a) XeOF4
(b) BrF5
(c) IF5
(d) ICl4-
Answer: (a)
Solution: XeOF4 has geometry of Sp3d2 and shape of square pyramidal

Question: Assertion: Loss of the electron from hydrogen atom results in nucleus (H+) of ~
1.5 × 10-3 pm size.
Reason: H+ does not exist freely and is always associated with other atoms or molecules.
Options:
(a) Both assertion and reason are correct but reason is not correct explanation
(b) Both assertion and reason are correct but reason is correct explanation
(c) Both assertion and reason are incorrect
(d) Assertion is correct and reason is incorrect
Answer: (b)
Solution: Loss of the electron from hydrogen atom results in nucleus (H+) of ∼1.5 × 10–3 pm
size. This is extremely small as compared to normal atomic and ionic sizes of 50 to 200pm.
As a consequence, H+ does not exist freely and is always associated with other atoms or
molecules. Thus, it is unique in behavior.

Question: Assertion: The magnetic Moment of [Fe(H2O)6 ]3+ and [Fe(CN)6 ]3– are 5.92 BM
and 1.74 BM respectively.
Reason: The oxidation state Fe is +3.

Options:
(a) Both assertion and reason are correct but reason is not correct explanation
(b) Both assertion and reason are correct but reason is correct explanation
(c) Both assertion and reason are incorrect
(d) Assertion is correct and reason is incorrect
Answer: (a)
Solution: water as ligand do not cause pairing in complex but CN- does

Question: If radius of ground state hydrogen is 51 pm, find out the radius of 5th orbit of Li2+
(closest integer)
Options:
(a) 170 pm
(b) 180 pm
(c) 120 pm
(d) 425 pm
Answer: (d)
Solution: Apply r = 51*5*5/3

Question: Identify the product formed in the following reaction.

Options:
(a)

(b)
(c)

(d)

Answer: (d)
Solution:

Question: Matrix match for detection of element

Column-I Column-II
(A) Nitrogen (P) AgX
(B) Sulphur (Q) (NH4)3PO4.12MoO3
(C) Phosphorous (R) Fe(SCN)3
(D) Halogens (S) Fe4[Fe(CN)6]3
Options:
(a) A-P, B-R, C-Q, D-S
(b) A-R, Q, B-P, C-Q, D-S
(c) A-S, B-R, C-Q, D-P
(d) A-Q, B-R, C-P, D-S
Answer: (c)
Solution: A-S, B-R, C-Q, D-P

Question: Consider the following reaction.

A2B3(g) ⇌ 2A(g) + 3B(g)
If the initial concentration of A2B3(g) is c, find the value of α
Options:
1
 K eq  5
(a)  4 
 27c 
1
 K eq  5
(b)  4 
 c 
1
 K 5
(c)  eq 4 
 108c 
1
 K eq  5
(d)  4 
 4c 
Answer: (c)
Solution:
1
 K eq  5
 4 
 108c 
 JEE-Mains-06-04-2023 [Memory Based]
[Morning Shift]

Mathematics

x 2 ( x sec 2 x + tan x )
Question:  ( x tan x + 1)
2
dx is equal to

− x2
Answer: + 2ln x sin x + cos x + c
x tan x + 1
Solution:

x 2 ( x sec 2 x + tan x )
 ( x tan x + 1)
2
dx

Integrating by parts
−1  −1 
I = x2  −  2x   dx
x tan x + 1  x tan x + 1 

− x2 x cos x
= + 2 dx
x tan x + 1 x sin x + cos x
− x2
= + 2ln x sin x + cos x + c
x tan x + 1

15
 1
Question: The coefficient of x 18
in the expansion of  x 4 − 3  is
 x 
Options:
(a) 14 C7
15
(b) C8
15
(c) C6
14
(d) C8
Answer: (c)
Solution:
15
 4 1
x − 3 
 x 
r
 1  15
Tr +1 = Cr ( x )
4 15− r
 − 3  = Cr ( −1) x
15 r 60−7 r

 x 
60 − 7 r = 18
 7 r = 42
r =6

Coefficient of x18 = 15C6

Question: The number of ways of distributing 70 distinct oranges among three children such
that each child gets atleast one orange is
Answer: 370 − 3 ( 2 70 − 2 ) − 3
Solution:
Number of ways = 370 − 3C1  270 + 3C2 170

= 370 − 3 ( 2 70 − 2 ) − 3

Question: Sum of first 20 terms of the series 5,11,19, 29, 41,..... is

Answer: 3250.00
Solution:
Let Sn = 5 + 11 + 19 + 29 + 41 + ...tn

tn = an 2 + bn + c

a+b+c =5
4a + 2b + c = 11
9a + 3b + c = 19
5a + b = 8
3a + b = 6
2a = 2
a =1
b=3
c =1

tn = n 2 + 3n + 1

Sn =  tn

n ( n + 1)( 2n + 1) 3n ( n + 1)
Sn = + +n
6 2
20  21  41 2  20  21
S 20 = + + 20
6 2
= 2870 + 630 + 20
= 3520
Question: Number of ways in which 20 chocolates can be given to 3 children such that each
gets atleast one is ____.
Answer: 19 C2
Solution:
x + y + z = 20
x, y, z  1
X + Y + Z = 17
X ,Y , Z  0
n + r −1 19
Cr −1 i.e. C2

Question: If 5 pairs of dice are thrown. Success is getting a sum 5. If the probability of
K
getting atleast 4 success is 11 , then the value of K is
3
Answer: 123.00
Solution:

(1, 4) , ( 2,3) , (3, 2) , ( 4,1)

4 1 8
p= = , q=
6 9 9
K
= P ( atleast 4 success ) = 5C4 p 4 q1 + 5 C5 p 5
311
K 1 8 1
11
= 5 4  + 5
3 9 9 9
K 40 1
 = +
311 310 310
K
 = 41
3
 K = 123

Question: If the ratio of the 5th term from the start to the 5th term from the end in the
n
 1 
expansion of  4 2 + 4  is 6 :1 . Find the 3rd term from start.
 3
Answer: 60 3
Solution:
n
4 1 
 2+ 4 
 3
 4

( )  1 
n−4
n 4
C4 2 4 
 3 =
6
n4

( 2 )  1 
n 4
4 1
C4 
4
3

( 2)
n −8
4

n −8
= 6
 1 
4 
 3

( 6)
n−8
4
= 6

( 6)
n−8
2
= 6

n = 10
2
 14   1 
8
10
C2  2    1 
   3 4 

1
= 45  4 
3
= 60 3

Question: Mean of 15 observations is 12 and its variance is 14. Mean of another 15

observations is 14 and variance  2 . Combined variance is 13. Find  2 .
Answer: 10.00
Solution:
Subtract each entry by 13

So 14 =
x i
2

− ( −1)   xi 2 = 2.25
2

15

& 2
=
y i
2

− 12   yi 2 = 15 ( 2 ) + 1
15

225 + 15 2 + 15
13 =
30
2
 8+ =0
2
 = 10
2
Question: A 2  2 matrix A is such that none of its elements is 0, and A2 = I . ‘ a ’ is the
sum of diagonal elements and ‘ b ’ is A . Find 3a2 + 4b2 .
Answer: 4.00
Solution:
2
 p q   p 2 + qr pq + qs 
A =
2
 = 
 r s   pr + rs qr + s 2 

( p + s) q = r ( p + s) = 0
p+s =0 p = −s

p 2 + qr = 1

So, a = 0, b = ps − qr

= − ( p 2 + qr ) = −1

So 3a2 + 4b2 = 4

dy
Question: If 2 x y + 3 y x = 20 . Find at (2, 2).
dx
Answer:
Solution:
2 x y + 3 y x = 20

f
= − x
dy
dx f
y

2 y x y −1 + 3 y x ln y
=− y
2 x  ln x + 3xyx − 1
2 + 3ln 2
=−
1ln 2 + 3
2 + ln 8
=−
ln 4 + 3

Question: If a1 , a2 , a3 ,...., an are in A.P.


d 1 1 
lim  + .... 
n →  a + a
n a3 + a2 
 2 1

Answer: 1.00
Solution:
a1 , a2 , a3 are in A.P.

d  1 1 
lim  + .... 
n → n  a + a a3 + a2 
 2 1

d  an − a1 
= lim  
n → n  d 

 a1 + ( n − 1) d d a1 
= lim  − 
n →
 n d nd 
=1

Question: f = 9 + 13sin x when x 0,   . Find the number of points where f is not
differentiable.
Answer: 25.00
Solution:

a + b sin x; x 0,  

Non-differentiable at 2b − 1 points
2b − 1 = 2  13 − 1 = 25

Question: Image of point P (1, 2,3) about the plane 2 x − y + 3z = 2 is Q , then the area of
PQR = ? where R = ( 4,10,12)
1531
Answer:
2
Solution:
 −1  −2  −3 −2 ( 7 )
= = =
2 −1 3 14
 −1  −2  −3
= = = −1
2 −1 3

Q ( −1,3,0)

1
Area =  PQ  PR
2

1531
=
2

2
1 1
Question: If 5 f ( x ) + 4 f   = + 3 , then 18 f ( x ) dx is:
x x 1

Answer: 10 log e 2 − 6
Solution:

1 1
5 f ( x) + 4 f   = + 3
x x

1
5 f ( x) + 4 f   .... (1)
 x
1
Take x =
x

1
5 f   + 4 f ( x) = x + 3 .... ( 2 )
 x

(1)  5 − 4  ( 2)
5
9 f ( x) = + 15 − 4 x − 12
x
 5
9 f ( x) = − 4x + 3
x
By integrating
2 2
5
9  f ( x ) dx =  − 4 x + 3 dx
1 1
x
2 2
10
2  9 f ( x ) =  − 8x + 6
1 1
x
2
8x2
= 10 ln x − + 6x
2 1

= (10ln 2 −16 +12) − ( 0 − 4 + 6)

= 10log e 2 − 6

Question: The sum of roots of x2 − 8x + 15 − 2 x + 7 = 0 is

Answer: 9 + 3
Solution:

x2 − 8x + 15 − 2 x + 7 = 0

( x − 3)( x − 5) − 2 x + 7 = 0
x  3 or x  5

x 2 − 8 x + 15 − 2 x + 7 = 0
x 2 − 10 x + 22 = 0

10  100 − 88
x=
2

= 5 3
Take intersection

x = 5+ 3
3 x5

− x2 + 8x −15 − 2 x + 7 = 0
x2 − 6 x + 8 = 0

( x − 2)( x − 4) = 0
x = 2, 4
x=4

So, sum of roots is 9 + 3

Question: ( P  Q )  ( R  Q) is equivalent to:

Options:
(a) ( P  R )  Q
(b) ( P  R )  Q
(c) ( Q  R )  ( P  R )
(d) ( R  P )  (Q  R )
Answer: (a)
Solution:

( P  Q)  ( R  Q)
 ( P  Q)  ( R  Q)

 ( P R)  Q

 ( P  R)  Q
 ( P  R)  Q

Question: Let a = 2iˆ + 3 ˆj + 4kˆ, b = iˆ − 2 ˆj − 2kˆ, c = −iˆ + 4 ˆj + 3kˆ and d is a vector

2
perpendicular to both b and c and a  d = 18 , then a  d is
Answer: 720.00
Solution:

(
d =  b c )
iˆ ˆj kˆ
= 1 −2 −2
−1 4 3

(
=  2iˆ − ˆj + 2kˆ )
a  d = 18
  ( 4 − 3 + 8) = 18
 =2

( )
2 2
a  d = a2d 2 − a  d

= 29  36 − 182
= 18 ( 58 − 18 )
= 18  40
= 720

n2 + 3n
Question: If 2n
C3 : nC3 = 10, then is equal to
n2 − 3n + 4
Answer: 2.00
Solution:
2n
C3 : n C3 = 10

2n ( 2n − 1)( 2n − 2 )
= 10
n ( n − 1)( n − 2 )

2 ( 2n − 1)  2 ( n − 1)
 = 10
( n − 1)( n − 2)
4n − 2 = 5n − 10
n =8
n2 + 3n 64 + 24 88
= = =2
n − 3n + 4 64 − 24 + 4 44
2