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Solutions to JEE Main - 5 | JEE-2023 (Gen 3)

PHYSICS
SECTION-1
1.(C) The work done by man is negative of magnitude of decrease in potential energy of chain.

m   3mg 
Wext  U  g  mg 
2 4 2 8
(1,1)
 
2.(B) W=  F .ds
(0,0)

Here ds = dxiˆ + dyjˆ + dz kˆ
(1,1) (1,1)
2  x3 y 2  5
 W=  ( x dx  ydy )  W=    = J
(0,0)  3 2  (0,0) 6
dU
3.(A) F=–  F = 3x2 – 12x
dx
dF
Now F = min.  =0  x=2m
dx
Ui + Ki = Uf + Kf
1 1
15 + × 2 × 80 = [– (2)3 + 6 × (2)2 +15] + × 2 × v2
2 2
2
v = 64
v = 8m/sec
4.(A) a x
Wnet = F. x
 a. x
 x2
 K  x2

5.(D) Apply impulse equation in x-direction

J cos60° = 4 × 4
[J = 32 N.S]

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6.(C) v  u 2  2 gh  5 gr  2 gr  3 gr

1 2
7.(C) Work done in changing speed from 0 to V is: W1   mV  0
2
work done in changing the speed from V to 2V is
1 1 1 W1 1
2
W2    m  2V   mV 2  3mV 2  
2 2 2 W2 3

dU
8.(A) From F = –
dx
U ( x) x x kx 2
0 dU = – 0 Fdx = – 0 ( kx ) dx  U (x) = – as U (0) = 0
2

9.(A) Due to conservation of momentum in the vertical direction , the man will begin to move up with speed
mu h  mu   h 
. The ball will reach the floor in time . In this time the man will move up by     . Total
M u  M  u
mh
distance of man from floor = h + .
M
10.(B)

a a
2m   m  2a 2m   ma
xcm  2 =a ycm  2 =0
3m 3m

11.(B) Let x be the maximum displacement of block downwards. Then from conservation of mechanical
energy:
decrease in potential energy of 2 kg block = increase in elastic potential energy of both the springs
1 2mg (2)(2)(10)
 mg x = (k1 + k2) x2 or x= = = 0.1 m
2 k1  k 2 100  300
(k1  k 2 ) x  mg
Acceleration of block in this position is a = (upwards)
m
(400)(0.1)  (2)(10)
=
2
2
= 10 m/s (upwards)

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12.(A)

m1v0 = m2v2 – m1v1 …(1)

v v
e= 1 2 =1 …(2)
v0
Solving (1) and (2)
2 2
 v1   m1  m2 
  = 
 v0   m1  m2 
13.(B) For the equation N  mg sin 
1
 mg cos   mg sin   tan     ...(i)
2
x2
From equation y =
20
dy x
 tan  = = ... (ii)
dx 10
From (i) & (ii) , x = 5 , so y = 1.25
14.(A) Normal reaction of each surface,

N N mg
 = mg cos   N = cos  ...(1)
2 2 2


 mgsin    k N  k N   ma  a  g sin   2 k cos  
15.(C)

f1max =  mg

f2 max = (3mg) =  mg
3

T= 0  Tension will be zero in all circumstances

16.(A) The mass of water reaching the wall per second = v Ad.
the momentum transferred per second = (vAd)v

 p per unit time = v2 Ad.
The flow of water , tangential to the wall does not contribute to the force.
Therefore, force = rate of change of momentum = v2 Ad.

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17.(A) F – 2T = 6a and T = 4 × 2a
 F – 16 a = 6a
F 6 1  4  2
 a=  a = 1m/s2  aCM = = 1.4 m/s2
22 10
18.(A) In this case spring force is zero initially F.B.D of A and B

19.(B)

20.(B) x 2  y 2  2
dx d
2x  0  2 [y = constant]
dt dt
d
dx dt v
or  
dt ( x / ) sin 

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SECTION-2
21.(2) For maximum compression, velocity of each block is same.
By momentum conservation.
1×2=1×v+1×v
v = 1 m/s
By energy conservation.
1 1 1 1 2
× 1 × 22 = × 1 × (1)2 + × 1 × (1)2 + kx
2 2 2 2
 x = 0.2 m = 2/10 m
X = 2m
22.(5)

 Vy   Vy  2eV y
   Vx  eVx    eVx 
 g   g  g
e  0.5
23.(9) At x = 5 m, U = 29 J and K = 20 J
 E = 49 J
Now, E = Umin + Kmax
49 = 20 + Kmax
Kmax = 29 J

24.(4) Area under P – x graph =  Pdx

v V
2  mv 3 
=  mv dv = 
 3 1

1

10 3
 (v  1) …(1)
73
1
from graph, area = × (2 + 4) × 10 = 30 …(2)
2
from (1) & (2)
10 3
(v – 1) = 30  v = 4m/s
73
25.(2) T  2  cos180  4T  VB  cos0  0
1
 VB  m / s
2

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Chemistry

SECTION-1
3
1.(C) For a real gas at critical conditions, Z1 
8
3
At Boyle’s temperature, Z 2  1  Z1  (Z 2 )
8
2.(D) According to the ideal gas equation, PV = nRT
For fixed quantity of a perfect gas:
1
(i) at constant temperature: P , nP  nK  nV
V
(ii) at constant pressure: VT, nV  nK ' nT
(iii) at constant volume: VT, nP  nK" nT

3.(C) BF3 ,SO3 , XeF4  planar and non-polar molecules

NF3 , NH 3  non planar and polar molecules

4.(B) (i) Stability of negative charge is greater over more EN atom thus order of stability
1 1 1 1
(A) N  N O  N  N  O
1 1
(D) N  CO  N  C  O
(ii) CH 3  C  O  and CH 3  C  O are identical and equally stable resonating structures
|| |
O O
(iii) When octet is complete then neutral state of atoms will be more stable than their charged
1 1
state thus for CO 2 , O = C = O is more stable than O  C  O

5.(A) B2 H 6 , AlCl3 and NO2 have central atoms in which total valence electrons are less than 8 thus these
  
given species are electron deficient. In BH 4 , NH 4 , CCl 4 , SiCl4 and NO3 their respective central
atoms have total 8 valence electrons thus these species are NOT electron deficient.

6.(A) Using Pf Vf  PA VA  PB VB
Put PA  2P, PB  P, VA  V, VB  2V, Vf  3V
2PV  2PV 4 (Pf ) Pf 4
 Pf   P For vessel B:  
3V 3 Pi P 3
a
7.(B) If b = 0 then Z m  1 
RTVm
1 a
Zm  y-axis  x-axis slope  
V(m) RT
At a temperature T, with increase in value of “a” slope decreases. So, correct option is B. For
H 2 (g) Z(m) can’t be less than 1.

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8.(C) (a) due to increase in number of lone pairs over the central atom, bond angle of the hydride
compounds with sp3 hybridization decreases, hence order of bond angle: CH 4  NH 3  H 2O
(b) due to significant H-bonding in NH3 and H 2O, order of boiling point: CH 4  NH3  H 2O
(c) According to M.O.T, bond orders are:
B2 O2 N2
(1) (2) (3)

9.(B) Structure of SNF3 is

 Zero lone pair over the central S atom

 S forming 2 pie bonds using its valence d orbitals.
3 3
10.(B) XeF6  sp d , polar
XeF2  sp3d , Non polar
SF6  sp3d 2 , Non polar
IF3  sp3d , polar
11.(D) Reduction in volume on cooling  VH O(g)  100ml
2

Reduction in volume on passing through an alkali solution  VCO  100ml

2
As reaction is
C x H y  (x  y / 4)O 2  x CO 2  y / 2H 2O
(50) (100) (100)
(1) (2) (2)
x2 y4
28
Hence C x H y  C2 H 4 , M  28, vapour density   14
2
8
12.(A) As TC  TB
27
27
 TB  [200]  27[25] kelvin  675 kelvin or (675  273)  402C
8
13.(B) H.O.M.O of O 2 is  *2p and LUMO of N 2 is  *2p .
Bond strength of B2  F2 . Lower bond strength of F2 is due to effective repulsion between lone pairs
of the bonded F atoms.

14.(B) Statement (B) is correct i e. VRMS  Vavg for all types of molecular speeds distribution
(A) Average molar thermal kinetic energy of an ideal gas depends only upon temperature (acc. To
K.T.G)
(C) Collisions between ideal gas particles are perfectly elastic
(D) Ideal gases have NO potential energy

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10 14 7
15.(C) PHe   He PTotal  (5.6)    1.75
32 8 4
 w/4 1/ 4 10 
He    
 w / 2  w / 4  w / 20 1/ 2  1/ 4  1/ 20 32 

16.(C)

Cl-atom: sp3 sp3 sp3

 bonds: Three Two One
  
Cl O Cl O Cl O
(d) (p) (d) (p) (d) (p)
Avg. Cl–O 7/4 5/3 3/2
bond order :
Magnetic Diamagnetic Diamagnetic Diamagnetic
nature:

17.(D) b  N A [4Vsphere ]

4
 6 1023[4  r 3 ]
3
 6  1023[16r 3 ]
 6 16  1023[4]3 1030 m3 / mole
 96  64  107 m3 / mole
 6.144  104 m3  0.6144 L / mole
18.(A) Using the given figure it can be concluded that molar mass of X is greater than that of Y. Thus V.D of
X > V.D of Y; and at a given temperature VMP (X)  VMP (Y), VRMS (X)  VRMS (Y) .
19.(C) According to M.O.T, both Assertion and Reason are true and the Reason is correct explanation of the
Assertion.
20.(D) For a real gas
8a a a
TC  VC  3b PC  TB 
27Rb 27b 2 Rb

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SECTION-2

21.(5) A H 2O molecule can be involved in 4 H bonds, thus x = 4, Anion of KO 2 is O 2 having 1 unpaired
electron, thus y = 1
 (x  y)  4  1  5
22.(4) Using Graham’s law of diffusion
n H 2 /t H
2
M oO2 32
   16  4
n O2 /t O M oH2 2
2

 n H 2  1, t H2  2sec. 
   t O 2  4sec.
 n O  1/ 2, t O  ? 
 2 2 
3 3
23.(7) K.E(avg)  RT  [8.314][273]  3.404kJ / mole
2 2
 x  3.404  2x  6.808
24.(2)  = charge × distance

 1.6  1010 esu  1.25  108 cm

 2  1018 esu  cm  2 Debye
(As 1018 esu × cm = Debye)
25.(9)
nRT P(N 2 )  Pair  1.23  1.20  x  2.25
P(N 2 ) 
V = 0.03 atm. Then
0.5[0.082][300] 4x = 9
 H (Hg)  0.03  75  2.25cm
10
 1.23atm

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Mathematics

SECTION-1
1.(A) 7bc or a7c or ab7 i.e. total of 1 9  9  8  1 9  8  9  1  81  72  72  225 numbers
2.(B) ADIMNR
Words start with A  5!
with D  5!
with I  5!
with MAD  3!
with MAI  3!
Next MANDIR i.e. at 3  120  6  6  1  373

7! 7!
3.(A) First arrange MTHMTCS in 
2! 2! 4
4!
Now arrange vowels AEAI in 8 gaps | M | T | H | M | T | C | S | in C4 
2!
7! 8! 4! 7!  7!
     210  7!
4 4! 4! 2 4!

4.(A) 4 digit numbers  4  4  3  2  96

5 digit numbers  5  4  3  2  1  120
Total = 216 integers.

5.(C) 100!  297  348  . . . .

32

 23  3   2  316  . . .

6.(B) There are 4 prizes and three students.

Since, each prizes can be given to any persons.
Required number of ways = 3 × 3 × 3 × 3 = 34 ways.
7.(A) Let n be the number of diagonals of a polygon.
n n  n  1
C2  n  44 ;  n  44 ; n2  3n  88  0  n  8 or 11 ; n =11
2
8.(B) Total number of numbers = 2×2×2×…..×10 times = 210.
9.(C) As the 5 days can sit in 5! ways, there are 6 places are vacant in which the girls can sit in 6P3 ways.
Therefore, required number of ways are 6 P3  5! .

10.(B) In LUCKNOW, number of letters = 7(all distinct), vowels (U, O) = 2, consonants (L, C, K, N, W) = 5
There are 7 letters so 7 places are required.
Number of even places are 3 and vowels are 2, so 2 vowels can be placed in 3 C2 . 2!  6 and 5
consonants can be placed in remaining places in 5! ways. Hence, number of words = 6.5!  720 .

11.(D) Required number of ways  11C8  165

{As captain already be chosen, now from 11 players 8 are to be chosen}

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12.(A) Vowels = A, A, I, U
6
Case 1 : =
2
6
Case 2 : = 2
2
6
Case 3 : = 2
2
6
Case 4 : = 2
2 2
 360  720  720  360  2160

13.(A) x 2  2ix  1  0

2i  2  4 2  i  1 i 1
x   
2 2 2 2
  3 3
 x  cos  i sin or cos  i sin
4 4 4 4
1 2187  2187  3
x 2187  2187
 2i sin or 2i sin  2i
x 4 4
2 2
14.(B) x 2

 y 2  1  4 x2 y 2  x2  y 2  1  
x4  y 4  1  2 x 2 y 2  2 x 2  2 y 2  4 x2 y 2  x 4  y 4  1  2x 2 y 2  2 x 2  2 y 2  x2  0
7
 2  
 2cos  2i sin cos    
2 2 2
15.(B)   27 cos7  cos  i sin 
 cos3  i sin 3  2 2 2
16.(C) Let z  x  i y

z  2 z 1  i  0  x  i  y  1  2  x  1  i y  0.

Equating imaginary parts, y  1  0 as z  1 is real

2
 2  x  1  i y   x  2  x  1  y 2   x 2
 
2 2
 2  x  1   1   x 2
 
  x  2 2  0  x  2

 z   2i  z  5

17.(C) If z  x  iy ,  x  12  y 2  x  2 and 0  y  2

7z
18.(A) Given, f  z   and z  1  2i
1  z2
7  1  2i  6  2i 6  2i 6  2i 1 i 6  4i  2 8  4i 1
 f  z       =  2  i
1  1  2i 
2 1  1  4  4i  4  4i 4 1  i  1  i  4 12  i 2 
4 2 2 
4 1 5 z
f z    ( z  1  2i , given  z  5 )
2 2 2

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19.(C) Let z1  r1  cos 1  i sin 1  , z2  r2  cos 2  i sin 2 

 z1  z2  z1  z2

 r1cos 1  r2 cos 2   i  r1 sin 1  r2 sin 2   r1 cos 1  ir1 sin 1  r2 cos 2  ir2 sin 2

 r12  r22  2r1r2 cos  1  2   r12 cos 2 1  r12 sin 2 1  r22 cos 2 2  r22 sin 2 2

 r12  r22  2r1r2 cos  1  2   r1  r2  r12  r22  2r1r2 cos  1  2   r12  r22  2r1r2
 cos  1  2   1  1  2  0
 1  2  arg  z1   arg  z2 

20.(B) x  a  b, y  a  b and z  a  b
Now,   
xyz   a  b  a  b2 a2  b , where    and   2

 xyz   a  b   a 2
  
 ab2  ab  b 2   a  b  a 2  ab  b 2  a3  b3

SPACE FOR ROUGH WORK

SECTION-2
This section contains Five (05) Numerical Value Type Questions. The answer to each question is an integer
ranging from 0 to 9 (both inclusive).
8
21.(3) Cr  7 C3  7 C2
8
Cr  7 C3  7C2
8
Cr  8C3  r  3

22.(9) 1      0
 1  2      (1  2  )3  3  1
3  3  5  3  3  3  2  2
(2)3  83  8

4!
23.(9) Case-1: 2 alike, 2 alike  2 C2  6
2!  2!
4!
Case-2: 2 alike, 2 different  2C1  6C2  360
2!
Case-3: All different  7C4  4!  840
 total words  6  360  840  1206

24.(5) For the given case if we try to mark chairs like C1 , C2 , C3 , ... C10 then first of the three persons can
take any one of the eight chairs C1 , C2 , ... C8
 total ways  8  3!  48  24  31

25.(6) Let 3  6 6 i  3  6 6 i  k  k 2  6  30  k  2 6 i  a  6

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