20-07-2025

3010CJA101021250023 JA

PART-1 : PHYSICS

SECTION-I(i)

1) Four identical particles A, B, C and D move with equal speed v, along the diagonal of a square
towards the centroid O of the square, beginning at the vertices. After the collision at O, A and C
retrace their path with speed 2v and B retraces its path with speed . Speed of the particle D

after the collision is

(A)
(B) v

(C)

(D) 2v

2) Magnitude of electric field and potential at corner (D) of a rhombus ABCD in case (i) and case (ii)

are respectively E1, V1 and E2, V2 :-

(A) E2 = E1 V2 = V1
(B) E2 = E1 V2 = 0
(C) E2 = V2 = 0
(D) E2 = V2 = V1

3) A particle having mass m, charge q enters a cylinder region having uniform magnetic field B in
the inward direction as shown. If the particle is deviated by 60° as it emerges out of the field then
what is the time spent by it in the field.

(A)

(B)

(C)

(D) It depends on the speed of particle.

4) For a wheatstone bridge shown; R1 > R2 and R3 = R4. The direction of current in wire AB is :

(A) From A to B
(B) No current flows in AB
(C) From B to A
(D) Data is not sufficient

SECTION-I(ii)

1) In a potentiometer wire experiment the emf of a battery in the primary circuit is 20V and its
internal resistance is 5Ω. There is a resistance box in series with the battery and the potentiometer
wire, whose resistance can be varied from 120Ω to 170Ω. Resistance of the potentiometer wire is
75Ω. The following potential differences can be measured using this potentiometer.

(A) 5V
(B) 6V
(C) 7V
(D) 8V

2) A particle of mass M and positive charge Q, moving with a constant velocity , enters a
region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field
extends from x = 0 to x = L from all values of y. After passing through this region, the particle

emerges on the other side after 10 milliseconds with a velocity . The correct
statements(s) is (are) :-

(A) The direction of the magnetic field is –z direction.

(B) The direction of the magnetic field is +z direction.

(C)
The magnitude of the magnetic field units.

(D)
The magnitude of the magnetic field is units.

3) If a source is connected to points a & b, the charge on all capacitors are equal. If it is connected

to points m & n, then also charge on all capacitors are equal.

(A) Cx = 3.5 µF
(B) Cy = 2 µF
(C) If source of voltage 6 volts is connected between m & b charge on Cy = 5µC.
(D) If source of voltage 6 volts is connected between a & n, the charge on Cx = 2µC.

4) Three conducting bodies A(solid sphere), B(thin spherical shell) and C(hollow sphere with
thickness 2R) are arranged as shown in the figure. A charge q is given to the inner sphere. Choose

the correct option(s).

(A)
When S1, S2 both are open capacity of the system is C1, then C1 is
(B) When S1 is closed but S2 is open capacity is C2, then C2 is 16πε0R
(C) When S2 is closed but S1 is opened capacity is C3, then C3 is 12πε0R
(D) When both S1 and S2 are closed. Capacity is C4, then C4 is 24πε0R
5) The figure contains an infinite slab having current per unit area of between the infinite
planes at x = –b and x = b. Slightly to the right of x = b, an infinite thin sheet is kept. It carries a

current per unit length . Choose the correct option(s) :

(A) The field due to sheet at any point is

(B)
The field due to sheet at any point is
The magnetic field due to slab at a point inside the slab must be independent of y co-ordinate
(C)
and y component of field must be odd function of x.
Magnetic field at a point (x, y) inside slab, due to slab is
(D)

6) Rod AB, BC, CD and DA for a square loop having current i, mass and length of each rod is m
and respectively, is situated in a uniform magnetic field B which is parallel to the plane of paper as
shown in the figure and it can rotate about fixed non conducting axis P1P2 perpendicular to and
rod AB, rod CD are attached with the axis, which lies in the plane of paper. At t = 0 the angle
between the plane of the loop and the magnetic field is 30o and point C is above the plane of the
paper (O is geometrical center of the loop). Then choose the correct option(s).
 Angular acceleration of the square loop at t = 0 is
(A)

Angular acceleration of the square loop at t = 0 is

(B)

(C)
Angular velocity when square loop rotated by 60o, is

(D)
Angular velocity when square loop rotated by 60o, is

SECTION-II(i)

Common Content for Question No. 1 to 2

Two charge particles each of mass ‘m’, carrying charge +q and connected with each other by a
massless inextensible string of length 2L are describing circular path in the plane of paper, each

0
with speed (where B is constant) about their centre of mass in the region in which an
uniform magnetic field exists into the plane of paper as shown in figure. Neglect any effect of

electrical & gravitational forces.

1) The magnitude of the magnetic field such that no tension is developed in the string will be ,
find N :-

2) If the actual magnitude of magnetic field is half to that of calculated in part (i) then tension in the

string is , where N is least integer. Find the value of N.

Common Content for Question No. 3 to 4

Initially the spring is at NLP. It is given the velocities as shown above. For the subsequent motion.

3) Maximum potential energy stored in spring is . Find the value of k.

4) Maximum speed of 2m is nv0. Find the value of n.

Common Content for Question No. 5 to 6

Torque acting on a current loop in uniform magnetic field is given by . But force on it is
zero. If it is free to rotate then it will rotate about an axis passing though its centre of mass and
parallel to . Then potential energy of loop is given by
A current carrying ring with its centre at origin and moment of inertia 2×10–2 kg–m2 about an axis
passing though its centre and perpendicular to its plane has magnetic
moment At time t = 0 an magnetic field is switched on.

5) Maximum angular velocity of ring is radian/second. Then N is ?

6) Angular acceleration of the ring at time t = 0 in rad/s2 is :-

SECTION-II(ii)

1) A neutral conducting ball of radius R is connected to one plate of a capacitor (Capacitance = C),
the other plate of which is grounded. The capacitor is at a large distance from the ball. Two point
charges, q each, begin to approach the ball from infinite distance. The two point charges move in
mutually perpendicular directions. The charge on the capacitor when the two point charges are at
distance x and y form the centre of the sphere is nq. Find n.

(Given x = y, 4π∈0 R, R = 2m)

2) A small ball of mass m carrying positive charge +Q is dropped in uniform horizontal magnetic

field B. The vertical depth of the deepest point of its path from initial position is Find k.

3) In the given circuit, the value of R so that thermal power generated in R will be maximum (in Ω).

PART-2 : CHEMISTRY
 SECTION-I(i)

1)
Product (A) is

(A)

(B)

(C)

(D)

2) Incorrect order of Nucleophilicity is ?

(A)

(B)

(C)

(D)

3) The major product of the following reaction is

(A)

(B)

(C)

(D)

4) The hydrocarbon that cannot be prepared effectively by Wurtz reaction is

(A)

(B)

(C)

(D)

SECTION-I(ii)
1) Choose the correct statement ?

(A) (P) & (Q) are chain isomers

(B) (P) is 1° Alcohol
(C) (P) & (Q) are identical
(D) (P) & (Q) are positional isomers

2) Consider the following statement and pick up the correct statement (s).

(A)
will react more readily than for SN1
reaction.

(B)

will react more readily than for SN1 reaction.

(C)

will react more readily than for SN1 reaction.

(D) SN1 reaction occurs in polar protic solvent.

3) Identify reactions correctly matched with their major products ?

(A)

(B)

(C)
(D)

4) When an aqueous solution containing MeCOOK and EtCOOK is electrolysed, possible product(s)
is/are:

(A) Me – Me

(B)

(C)

(D) CO2

5) C A B
B & C are identical when A is –

(A) HC ≡ CH
(B) CH3 – C ≡ CH
(C) CH3 – C ≡ C – CH3
(D) CH3 – CH2 – C ≡ CH

6) Choose all alkane that give only one type of monochloro derivative upon reaction with chlorine in
sun light.

(A)

(B)

(C)

(D)

SECTION-II(i)

Common Content for Question No. 1 to 2

Kolbe electrolysis method is a general method of preparation of substituted hydrocarbons from the
substituted carboxylic acids by the use of the electric discharge method where carbon dioxide gas is

released. (i) (ii) (iii)

(iv) (v)

(vi) (vii)

(viii)

1) Out of above how many compounds will produce benzene as the major product?

2) Out of above how many compounds will produce cycloalkane as major product by KES?

Common Content for Question No. 3 to 4

Dehydration of alcohol is an elimination reaction, which may proceed via formation of carbocation
intermediate & carbocation rearrangement is also considered if required. Alkene is the product of
the reaction where usually more stable alkene is major product.

Major product

3) Number of α-H present in major product :

4) Find the number of carbocation formed as intermediate in the reaction.

(consider rearrengment)

Common Content for Question No. 5 to 6

Nucleophilic aliphatic substitution reaction is mainly of two types: SN1 and SN2. The SN1 mechanism
is a two step process. Reaction velocity of SN1 reaction depends only on the concentration of the
substrate. Since product formation takes place by the formation of carbocation, optically active
substrate gives (+) and (–) forms of the product. In most of the cases the product usually consits
of 5-20% inverted product and 80-95% racemised species. The more stable the carbocation, the
greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent,
the greater is the proportion of inversion. (i) (ii)

(iii) CH3 – Br

(iv) C6H5–CH2–Br (v) CH2 = CH– CH2–Br (vi)

5) How many compounds will give SN1 reactions predominantly?

6) How many compounds will give SN1 and SN2 reactions with considerable rate?

SECTION-II(ii)

1) mixture of Alkene
Identify total number of alkene formed as product.

2) How many alkyl halide gives SN reaction faster then chlorocyclohexane with NaOH ?
2

3) Total number of monobrominated product obtained in the given reaction is : CH2=CH–CH2–Ph

 PART-3 : MATHEMATICS

SECTION-I(i)

1) A function y = g(x) defined in (–1,3] and its graph is shown below.

A function y = h(x) is defined below as

h(x)
choose incorrect option

(A) Jump of discontinuity of function y = h(x) at x = 0 is 1

(B) Function y = h(x) is continuous at x = 2
(C) Range of function y = h(x) is (0,2]
(D)

2) , given that f'(2) = 6 and f'(1) = 4

(A) Does not exist

(B) Is equal to –3/2
(C) Is equal to 3/2
(D) Is equal to 3

3) Let x = tcost, y = t + sint, then at , is

(A)

(B)

(C) –2
(D) –1

4) The sum of coefficients of all the even powers of x in the expansion of (x3 – 3x2 + 3x – 1)11 is
(A) 811
(B) –2.810
(C) 4.810
(D) –4.810

SECTION-I(ii)

1) , , , then which of the

followings does not holds ?

(A)
h is continuous at
h has an irremovable discontinuity at
(B)

(C)
h has a removable discontinuity at

(D)

2) Let if derivative of f(x) with respect to

g(x) at x = 1 exists and is equal to k, then

(A)
(B)

(C)

(D)

3) Let f(x) = then

(where {.} denotes the fractional part function)

(A)
=
(B) f is discontinuous at x = 0

(C)
=
(D) f is continuous at x = 0

4) If (where c ≠ 0), then has the............ equal to :

(A)

(B)

(C)

(D)

5) The functions u = ex sin x ; v = ex cos x satisfy the equation

(A)

(B)

(C)

(D)

6) In the expansion of

(A) the number of irrational terms is 19

(B) middle term is irrational
(C) the number of rational terms is 2
(D) 9th term is rational

SECTION-II(i)

Common Content for Question No. 1 to 2

Consider ƒ(x) = x4 – 3x3 + 2 and h(x) = be a rational function.

(i) infinite

(ii)
(iii) h(x) is continuous ∀ x ∈ R – {1, 2}
(iv) g(x) does not have any repeated factors.

1) Number of solutions of equation |g(|x|)| = is -

2) Number of points where P(x) = sgn(g(x)) is non differentiable in R is (where sgn(x) denotes
signum function of x)
Common Content for Question No. 3 to 4
Consider a function y = ƒ(x) defined parametrically by , y = 2sint

3) ƒ'(0) = kℓn(2), then k is equal to -

4) Number of values of t satisfying is equal to

Common Content for Question No. 5 to 6

If n and x are positive integers then by binomial theorem (1 + x)n = 1 + nC1x + nC2x2 + ... + nCnxn. The
R.H.S. can be written in several ways for instance (1 + x)n = 1 + (multiple of x) = 1 + λx, where λ is
an integer. So we can say (1 + x)n – 1 is divisible by x. Similarly (1 + x)n – nx – 1 is divisible by x2.

5) If 7103 is divided by 25 then remainder will be -

6) If n ≥ 5 then the expression 8n – 7n – 1 must be divisible by -

SECTION-II(ii)

1) Let f(x) be a differentiable function such that , x, y ∈ R &

f'(0) = 0, then f'(10) is equal to

2) If y = tan–1 + tan–1 + tan–1 + tan–1 + ..... to 10 terms.

Find (f ' (0) + 1)–1.

3) If Cr ≡ 25Cr and C0 + 5.C1 + 9.C2 + .... + (101).C25 = 225.k, then k is equal to _____.
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I(i)

Q. 1 2 3 4
A. A B C C

SECTION-I(ii)

Q. 5 6 7 8 9 10
A. A,B,C A,C B,C A,B,C,D A,C,D A,C

SECTION-II(i)

Q. 11 12 13 14 15 16
A. 1.00 2.00 3.00 3.00 50.00 1250.00

SECTION-II(ii)

Q. 17 18 19
A. 2.00 1.00 3.00

PART-2 : CHEMISTRY

SECTION-I(i)

Q. 20 21 22 23
A. B A B B

SECTION-I(ii)

Q. 24 25 26 27 28 29
A. B,D A,C,D A,C,D A,B,C,D A,C A,B,D

SECTION-II(i)

Q. 30 31 32 33 34 35
A. 2.00 4.00 7 3.00 3 2

SECTION-II(ii)

Q. 36 37 38
A. 5.00 5.00 4.00

PART-3 : MATHEMATICS
 SECTION-I(i)

Q. 39 40 41 42
A. C D B D

SECTION-I(ii)

Q. 43 44 45 46 47 48
A. A,C,D A,B,C,D A,B,C A,B,C A,B,C,D A,B,C,D

SECTION-II(i)

Q. 49 50 51 52 53 54
A. 6.00 2.00 1.00 2.00 18.00 49.00

SECTION-II(ii)

Q. 55 56 57
A. 20.00 101.00 51.00
 SOLUTIONS

PART-1 : PHYSICS

1)

Initial momentum of sys = 0.

Final momentum com be zero only when D retraces its path with .

2)

Answer : E2 = E1 V2 = 0

3) For complex circular motion

Since it is deflection by 60°

4)

As
∴ VB > VA ⇒ current from B to A
5)

6)

Answer : (AC)

7)

⇒ 10Cx + 2CxCy = 10 Cy + 5CxCy

10 (Cx – Cy) = 3CxCy
10 (Cx + Cy) = 7CxCy

Cy = 2
Cx = 5

(C)

8)

Answer : (ABCD)

9) Due to sheet .
The slab is symmetrical under translation in y so field is independent of y.

Also slab is symmetric under rotation by 180º around Z axis, so y component of field is odd
function of x. conside the ampere loop shown in diagram
2 Bh =

10)

Answer : (AC)

11)

T + qvB =
T = 0; ⇒

12)

T+ =

T= – =

13)

Spring block problems are similar to head-on elastic collision.

14)

Spring block problems are similar to head-on elastic collision.

= 2v0.
∴ v'1 = 2vc – v1 = 4v0 – 4 v0 = 0
∴ v'2 = 2vc – v2 = 4v0 – v0 = 3v0

15)

Answer : 50

16) New Ans. 1250 by Abhishek Jain sir

17) Let a charge Q be induced on capacitor plate connected to the ball. Induced charge on the
ball is –Q.
∴ Potential at the centre of the ball will be

V=
∴ Potential difference across capacitor plates = V
 ⇒

18)

Answer : 1

19) Req = Rin

PART-2 : CHEMISTRY

20)

Answer : (B)

21)
(Sterically hindered nucleophiles are bad nucleophiles)

22)

Answer : (B)

23) Wurtz reaction is not good method for synthesis of odd number alkane.

24) Explanation - Question is asking about correct regarding following reaction products
statement.
Concept -

A. Oxymercuration–Demercuration of Alkene
B. Hydroboration–Oxidation of Alken

Solution -
They are position isomers of each other.
Final Answer - Option (4) is correct

25) (A) EWG at ortho and para Position increase aromatic nucleophilic substitution.

(B) is wrong as is highly unstable.

(C) Same reason as ‘a’.
(D) SN1 reaction occurs in polar protic solvent with high dielectric constant.

26) Explanation - Question is asking about Correctly matched with their major product.
Concept -

A. Hydroboration oxidation of alkene

B. Hydroboration Oxidation of alkyne
C. Oxymercuration of of alkyne
D. Intermolecular electrophilic addition (Via Oxymercuration - demer curation)

Solution -

(1)
Hydroboration oxidation is addition of water molecule according to Anti markovnikov rule and
syn addition.

(2)
(3)

(4)

Final Answer - Option (1),(3) and (4)

28) When both are symmetrical

(A)

(C)
30) iii, vi

31) ii, iv, vii, viii

32)

Answer : 7

33)

34)
Answer (A)
Which of the following compounds will give SN1 reactions predominantly?
 35) Answer : (3)

36) Solution/Explanation/Calculation:
Explanation:
We have to find total no. of possible products formed for given reaction:
Concept:

Total products = 2 + 2 + 1= 5

37) II, III, IV, V, VI

38)

PART-3 : MATHEMATICS

39)
40) Here,
[∵ f'(2) = 6 and f'(1) = 4, given]
ApplyingL' Hospital's rule,

= [using f'(2) = 6 and f'(1) = 4]

41)

Now put

42)

43) h(x)

44)

If it is differentiable
then 1+ a = 2+b ⇒ a – b = 1
& 3 + b = 1 ⇒ b = –2
 a = –1
(B) a + b = –3

(C)

(D)

45) = =

= =

= .1.

= = =
f(x) is discontinuous at x = 0

46) ..........(1)

................(2)
from 1 + 2

On putting value of C, other form of can be arrived.

47)

u = ex sin x, v = ex cos x

=v(ex cosx + ex sin x) – u(ex cos x – ex sin x)

= ex sin x (v + u) + ex cos x (v – u)
= u(v + u) + v(v – u) = v2 + u2

again = ex sin x + ex cos x

= ex sin x + ex cos x + ex cos x – ex sin x

 = 2v
similarly, other options can be checked.

48)

For rational terms

& , where ⇒
∴ no. of rational terms = 2
∴ no. of irrational terms = 19

49) graph of |g(|x|)|

6 solutions

50) by (i) degree of g(x) must be less than 4

by (iii) g(x) must have x – 1 and x – 2 in its factors
by (iv) No other factor of g(x) will be possible as it can not be a linear factor or a repeated
fector
∴ Let g(x) = a(x – 1)(x – 2)

h(x) =

⇒ a=2
∴ g(x) = 2(x – 1)(x – 2)
= 2x2 – 6x + 4
51)
when x =0 ⇒ t = 0
⇒ ƒ'(0) = ℓn2

52)

Now,

⇒ ⇒

Plotting the graphs of y = |t| & , we get

⇒ Number of solutions = 2

53) 7103 = 7(49)51 = 7(50 – 1)51

= 7{5051 – 51C15050 + 51C25049 +...+ 51C5050 – 1}
= 7{5051 – 51C15050+51C25049 +...+ 51C5050}+ (–7 – 18 + 18)
= 7{5051 – 51C15050 + 51C25049 +...+ 51C5050} – 25 + 18
= (multiple of 25) + 18 so remainder is 18

54) 8n – 7n – 1 = (7 + 1)n – 7n – 1
= (7n+nC17n–1+...nCn–272+nCn–1–17+1) –7n – 1
= (7n+nC17n–1+...+nC272) + 7n+1 –7n – 1
= 49λ (where λ is an integer)
so 8n – 7n – 1 is divisible by 49.

55)
diff. w.r.t. ‘y’ we get.

Also, f'(0) = 0 So, put y = 0

⇒ ⇒
⇒ f(0) = 0 ⇒
⇒
∴
56) y = tan–1 + tan–1 + tan–1 ---------

y = tan–1 + tab–1 + tan–1 ----------

–1 –1 –1 –1 –1 –1
y = (tan (x+1) –tan x) + (tan (x+2) – tan (x+1)) + (tan (x+3) – tan (x+2)) -----------
y = tan–1 (x+n) – tan–1 x

n = 10, at x = 0 we get

f'(0) =
→ (f' (0) + 1)–1 = 101 Ans.

57) S = 1.25C0 + 5.25C1 + 9.25C2 + .... + (101)25C25

S = 10125C25 + 9725C1 + .......... + 125C25
2S = (102) (225)
S = 51(225)