04-05-2025

1001CJA106216250076 JA

PART-1 : PHYSICS

SECTION-I (i)

1) AB is a solid cylinder of radius a0 and length L. Resistivity of the material varies with x-coordinate

from end A as . An ideal battery of emf E is connected from end A and B as shown.

If power gradient (i.e. , from end A) is , then find N

(A) 2
(B) 4

(C)

(D) 6

2) Two identical spherical masses are kept at some distance as shown. Potential energy when a mass

m is taken from surface of one sphere to the other.

(A) Increase continuously

(B) Decrease continuously
(C) First increase then decreases
(D) First decreases then increases

3) Moment of inertia of the given isosceles triangular plate in figure-1, about an axis passing through
‘O’ and perpendicular to plate is I0. Find the moment of inertia of the given plate (lamina) in figure-2,
about an axis passing through ‘C’ and perpendicular to plane of the plate as shown in the figure.
(Mass per unit area of both plate is same)
(A) 3.25 I0
(B) 6.5 I0
(C) 3.5 I0
(D) 3.75 I0

4) Consider the shown uniform solid insulating sphere with a short and light electric dipole of dipole
moment (embedded at its centre) placed at rest on a horizontal surface. An electric field is
suddenly switched on in the region such that the sphere instantly begins rolling without slipping.
Speed of the sphere when the dipole moment becomes horizontal for the first time is (m = Mass of

the sphere)

(A)

(B)

(C)

(D) zero

SECTION-I (ii)

1) Consider the circuit shown. Resistance connected between terminal 'A' and 'B' is 20Ω and

ammeter is ideal. Then select the correct statement :-

(A) When switch 'S1' and 'S4' are closed, current reading of ammeter is 3 amp.

(B)
When switch 'S3' and 'S2' are closed, current reading of ammeter is amp.
(C) When only 'S2' is closed, reading of ammeter is zero.
(D) When all the switch are close, reading of ammeter is 0 amp.

2) The minimum and maximum distances of satellite from the center of the earth are 2R and 4R
respectively, where R is the radius of earth and M is the mass of the earth.

(A) Radius of curvature of the path of the particle at apogee is 8R/3

(B) Radius of curvature of the path of the particle at apogee is 2R/3
(C) Radius of curvature of the path of the particle at perigee is 4R/3
(D) Radius of curvature of the path of the particle at perigee is 8R/3

3) The expression for the angular momentum of a point mass about origin is given by

where a is a certain constant and ‘t’ is time in seconds.

(A)
The torque acting on the mass as a function of time is .
(B) The mass can pass through the point (0, 0, 4).
If the position of the particle at t = 1s is (0, 2, 0), the x component of its momentum at t = 1 is
(C)
.
If the position of the particle at t = 1s is (0, 2, 0), the z component of its momentum at t = 1 is
(D)
.

4) A block of mass M = 4 kg of height ‘2b’ and breath ‘b’ is placed on rough plank of same mass M. A
light inextensible string is connected to upper end of the block and passed through a light smooth

pulley. Coefficient of friction between block and plank is μ = 0.3.

The maximum value of mass of block A so that block B neither toppll nor slide on C is
(A)

If m = 1 kg then friction between block and plank is

(B)

If m = 1 kg then distane of line of action of normal between block and plank and centre of block
(C)
is .
(D) Impulse given by force of gravity on the block B is zero in given non-zero time interval

5) A hollow spherical quarter shell of non – conducting nature and negligible thickness and uniform
surface charge density “σ” on its outer surface. It’s radius s r and center of curvature as shown is at
origin “O”. Mark the CORRECT statement(s).

(A) Solid angle subtended by the quarter shell at “O” is π sr.

(B)
Solid angle subtended by the quarter shell at “O” is sr.
Magnitude of electric field at “O” due to the quarter shell is
(C)

Magnitude of electric field at “O” due to the quarter shell is

(D)

6) A small dipole of dipole moment p is placed at a distance 2R from the centre of a neutral
conducting sphere of radius R. The direction of dipole is towards the centre of the sphere. A tangent
is drawn from the centre of dipole to the sphere which meets the sphere at point A.

The potential at point A is

(A)

(B) The potential at point A is

(C)
The potential at point A due to induced charges is
The potential at point A due to dipole is
(D)

SECTION-III

1) A prism is made of wire mesh with each side having equal resistance R. A battery of 6V and zero
resistance is connected across E and F as shown in the figure. The current in the branch AD (in
Ampere) if R is equal to 5Ω is α, then find 10 α.

2) The current supplied to the circuit by 10 V cell is I ampere. Assume both cells as ideal find 4I.

3) Two planets, A and B are orbiting a common star in circular orbits of radii RA and RB, respectively,

with RB = 2RA. The planet B is times more massive than planet A. The ratio of angular
momentum (LB) of planet B to that of planet A(LA) is closest to integer __________.

4) A semicircular disc of radius R and mass M is pulled by a horizontal force F so that it moves with

uniform velocity. The coefficient of friction between disc and ground is . If the angle

radians, then find the value of n.

5) If the electric field in a region is given as E = y2 + 2xy and the potential is assumed 4 Volts at
the origin, find the potential, in Volts, at the point (2,1,9). All values are in SI units.

6) A regular pyramid of square is made of eight identical wires each resistance 12 Ω shown in the
figure. The terminals A and B are the mid-points of the corresponding sides. Find equivalent
resistance between the terminals A and B in ohms.

7) A small asteroid is approaching a planet of mass M and radius R from a large distance. Initially its
velocity (u) is along a tangent to the surface of the planet. It fall on the surface making an angle of

30° with the vertical. The magnitude of u is . Find x.

8) In the figure two concentric conducting shells of radius R & 2 R are shown. The inner shell is
charged with Q and the outer shell is uncharged. The amount of energy dissipated when the shells

are connected by a conducting wire is Calculate value of n =

PART-2 : CHEMISTRY

SECTION-I (i)

1) Which of the following statement is not correct regarding 'Ferrocene'?

(A) It has a sandwich - like arrangement.

(B) It is stable due to aromatiicity in the molecule.
(C) It has EAN equal to next nearest noble gas configuration.
(D) It is synthesized by reacting cyclopentadienyl anion with an iron(III) salt.
2)
Which compound reacts with Ag powder to give X

(A) H3C – CCl3

(B) CHCl3
(C) CH3–CH2Cl
(D) H3C–C ≡ C–CH2Cl

3) 30.11 × 1022 unit cell of K2S is dissolved in 10 litre water. K2S has antifluorite structure. The
freezing point of the solution becomes [Kf of water is 1.86 ºC kg/mole, = 273K]

(A) 272.78K
(B) 271.884 K
(C) 272.87K
(D) 272.9K

4) Calculate KC for reaction A(g) + B(g) 2C(g).

If 0.5 mole of A, 1.25 moles of B and 0.5 moles of C are placed in one litre flask and allowed to reach
equilibrium & final concentration of C is 1M

(A) 1
(B) 2
(C) 4
(D) 1.5

SECTION-I (ii)

1) Incorrect options among the following are :

+
(A) [CuCO)3] shows stronger synergic bonding compared with [Ni((CO)4]
(B) [Pt(NH3)2(Br)2(Cl)2] exhibits geometrical isomerism as well as optical isomerism.
(C) When excess of KCN is added to aqueous solution of copper sulphate K2[Cu(CN)4] is formed.
(D) [Co(NH3)4(NO2)2]Cl exhibits linkage, ionization and geometrical isomerism.

2) Select the incorrect statement(s) among the following.

(A) Stability of is greater than

(B) NF3 > NCl3 > NBr3 (Order of lewis basic strength)
(C) Basicity of O(SiH3)2 is greater than O(SiF3)2
(D) Thermal stability of PH3 is greater than SbH3

3) Product(s) obtained during complete hydrolysis of

CH3 – CH = CH – O – CH2 – CH3

(A) CH3 – CH2 – CH2 – OH

(B) CH3 – CH = O
(C) CH3 – CH2 – OH
(D) CH3 – CH2 – CH = O

4) Correct statement(s) is/are regarding given compound

(A) Basic strength order I > II > III

(B) Acidic strength order I > II > III
(C) III compound having all inductive, mesomeric & hyperconjugation effects.
(D) All I. II & III are heterocyclic compounds

5) For a zero order reaction which of the following statements are true?

(A) The rate is independent of the temperature of the reaction

(B) The rate is independent of the concentration of the reactants
(C) The half life depends on the concentration of the reactants
(D) The rate constant has the unit mole Lit–1 sec–1

6) Which of the following statement(s) for crystal having Schottky defect is/are correct.

Schottky defect arises due to absence of cations & anion from positions which they are expected
(A)
to occupy.
(B) The density of crystal having Schottky defect is smaller than that of perfect crystal.
(C) Schottky defect are more common in co-valent compound with higher co-ordination number.
(D) The crystal having Schottky defect is electrically neutral as a whole.

SECTION-III

1) Bond order of molecule and CO molecule are 'x' and 'y' respectively.
Report for :- x – y

2) Number of enantiomeric pairs for the complex of type [M(AB)(CD)ef] are.

(Fill your answer as sum of digits till you get the single digit answer)

3) The Co-ordination number of copper in cuprammonium sulphate (Switzer's salt) is 'a' and when it
is dissolved in water the number of ions produced are 'b'.
Report for : a + b
4) Which is the strongest nucleophilic site in the following species ?

5) Consider the following compound (X):

If 0.5 moles of the compound (X) is heated. How many moles of CO2 are produced?

6) Identify total number of SN1 reactions

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii) (ix)

7) For a chemical reaction aA bB

The ratio of ‘a’ and ‘b’ is

8)

Liquids A and B form ideal solution over the entire range of composition. At temperature T,
equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature,
a new solution of A and B having mole fractions xA and xB respectively, has vapour pressure of 22.5

Torr. If the value of in the new solution is Y. Then the value ‘Y–10’ is :
(Given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

PART-3 : MATHEMATICS

SECTION-I (i)

1) a1 is the greatest value of f(x), where (Where [.] denotes the greatest integer

function) and , then is

(A) 1
(B) e2
(C) ℓn2
(D) none of these

2) Given the equation f''(x) = 0; f, f', f'' are continuous in a ≤ x < b and curve y = f(x) crosses the x-
axis at least in 3 distinct points in [a, b] then which of the following represents the minimum number
of roots number of roots of f''(x) = 0 in (a, b)

(A) 2
(B) 1
(C) 4
(D) None of these

3) The angle of intersection of the curves y = (1 + x)cos x + sin x and y = (x2 + x + 2) at (h, k) where
h and k are integers, is θ, then tan θ equals :

(A)

(B)

(C) 1
(D) 0

4) The graph of the function y = f(x) has a unique tangent at (πa, 0). If

, then is equal to :

(A)
(B)

(C) 4

(D)

SECTION-I (ii)

1) Let f(x) = x3 – x2 + x + 1 and then

(A) g(x) is discontinuous at x = 1

(B) g(x) is continuous at x = 1
(C) g(x) is differentiable at x = 1
(D) g(x) is non-differentiable at x = 1

2) Which of the following functions are twice differentiable at x = 0 ?

(A) f(x) = x|x|

(B)
g(x) = [x2]tan–1x – {x2}cot–1x – [x2]
(C) h(x) = |sin2x|

(D)

3) The equation has :

(A) One positive root

(B) One negative root
(C) None of the root ∈ (–1, 0)
(D) Three real roots

4) The equation x3 + 2x2 + x + k = 0 has three real roots in which two roots are equal, then value of
'k' is/are

(A) 0

(B)

(C)

(D) None of these

5) From (1, 2), tangents are drawn to the curve y2 – 2x3 – 4y + 8 = 0. Then
(A) Sum of x-coordinates of points of contact is zero
(B) Sum of x-coordinates of points of contact is 4
(C) Sum of y-coordinates of points of contact is zero
(D) Sum of y-coordinates of points of contact is 4

6) Given three distinct positive constants a, b, c we want to solve the simultaneous equations ax +
by =
bx + cy =

There exists a combination of values for a, b, c such that the above system has infinitely many
(A)
solutions (x, y).
There exists a combination of values for a, b, c such that the above system has exactly one
(B)
solution (x, y).
Suppose that for a combination of values for a, b, c the above system has NO solution. Then 2b
(C)
< a + c.
(D) Suppose 2b < a + c. Then the above system has NO solution.

SECTION-III

1) The number of the positive integral solutions of the equation is 'n', then is (where [.]
denotes the greatest integer function)

2) Let f(x) + f(y) = for all x, y ∈ R, xy ≠ 1 and , then the value of

is

3) Let f(x) be a function such that for all x ∈ R, then f is periodic with

period T. Then value of is

4) Let ; then the value of A is

5) For real numbers x, let Tx be the triangle with vertices (5, 53), (8, 83) and (x, x3) in R2. Over all x in
the interval (5, 8), the area of the triangle Tx is maximized at x = , for some positive integer n.

Then the value of is

6) For some a > 1, the curves y = ax and y = loga(x) are tangent to each other at exactly one point.
Then the value of |ℓn(ℓna)| is equal to
7) The functions F(x) = sinx and G(x) = cosx are defined for every real number x. Cauchy' theorem

applied to F and G on the interval [a,b], 0 < a < b < π, gives with c ∈ (a,b).

For and , then the value of 2c – 1 is equal to

8) If y = tan–1 + tan–1 where x ∈ . If where k and λ ∈ N. Then

is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-III

Q. 11 12 13 14 15 16 17 18
A. 4 9 8 6 2 8 6 4

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 19 20 21 22
A. D A B C

SECTION-I (ii)

Q. 23 24 25 26 27 28
A. A,C A,B C,D A,B,C,D B,C,D A,B,D

SECTION-III

Q. 29 30 31 32 33 34 35 36
A. 0 1 6 4 2 5 2 9

PART-3 : MATHEMATICS

SECTION-I (i)

Q. 37 38 39 40
A. C B A A

SECTION-I (ii)

Q. 41 42 43 44 45 46
A. B,D B,C,D A,B,C A,C B,D A,B,C
 SECTION-III

Q. 47 48 49 50 51 52 53 54
A. 8 5 6 5 4 1 3 5
 SOLUTIONS

PART-1 : PHYSICS

3)
IC ⇒ moment of inertia about an axis passing through ‘C’ and perpendicular to the plane.

4)

or

or

5)
V = IR ⇒ 60 = I(20) ⇒ I = 3 amp.
When S1 & S4 are closed
When S2 & S3 are closed

By applying nodal analysis at junction x

+ =0
2x – 120 + x – 60 + 2x = 0
⇒ x = 36
I=
(C) When only S2 is closed, circuit will be balanced wheatstone bridge current in ammeter will
be zero.
(D) When all the switch are closed branch 'AB' is short circuited and current will become zero.

6) (a) Applying conservation of angular momentum

Mv1(2R) = mv2 (4R)

V1 = 2v2 ……….(1)
From C.O.E

……………..(2)
Solving Eqs. (1) and (2)

V2 = , V1 =
(b) if r is the radius of curvature at point B

V2 = (putting value of v2)

7)

Particle can never be found at point (0, 0, 4) m

At t = 1s, ;

9)
10)
Potential at centre and at any point on the sphere will be same So,

11) by symetring, potential at A and potential at B will be equal.

Similary V C = VD
Hence, iAB = 0
iCD = 0
KVL for EADFE
+6 – i2 × 3 × 5 = 0

⇒
10α = 4

12)
applying KCL at junction A & B ;

&

we get, &

13)

0
L = mv R =

here M is mass of star

14)

F = μmg
The force F and friction force will form a couple.
Balancing torque about centre of mass
⇒

15) = = = =2

16)
Due to symmetry
no current will flow in CD & DE. So can be removed. So circuit will become

Req = 8Ω

Req= 8W
 17)
The velocity (V) makes 30° angle with the radius (normal) near the surface.
Conservation of angular momentum (about centre O)
mVR sin 30° = muR
∴ V = 2u …(i)
Energy Conservation

18) Energy dissipated = Uinitial –Ufinal

PART-2 : CHEMISTRY

21)

Antifluorite structure = Anion + Cation

FCC lattice point Tetrahedral voids
S2– K⊕
No. of molecules in an unit cell = 4
No. of unit cell = 30.11 × 1022 unit cell
Total number of molecules = 30.11 × 4 × 1022 molecules

Moles of molecules =
Van’t Hoff factor = i = 3 as it is strong electrolyte.
K2S → 2K+ + S2–
ΔTf = Kf.m.i
ΔTf = 1.86 × 0.2 × 3 10 litre of water = 10 kg of water
ΔTf = 1.116 Density = 1 g/mL = 1 kg/litre

273 – Tf = 1.116 Molality = = 0.2

Tf = 271.884

22) A(g) + B(g) 2C(g)

0.5 1.25 0.5
 +0.5
= 0.25 M = 1M = 1M

32) The electron pair at position (4) is not delocalised other are delocalised, hence position 4 is
strongest nucleophile.

33)

One mole of the above compound (X) on heating produces 4 moles of CO2.
ence, 0.5 M of X produce 2 moles of CO2.

36)

Given : →
1. → A and B are in equimolar solution.
= 0.5
= 45 Torr, = 20 Torr

45 = 0.5 × 20 + 0.5
45 – 10 = 0.5
= 70 Torr
2. PT = 22.5 Torr
PT = XA + XB , XA + XB = 1 XA = 1– XB
PT = (1–XB)
22.5 = 20 – 20XB + 70 XB
2.5 = 50 XB
XB = 0.05
XA= 1–0.05 = 0.95

y – 10 = 19 – 10 =

PART-3 : MATHEMATICS

37) a1 = 1

a2 = 1 –

a3 = 1 –
38) Let c, d, e be the three points where y = f(x) crosses x-axis. Then, f(c) = f(d) = f(e) = 0
Assuming a < c < d < e < b. The function f satisfies Rolle's theorem in two intervals (c, d) and
(d, e).
Since f and f ′ are continuous and f(c) = f(d) = 0
So, there exists, at least one point in the interval (c, d) and (d, e) such that derivative is zero
Let, C1 ∈ (c, d) such that f ′ (C1) = 0 and C2 ∈ (d, e) such that f ′ (C2) = 0.
Now the function f ′ satisfies Rolle's since f ′, f ″ are continuous and f ′ (C1) = f ′ (C2) = 0
So, by Rolle's theorem, there exists a number C3 in between C1 and C2 such that f ″ (C3) = 0
⇒ Minimum one root C3 of the equation f ″ (x) = 0 lies in the interval (a, b)

39) At x = 0, y = 1 for both curves intersect

Now y = (1 + x)cos x + sin x

∴ = 2 at x = 0

And y = (x2 + x + 2)

at x = 0

Now m1 = 2, m2 =

∴ tan θ =

40)

41) f(t) = t3 – t2 + t + 1
f ′ (t) > 0
∴ max. {f(t) : 0 ≤ t ≤ x} = f(x)

∴ g(x) =
LHL = f(1) = RHL = 2
f ′ (1–) = 2
f ′ (1+) = –1
∴ f(x) is continuous but non-differentiable at x = 1

42) ⇒ not differentiable

g(x) ⇒ ⇒ twice differentiable
h(x) = sin2x → differentiable

k ′ (x) =
twice differentiable

43) f ′ (x) = – 7 + 2 cos x < 0 ∀ x ∈ R

44)

45) Curve is (y – 2)2 = 2x3 – 4

∴ Equation of tangent at (x1, y1)

y – y1 = (x – x1)
As it passes through (1, 2)

⇒ 2 – y1 = (1 – x1)
2 2
or (y1 – 2) = 3x1 (x1 – 1)
or [As (x1, y1) lies on the curve]
or
or (x1 + 1) (x1 – 2)2 = 0
∴ x1 = –1 or 2
But x1 ≠ –1 (it is clear from equation (1))
∴ x1 = 2
∴ y1 = 2 ±
Points of contact are and
∴ sum of x-coordinates of points of contact is 4
∴ sum of y-coordinates of points of contact is 4

46) Each of the given equations defines a line in the XY plane. (a) One can arrange both lines

to be identical by having each equation a scalar multiple of the other, e.g., a = 1, b = ,c=

. (b) There is a unique solution when the two lines are distinct and not parallel. (c) The two
lines are given to be parallel. So slopes are equal, i.e., b2 = ac. Thus b is the geometric mean of

a and c, so b < the arithmetic mean . (Recall that a, b, c are distinct and positive.) (d) is
absurd. Just ensure b2 ≠ ac.

47)

, Solution is possible only when 11I < 9I + 9

Total number of solution

48) Given, f(x) + f(y) = f

Putting x = 0, y = 0, we get f(0) = 0 ....(i)
and putting y = –x, we get
f(x) + f(–x) = f(0) = 0
∴ f(x) = –f(–x) ....(ii)

Now, f ′ (x) =

[from Eq. (ii)]

∴
and f(x) = 2tan–1x + c
f(0) = 0 + c = 0
∴c=0
∴ f(x) = 2tan–1x
49) f(x) + f(x + 2) = f(x + 1) ....(1)
x→x+2
f(x + 2) + f(x + 4) = f(x + 3) ....(2)
(1) + (2)
f(x) + 2f(x + 2) + f(x + 4)
= {f(x + 1) + f(x + 3)}
= 3f(x + 2)
f(x) + f(x + 4) = f(x + 2) ....(3)
x→x+2
f(x + 2) + f(x + 6) = f(x + 4) ....(4)
(3) + (4)
f(x) + f(x + 6) = 0
f(x + 6) = –f(x)
f(x + 12) = –f(x + 6)
= f(x)
period T = 12

50)

L = 50 – L

∴L= = 25

51) The area of a triangle is half the product of its base length and its height, where the height
is the distance between the base line and the opposing vertex. No matter the value of x in the
open interval (5,8), the triangle Tx has a fixed base between the fixed points (5,53) and (8.83);
only the vertex opposing this fixed base of the triangle is allowed to vary. Therefore,
maximizing the area of Tx is equivalent to maximizing the distance from the point (x,x3) to the
base. Since the point (x,x3) is a continuously differentiable function of x, its distance from the
base is also a continuously differentiable function of x, so this distance is maximized at a given
point x = if and only if the line tangent to the function y = x3 at x = is parallel to the
3 3
base line connecting the points (5.5 ) and (8.8 ). The slope of this base line is

, and the slope of the tangent line of the function

3
y = x at x = is simply the derivative of this function evaluated at x = which equals
, or 3n. Hence, 3n = 129, so n = 43.

52) Since the functions ax and loga(x) are inverses, by symmetry their point of tangency occurs
on the line y = x. Let the intersection point be (z, z); at this point, again by symmetry, az =
loga(z), and

.
Substituting az = loga(z) into az ℓn(a) = 1, we obtain ℓn(a) loga(z) = 1, so ℓn(z) = 1, so z = e.

Substituting z = e back into the equation , we have , so a = e1/a and

lℓ(ℓn(a))| = 1 as desired.

53) The equality reduces to or

Hence, which shows that . Therefore, we see that

54)