31-08-2025

5312CJA101021250010 JA

PHYSICS

SECTION-I

1) Two surface OABC and OCDE lies in the plane of xy and yz as shown in the figure. A charged

particle 'q' lies in the space at a point P. if

coordinates of 'P' is (a − Δr, a − Δr, Δr) and a >> Δr, then flux passing through surface OABC is
(A)

coordinates of 'P' is (a − Δr, a − Δr, Δr) and a >> Δr, then flux passing through surface OCDE
(B)
is
coordinates of 'P' is (a + Δr, a + Δr, −Δr) and a >> Δr, then flux passing through surface
(C)
OABC is
coordinates of 'P' is (a + Δr, a + Δr, −Δr) and a >> Δr, then flux passing through surface OCDE
(D)
is

2)

If a satellite orbits as close to the earth's surface as possible :-

(A) its speed is maximum.

(B) time period of its rotation is minimum.
(C) the total energy of the 'earth plus satellite' system is minimum.
(D) the total energy of the 'earth plus satellite 'system is maximum.

3) Two identical moving coil galvanometer have 10 Ω resistance and full scale deflection at 2 µA
current. One of them is converted into a voltmeter of 100 mV full scale reading and the other into an
Ammeter of 1 mA full scale current using appropriate resistors. These are then used to measure the
voltage and current in the Ohm's law experiment with R = 1000 Ω resistor by using an ideal cell.
Which of the following statement(s) is/are correct? [JEE-Advanced 2019]
(A) The measured value of R will be 978Ω < R < 982Ω.
(B) The resistance of the Voltmeter will be 100 kΩ.
(C) The resistance of the Ammeter will be 0.02Ω ( round off to 2nd decimal place)
If the ideal cell is replaced by a cell having internal resistance of 5Ω then the measured value
(D)
of R will be more than 1000 Ω.

4) The circuit shown in the diagram is in steady state with the switch open. When the switch is
closed, which of the following will not change immediately?

(A) Potential difference across capacitor C.

(B) Current through the inductor L.
(C) Potential difference across resistance R1.
(D) Current through the resistance R2.

5) The torque on a body about a given point is found to be equal to × , where is a constant
vector and is the angular momentum of the body about that point. From this it follows that :-

(A)
is perpendicular to at all instants of time
(B) the component of in the direction of does not change with time
(C) the magnitude of does not change with time
(D) does not change with time

6) The disc of radius r is confined to roll without slipping at A and B. If the plates have the velocities

shown, then

(A) linear velocity v0 = v

(B)
angular velocity of disc is

(C)
angular velocity of disc is
(D) None of these

SECTION-II
1) A solid cylinder initially at rest of mass 3 kg and radius 1 m rolls down on inclined plane. Friction

force (in N) acting on it will be :- [g = 10 m/s2]

2) An electric field, passes through the surface of 4 m2 area having unit vector

. The electric flux for that surface is _________ V m.

3) A small ball of mass 2 × 10–3 kg having a charge of is suspended by a string of length 0.8m.
Another identical ball having the same charge is kept at the point of suspension. The minimum
horizontal velocity (in ms–1) (Take: g= 10m/s2) which should be imparted to the lower ball so that it

can make complete revolution is Find x.

4) The density of the core of a planet is ρ1 and that of the outer shell is ρ2, the radii of the core and
that of the planet are R and 2R respectively. The acceleration due to gravity at the surface of the

planet is same as at a depth R. Find the value of .

5) Two bulbs, each of rating 10 V, 50 W, are connected as shown. To make the bulbs glow at their
full intensity, find the emf (in Volt) of the battery connected.

6) A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its
plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab (∈r = 6)
is inserted between the plates. The change in its potential energy after inserting the dielectric slab is
_______× 10–12 J.

SECTION-III

1)

In the given circuit find energy stored in capacitors in micro Joule.

2) In the figure capacitors and their connection is shown.

Then final charge on capacitor between A & B will be :

3) Find n if the total power dissipated by the circuit is 6n watts.

4) An Earth satellite revolves on a circular orbit at a linear speed v0. A missile device increases the
absolute value of this speed to 1.5v0, and the satellite goes into an unknown path as shown (Fig).
With what speed will the satellite leave the gravitational field of earth ? The resistance of the cosmic

dust is not taken into account. Mass of earth is M. If your answer is , fill α in OMR sheet.

5)

A dipole of dipole moment is placed at point A(2,–3,1). The electric potential due to
this dipole at the point B(4,–1,0) is (ab) × 109 volt here 'a' represents sign (for negative answer
select 0 for positive answer select 1. Write the value of (a+b)2. The parameters specified here are in
S.I. units.

6) In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times

as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias will be x :

1. The value of x will be _____.

CHEMISTRY

SECTION-I
1)
R can be obtained as a product in reaction :

(A)

(B)

(C)

(D)

2)

Which of the following is/are correct?

(A)
(Rate of SN1)

(B)
(Rate of SN2)

(C)
(Rate of electrophilic addition reaction)

(D)

(Rate of dehydration of alcohol)

3) Which of the following statements is/are correct

(A) All adiabatic process are iso-entropic process

(B) All spontaneous process must be exothermic in nature
(C) Reversible isothermal process for an ideal gas is iso-enthalpic process
(D) Heat capacity of any substance undergoing adiabatic process is zero

4) Which of the following are intensive as well as state functions?

(A) internal energy

(B) molar entropy
(C) heat
(D) molar enthalpy

5) Correct statement among the following :

(A) F– > Cl– > Br– > I– (Nucleophilicity order in polar protic solvent)
(B) F– > Cl– > Br– > I– (Basicity order in polar protic solvent)
(C) I– > Cl– > Br– > I– (Leaving group tendency)

(D) H3C– > > > F– (Nucleophilicity order)

6) Which combination of reagents will not bring about the following conversion?

⊕
(A) MeMgBr/H , H2SO4/D, HBr/H2O2, hv
⊕
(B) MeMgBr/H , H2SO4/Δ, HBr
⊕
(C) MeMgBr/H , HBr/CCl4
(D) MeMgBr, NH4Cl

SECTION-II

1)

The effective atomic number for the complex ion [Pd(NH3)6]4+ is:

2) 1 mol of an ideal diatomic gas is expanded by heating from state-I (10 litre, 500K) to (20 litre,
800K) against constant pressure 2 bar. Find heat supplied in joule. (Given R = 25/3 J/mole/K)

3) absorbs light of wavelength 498 nm during a d-d transition. The octahedral splitting
energy for the above complex is................ (Round off to the nearest integer).

4) Find the heat absorbed by an ideal gas (in kJ) when it follows the graph (given below) during an

isothermal expansion [ln2 = 0.7] [Take : 1 litre-bar = 0.1 kJ]

5) Which of the following will give same Markovnikov and anti-Markovnikov product ?

6) How many of the following can be prepared by Williamson's ether synthesis in good yield

SECTION-III

1) Number of alkyl halide which has higher rate of solvolysis than 3-chloropropene.

2) The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane

using alcoholic KOH is :
3) The boiling point of a solution of 0.512 gm of napthalene, C10H8, in 24 gm of chloroform is 0.5 °C
higher than that of pure chloroform. The molal boiling point elevation constant of chloroform is (in
K-kg/mol)

4) Number of correct statements are-

(i) Azeotrope obey Raoult's law
(ii) Minimum boiling point azeotrope is formed if ΔHmix < O
(iii) ΔHvap for ideal aqueous solution containing non-volatile solute is same as ΔHvap for pure water
(iv) When azeotrope is vapourised, composition of vapour formed is same as solution.

5)

How many Ether are possible for molecular formula C4H10O.

6)

8 moles of linear diatomic ideal gas at 300 K is expanded reversibly and adiabatically from 1 L to 32
L.
Calculate work-done (in kcal) by the gas in this process? use: (R = 2 cal mol-1 K-1)

MATHEMATICS

SECTION-I

1) A line L1 passes through a point P in I quadrant on y = x at a distance from origin and point

Q(even prime, 0). Another line L2 passes through perpendicular to L1, then

(A) slope of L1 is –1

(B)
initial ordinate of L2 is
area of triangle formed by L1, L2 and x = 0 is
(C)

area of triangle formed by L1, L2 and x = 0 is

(D)

2) Locus of P(x, y) is y – y1 = m(x – x1). If

(A) x1 and y1 are constant and m is variable then it represents concurrent lines.
(B) x1 and m are constant and y1 is variable then it represents parallel lines.
(C) x1 and y1 are constant and m is variable then it represents parallel lines.
(D) x1 and m are constant and y1 is variable then it represents concurrent lines.
3) If , then (where c is constant of integration)

(A)

(B)

(C)

(D)

4)

If (where C is constant of integration), then :–

(A) g(2) – g(1) = 3

(B) g(3) – g(2) = 23
(C) g(3) – g(2) = 5
(D) g(2) – g(1) = 7

5) If = ƒ(x) + C where ƒ(0) = and ƒ (C is constant of integration), then

(A)
The minimum value of |ƒ(x)| is
(B) The minimum value of |ƒ(x)| is 0

(C)
The value of ƒ(π) is
(D) The value of ƒ(π) is 0

6)

The value of the integral is equal to

(A) cot–1(e–x)+c
(B) log (ex – e–x)+c
(C) tan–1 (ex) + c
(D) tan–1 (e–x)+c

SECTION-II

1) In ΔABC, if A(0,0), B(9,–6) & C(3,2) and internal angle bisector from vertex A meets side BC at D,
then BD is -
2) For ℓ, m, n , p and q are the value of ℓ and m respectively such that equation ℓ(13 + 11n) +
m(19 + 17n) - 4 (8 + 7n) = 0 is satisfied then (p + q) equals

3) If where e is napier's constant then value of is

4) If (where c is constant of integration), then (a + b) is

equal to :

5) The number of real values of α satisfying the equation

is

6) If = (where C is constant of integration), then λ2 +

μ2 is equal to

SECTION-III

1) Area of the triangle formed by the line x + y = 3 and the angle bisectors of the pair of lines x2 – y2
+ 2y – 1 = 0 is ?

2) Number of integral values of 'a' for which the point P(a2, a) lies in the region corresponding to the
acute angle between lines 2y = x and 4y = x, is

3) The value of k for which the lines 3x + 4y = 5, 5x + 4y = 4 and kx + 4y = 6 meet at a point is

4) If

5) Let , then k2 + ℓ2 is equal to

6) Let and ƒ(x) passes through (π,0) then the number

of solutions of the equation ƒ(x) = x3 in x ∈ [0,2π] is
 ANSWER KEYS

PHYSICS

SECTION-I

Q. 1 2 3 4 5 6
A. A,B,C,D A,B,C A,C A,B,C A,B,C A,C

SECTION-II

Q. 7 8 9 10 11 12
A. 5.00 12.00 8.00 7.00 24.00 750.00

SECTION-III

Q. 13 14 15 16 17 18
A. 6 0 6 2 4 9

CHEMISTRY

SECTION-I

Q. 19 20 21 22 23 24
A. A,B,C,D B,C,D C,D B,D B,C,D B,C,D

SECTION-II

Q. 25 26 27 28 29 30
A. 54.00 8250.00 4.00 7.00 4.00 4.00

SECTION-III

Q. 31 32 33 34 35 36
A. 5 5 3 2 3 9

MATHEMATICS

SECTION-I

Q. 37 38 39 40 41 42
A. A,B,C A,B A,D A,B B,C A,C

SECTION-II

Q. 43 44 45 46 47 48
A. 7.50 2.00 0.33 to 0.34 0.00 1.00 3.00
 SECTION-III

Q. 49 50 51 52 53 54
A. 2 1 1 5 8 3
 SOLUTIONS

PHYSICS

1) Using symmetry if charged particle lies at P1 then ϕOCDE =

and
If the charge particle lies at P2 then

and

2)

T ∝ r3/2 r↓ T↑
L = mvr constant r↓ v↑

E= r↑ E↑

3)
0.1 = 2 × 10–6 (10 + RV)
∴ RV = 49990 Ω

2 × 10–6 × 10 = 10–3 RA
y 50000 = (x – y) 1000
∴ 51y = x

Reading = ≈980

4)

Charge of capacitor and current through inductor does not charge instantly.
5)

is perpendicular to and

does not change magnitude of L.

angle between and is constant
rotates abovt in a cone-shape structure.

6) vA = w0R – v0 = v
ω0R – v0 = v ...(i)
vB = ω0 R + v0 = 3v ...(ii)
From equation (i) and (ii)
2ω0R = 4V
⇒ ω0R = 2v

0
ω =
From (1) v0 = v

7) mgsinθ – Fr = mxa

a = αR
Solving these equation

8)

9)

For the ball just to complete the circle, the tension must vanish at the topmost point, i.e., T2 =0
From Newton's Second Law,
By Law of Conversation of Energy

10) Let m1 be the mass of the core and m2 the mass of outer shell
gA = gB (given)

Then, =

∴ 4m1 = (m1 + m2)

or = πR3.r1 + r2
∴ 4ρ1 = ρ1 + 7ρ2

∴ =
11)
I2 = 2 × 50
I = 5A ; 20V = I'R ; I' = 1A

, E – 6r = 20
E = 24 volts

12) Before inserting dielectric capacitance is given

C0 = 12.5 pF and charge on the capacitor Q = C0V
After inserting dielectric capacitance will become ∈rC0.
Change in potential energy of the capacitor
= Ei – Ef

= =

Using C0 = 12.5 pF, V = 12 V, ∈r = 6

= 750 pJ = 750 × 10–12J

13)

Ceq = 2 + 2 + 2 = 6 μF

Energy stored =

14)

Final potential difference = zero

Final charge = Zero
15)

Power dissipated = = 36 watt

16) ... (1)

... (2)

17)

The potential of at a point due to dipole can be defined by V = where is the positive
vector of the point with respect to the dipole. Here

= – 2 × 109 volts
⇒ a = 0, b = 2 ⇒ (a + b)2 = (0 + 2)2 = 4
 18)

CHEMISTRY

19)

20)

Ans.-2,3,4

21)

Ans.-3,4

22) The quantity per mol is intensive property But molar heat capacity is also a path function

23)

Ans.-2,3,4

24) (A)

(B)

(C)
(D)

25)

Ans.-54

26) As ideal gas

ΔU = nCvΔT

=1× × (800 – 500)

=
= 6250 J
W = –Pext. (V2 – V1)
= –2 · (20 – 10) bar-L
= –20 × 100 joule
By 1st law ΔU = q + W
6250 = q + (–2000)
q = 8250 Joule

27) Ans. (4)

28) w = –nRT ln
= –100 (100) ln 2
= –(100) (100) (0.7)
= – 7000 J
= – 7 kJ
ΔU = 0
q = – w = 7 kJ

29)

Answer (4)

30)
31)

Ans.-5

32)

33) ΔTb = Kb × m

Kb = 3
 34) (i) Incorrect : → Azeotropes are non-ideal solutions so do not obey Rault’s law.
(ii) Incorrect : → for minimum boiling azeotropes ΔHmix > 0.
(iii) Correct : → addition of non-volatile solute decreases vapour pressure However ΔHvap
remains unaffected.
(iv) Correct : → In case of azeotropes, composition of vapours formed is same as solution after
vapourisation.
Correct statement =

35)

36)

Ans.-9

MATHEMATICS

37) L1 : y = (x – 2) ⇒ x+ y = 2

L2 :

Slope of L1 = –1,

initial ordinate of L2 is

area of triangle ⇒

38) If (x1, y1) are constant and m is variable

⇒ Concurrent lines
If x, m are constant and y1 is variable
⇒ Parallel lines
39)

40)
g(2) – g(1) = 4 – 1 = 3
g(3) – g(2) = 27 – 4 = 23

41) Multiply & divide by sec2x

put 3secx + 4tanx = t

(3secxtanx + 4sec2x)dx = dt

since ƒ(0) = and ƒ

ƒ(x) =

|ƒ(x)|min = 0 & ƒ(π) =

42)
Let

43)

44) (13l + 19m - 32) + n (11l + 17m - 28) = 0

Let (13x + 19y - 32) + n (11x + 17y - 28) = 0
Equation is family of line passes through (1,1)
p+q=2

45)
0
= e(1 – 3 + 6 – 6) – e (–6)
= –2e + 6

46)

Let

47)
⇒ α (2α + 3) (4α + 1) = 0

∴ .

But (Rejected)

48)
 λ = 1, ⇒ λ2 + μ2 = 3

49)

x2 – y2 + 2y – 1 = 0
⇒ x2 – (y – 1)2 = 0
⇒ (x + y – 1)(x – y + 1) = 0

Therefore, the pair of lines is x + y – 1 = 0 and x – y + 1 = 0 and their angle bisectors are

That is y = 1 and x = 0. Therefore, the sides of the triangle are x = 0, y =

1 and x + y = 3. The vertices are (0, 1), (0, 3) and (2, 1). Hence the area of the triangle is

50)

∴
2<a<4
a=3

51)
3(–24 + 16) – 4(–30 + 4k) – 5(20 – 4k) = 0
–24 + 120 – 16k – 100 + 20k = 0
4k – 4 = 0 ∴ k = 1

52)

53)
using partial fractions
 Let
⇒ 4 + 3x = A(x2 + 2x + 2) + (Bx + C)(x – 2)
Put x = 0, 2, –1 to get
A = 1, B = C = –1

⇒ ⇒ I

= ⇒ k2 + ℓ2 = 8

54)

2
ƒ(x) = x sin2x + c
ƒ(π) = 0 ⇒ c = 0
ƒ(x) = x3sin2x
Number of solution of ƒ(x) = x3
⇒ x3sin2x = x3 ⇒ and sin2x = 1

⇒
Total = 3