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JEE Advanced Mock Test - 3 (Paper II) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 3 (Paper II)

JEE Advanced Mock Test - 3 (Paper II) for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Advanced Mock Test - 3 (Paper II) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 3 (Paper II) MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 3 (Paper II) below.
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*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper II) - Question 1
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A resistance R of thermal coefficient of resistivity = α is connected in parallel with a resistance = 3R, having thermal coefficient of resistivity = 2α. Find the value of αeff.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 1

For parallel combination

From (1) and (2)

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper II) - Question 2
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Two cars A and B are approaching each other as shown in figure.

At t = 0, the velocities of A and B are v1 and v2 and they are at x = 0 and x = x0, respectively. The car A is moving with constant velocity while B is moving with constant acceleration a, whose direction is shown. If Xrel, Vrel and arel are displacement, velocity and acceleration of A w.r.t. B respectively, then mark out the correct options.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 2

a1 = 0

a2 = −a

arel = a1 − a2

= 0 − (−a)

= a

arel is constant w.r.t time.

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*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper II) - Question 3
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Suppose the potential energy between electron and proton at a distance r is given by . Application of Bohr's theory to hydrogen atom in this case shows that

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 3

By Eqs. (ii) and (iii), we get

Total energy = 1/2(potential energy)

=

Total energy ∝ n6

Total energy ∝ m−3

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 4
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For the circuit shown in the figure, find the current (in ampere) through the source.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 4

Peak current through

Peak current through R2

Phase difference between I1 and I2 is π/2

∴ Peak current through the source is

∴ I = 5 A

RMS current = 3.53 A

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 5
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Suppose the potential energy between an electron and a proton at a distance r is given by Ke2 / 3r3. Application of Bohr's theory to hydrogen atom in this case shows that energy in the nth orbit is proportional to nx then x=

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 5

By (ii)ii and (iii)

Total energy = 1/2 (potential energy)

=

Total energy ∝ n6

Total energy ∝ m−3

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 6
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A disc of radius R rotating with an angular velocity ω0 is placed on a rough horizontal surface. If initial velocity of centre of disc is zero, then velocity of centre of disc when it ceases to slip, comes out to be ω0R/k, value of k is _____.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 6

Let v be the velocity of centre of mass of disc.

Condition for pure rolling i.e., when slip ceases is v = Rω

From conservation of angular momentum,

Hence, the value of k = 3.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 7
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A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 7

Velocity of sound in the string = v = ; where T is the tension and μ is the mass per unit length.

Distance between successive nodes,

λ/2 = 5 cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 8
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A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 μ, then N is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 8

From the figure:

F1 = mg(sinθ − μcosθ)

F2 = mg(sinθ + μcosθ)

Given, F2 ​= 3F1

∴ sinθ + μcosθ = 3sinθ − 3μcosθ

∴ N = 10 μ = 5 N

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 9
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Steel wire of length L' at 40°C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is 10-5/°C, Young's modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of 'm' in kg is

(Round off up to 2 decimal places)

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 9

Youngs modulus

We know that

Also

From (i) and (ii)

= 3.14

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 10
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Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be ____ K.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 10

Heat loss = heat gain
n1 CP1 (T - 400) = n2 CV2 (700 - T)
7(T - 400) = 3(700 - T)
10 T = 4900
T = 490K

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 11
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Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then, magnitude of total work done by the gases till the time they achieve equilibrium will be kR. Find the value of k.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 11

n1 CP1 (T - 400) = n2 CV2 (700 - T)
7(T - 400) = 5(700 - T)
12 T = 6300

w1 = n1 R ΔT1
= 2(R) 125 = 250 R
w2 = n2 R ΔT2
= 2 (R) (-175) = -350 R
wnet = -100 R
Magnitude = 100R = kR
k = 100

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 12
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The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is BR/t. Find the value of t.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 12

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 13
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The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is . Find the value of p.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 13



JEE Advanced Mock Test - 3 (Paper II) - Question 14
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A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The speed of the first ball is
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 14
Let the speeds of the two balls, 1 and 2, be v1 and v2, respectively. Since the speed of the second ball is half of the first, let

v1 = 2v ...(1)

v2 = v ...(2)

If y1 and y2 are the maximum heights reached by the balls 1 and 2, respectively, then

Since it is given that y1 − y2= 15 m

Substituting in equation

v1 = 20 m s−1

JEE Advanced Mock Test - 3 (Paper II) - Question 15
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A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and after a time interval (less than 22 seconds). The later ball is thrown at a velocity of half the first. At t = 2 s, both the balls reach their maximum heights. At this time the vertical gap between first and second ball is +15m.

The time interval between the throw of balls is
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 15

Let the speeds of the two balls, 1 and 2, be v1 and v2, respectively. Since the speed of the second ball is half of the first, let

v1 = 2v ...(1)

v2 = v ...(2)

If y1 and y2 are the maximum heights reached by the balls 1 and 2, respectively, then

Since it is given that y1 − y2= 15 m

Substituting in equation

v1 = 20 m s−1

v2 = 10 m s−1

When maximum heights are reached,

Given that both the balls reach the maximum heights at the same time.

If t2 is the time taken by the ball 2 to cover a distance of 5 m, then from

Since t1 (time taken by ball 1 to cover distance of 20 m) is 2 s, time interval between the two throws,

= t1 − t2 = 2 − 1 = 1 s

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper II) - Question 16
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Which of the following statements is/are true for P4S3 molecule -
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 16

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JEE Advanced Mock Test - 3 (Paper II) - Question 17
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The equilibrium constant Kp for this reaction at 298 K is . Find the value of k.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 17

Given: 2α = βequilibrium

So, α = βequilibrium/2

Total moles at equilibrium – (1 + α) = (1 + (βeq/2))

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 18
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For the given reaction, KC < n. Find the value of n.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 18

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JEE Advanced Mock Test - 3 (Paper II) - Question 19
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The total number of N-atoms present in compound R is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 19

On oxidation, the alkyl side chains of o-dipropyl benzene will be oxidised to carboxylic acid groups and will yield phthalic acid.
Phthalic acid on heating with NH3 yields phthalimide, which in the presence of Br2/KOH yields o-phenylenediamine (R).
This is the Hoffmann bromamide degradation reaction.

So, the total number of N-atoms present in compound R is 2.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 20
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The sum of N- and O-atoms present in compound T is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 20

Oxidation of o-dipropylbenzene (O) yields phthalic acid (P), which reacts with NH3 to form phthalamide (Q).
(Q) on heating forms phthalimide (S) with the elimination of NH3.
(S) on further treatment with ethyl 2-bromopropanoate followed by hydrolysis yields alanine (T).

So, the sum of N- and O-atoms present in compound T is 3.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 21
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The orthocentre of the triangle F1MN is . Find the value of k.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 21


Foci of ellipse = (±1, 0)
The equation of a parabola having vertex (0, 0) and focus (1, 0) is y2 = 4x. … (2)

From equations (1) and (2), 
⇒ 2x2 + 9x - 18 = 0
⇒ x = 3/2, -6 (rejected

The equation of the altitude from vertex  

Put y = 0.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 22
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If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is m : n. Find the sum of m and n.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 22

The equation of the tangent at point  to the ellipse is 
Put y = 0.
 R is (6, 0).
The equation of the normal to the parabola at point

Put y = 0.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 23
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The probability that x1 + x2 + x3 is odd is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 23

(1) If x1 = 1, x2 and x3 can be selected in 4 + 3 + 4 + 3 + 4 = 18 ways
(2) If x1 = 2, (x2, x3) can be selected in 3 + 4 + 3 + 4 + 3 = 17 ways
(3) If x1 = 3, (x2, x3) can be selected in 18 ways

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper II) - Question 24
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The probability that x1, x2 and x3 are in an arithmetic progression is

Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 24

The favourable cases are (1, 1, 1), (1, 2, 3), (1, 3, 5), (1, 4, 6), (2, 2, 2), (2, 3, 4), (2, 4, 6), (3, 3, 3), (3, 4, 5), (3, 5, 7), (3, 2, 1).
Hence, answer is 11/105.

JEE Advanced Mock Test - 3 (Paper II) - Question 25
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The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10°C. Now, the end P is maintained at 10°C, while the end S is heated and maintained at 400°C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 x 10–5 K–1, the change in length of the wire PQ is
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 25

From given data:


As a function of x, T(x) = 10 + 130 x
ΔT(x) = T(x) - 10 = 130 x
Extension in a small element of length dx is
dℓ = αΔT(x)dx = 130 α xdx
Net extension

JEE Advanced Mock Test - 3 (Paper II) - Question 26
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The switch in circuit shifts from 1 to 2 when VC > 2V/3 and goes back to 1 from 2 when VC < V/3. The voltmetre reads voltage as plotted. What is the period T of the wave form in terms of R and C?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 26

During time 't2', capacitor is discharging with the help of three resistors of resistance 'R' each.
Therefore, q = q0 e(-3t/RC)
[∵ Q = CV]
V = V0 e(-3t/RC)
As
V0 = 2V/3, V = V/3
t2 = (RC/3).loge2
During time 't1', capacitor is charging with the help of battery.
Therefore, q = q0(1 - e(-3t/RC)) or V = V0(1 - e(-3t/RC))
As V0 = 2V/3, V = V/3
t1 = (RC/3).loge2
T = t1 + t2 = (2RC/3).loge2

JEE Advanced Mock Test - 3 (Paper II) - Question 27
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Consider the cell:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0, M) | Cu(s)
If we want to make a cell with a more positive voltage using the same substances, we should
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 27

Cd(s) + Cu+2(aq) Cd+2(aq) + Cu(s)
According to Nernst equation,

According to the equation, decrease in the concentration of Cd+2 or increase in the concentration of Cu+2 would increase the voltage.
Hence, option (4) is correct

JEE Advanced Mock Test - 3 (Paper II) - Question 28
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The van der Waals' constants for three different gases are given in the table below:

Which of the following interpretations is incorrect?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 28

This statement is incorrect because the intermolecular attraction will be maximum for 'Y' as it has the highest value of 'a'.
Hence, option 4 is the answer.

JEE Advanced Mock Test - 3 (Paper II) - Question 29
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If f(x) = x tan-1x, find the value of f'(1)
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 29

Given: f(x) = x tan-1x

Using first principle,

JEE Advanced Mock Test - 3 (Paper II) - Question 30
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If the tangent at any point on the curve y = f(x) intercepted between the point and the x-axis is of length 1, find the equation of the curve.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper II) - Question 30


Again put cosθ = t
-sinθ dθ = dt

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