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JEE Exam  >  JEE Tests  >  Mock Tests for JEE Main and Advanced 2025  >  JEE Advanced Mock Test - 2 (Paper I) - JEE MCQ

JEE Advanced Mock Test - 2 (Paper I) - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Advanced Mock Test - 2 (Paper I)

JEE Advanced Mock Test - 2 (Paper I) for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Advanced Mock Test - 2 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 2 (Paper I) MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Mock Test - 2 (Paper I) below.
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JEE Advanced Mock Test - 2 (Paper I) - Question 1
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An ideal gas has adiabatic exponent γ. In some process, its molar heat capacity varies as C = α/T , where α is a constant. Work performed by one mole of gas during its heating from To to nTo will be

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 1

JEE Advanced Mock Test - 2 (Paper I) - Question 2
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Two non-conducting solid spheres of radi R and 2R having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R from centre of the smaller sphere is zero. The ratio ρ1/ρ2 can be

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

At point P, if resultant electric field is zero, then

At point Q, if resultant electric field is zero, then

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JEE Advanced Mock Test - 2 (Paper I) - Question 3
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A thin uniform rod of mass M and length L is hinged at its upper end and released from rest in a horizontal position. Find the tension at a point located at a distance L/3 from the hinge point, when the rod becomes vertical.

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 3
From energy conservation

Loss of P.E. = Gain of K.E.

for centripetal force is

from (1) and (2)

T = 2 Mg

JEE Advanced Mock Test - 2 (Paper I) - Question 4
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Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 4
Work done

if x is the distance of mass 0.3 kg from the centre of

mass, we will have,

I = (0.3)x2 + (0.7)(1.4 − x)2

For work to be minimum, the moment of inertia (I) should be minimum,

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper I) - Question 5
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The temperature drop through a two-layer furnace wall is 900oC. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature θ of the interface?
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 5
H = rate of heat flow

Now, we can see that θ can be decreased by increasing thermal conductivity of outer layer (K0) and thickness of inner layer (li).

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper I) - Question 6
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In a head-on elastic collision of two bodies of equal masses
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 6
As momentum is conserved,

and masses are equal.

as collision is perfectly elastic. Therefore,

coefficient of restitution, e = 1 ⇒ vapproach = vseparation

By solving this,

Hence, the velocities are interchange including their magnitude(means speeds),

u1 = v2 and v1 = u2.

Multiply velocities with mass,

Hence, momentums are also interchanged.

Since speeds are exchanged. Hence, the faster body slows down and the slower body speeds up.

*Multiple options can be correct
JEE Advanced Mock Test - 2 (Paper I) - Question 7
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Suppose two particles, 1 and 2, are projected in vertical plane simultaneously.

Their angles of projection are 30o and θ, respectively, with the horizontal, Let they collide after a time t in air. Then
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 7
If they collide, their vertical component of velocities should be same, i.e

100 sin θ = 160 sin 30o ⇒ sin θ = ⅘

Their vertical components will always be same. Horizontal components:

160 cos 30 = 80√3 m/s

and

100 cos θ = 100 x 3/5 = 60 m/s

They are not same, hence their velocities will not be same at any time. So θ = sin-1(4/5) and they will have different speed just before the collision. is correct.

x = x1 - x2 = 160 cos 30o t - 100 cos θ t

⇒ x = (80 – √3 − 60)t

Now t < />x → collide in air

Since their times of flight are the same, they will simultaneously reach their maximum height. So it is possible to collide at highest point for certain values of x.

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper I) - Question 8
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An air-filled parallel plate capacitor having circular plates has a capacitance of 10 pF. When the radii of the plates are increased two times, the distance between them is halved and if a medium of dielectric constant k is introduced, the capacitance increases 16 times. The value of k is (answer in integer)

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 8

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JEE Advanced Mock Test - 2 (Paper I) - Question 9
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Two parallel identical plates carry equal and opposite charges having a uniform charge of 88.9 μC. Positive plate is fixed on the ceiling of a box and the negative plate has to be suspended. If the area of the plates is 6.35 sq. m and 'm' is the mass of the negative plate, then the value of m in kg, is

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 9
Force of attraction between the plates = Weight of the negative plate for it to be suspended

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JEE Advanced Mock Test - 2 (Paper I) - Question 10
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A solid sphere of radius R has a charge Q distributed in its volume with a charge density p = kra, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, find the value of a.

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 10

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JEE Advanced Mock Test - 2 (Paper I) - Question 11
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A point source of light S is placed a distance 10 cm in front of the center of a mirror of width 20 cm suspended vertically on a wall. An insect walks with a speed 10 cm/s in front of the mirror along a line parallel to the mirror at a distance 20 cm from it as shown in the figure. Find the maximum time (in seconds) during which the insect can see the image of the source S in the mirror.

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 11
Insect can see the image of source S in the mirror, so far as it remains in field of view of image overlapping with the road.

Shaded portion is the field of view, which overlaps with the road upto length PQ. By geometry we can see that, PQ = 3AB = 60 cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper I) - Question 12
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One end of a spring of natural length 2 m and spring constant k = 100 N/m is fixed at the ground and the other is fitted with a smooth ring of mass 1 kg which is allowed to slide on a horizontal rod fixed at a height 2 m (diagram). Initially, the spring makes an angle of 37o with the vertical when the system is released from rest. Find the speed (in m/s) of the ring when the spring becomes vertical. (Take sin(37o) = 3/5)

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 12

x = extension in spring

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JEE Advanced Mock Test - 2 (Paper I) - Question 13
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A bulb is placed at a depth of 2√7 m in water and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum radius(in meters) of the disc?

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 13
As Shown in figure, light from bulb will not emerge out of water if at the edge of disc,

i > θC

or sin i > sin θC ...(1)

Now if R is the radius of disc and h is the depth of bulb from it

So Eq. (1) becomes

…(2)

Here h = 2√7m and μ = (4/3)

So

So radius of disc = R = 6 m

JEE Advanced Mock Test - 2 (Paper I) - Question 14
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Directions: The question is based on the following paragraph.

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless, and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at distance L from the wall. The disk rolls without slipping with velocity . The coefficient of friction is μ.

The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 14
For translational motion we have

For rotational motion we have

As the disc is rolling without slipping then

It is directed toward the equilibrium position or we can write as

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper I) - Question 15
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2.616 g of an element (M) is heated with NaOH and NaNO3 to produce Na2MO2 and NH3. Ammonia produced is absorbed in 100 mL of 1 N H2SO4. Excess of the acid is back titrated with NaOH solution and required 80 mL of 0.25 M NaOH solution up to the equivalence point. What is the mass (in ‘g’) of hydrogen gas produced when 20 moles of ‘M’ is treated with excess of NaOH?
[Atomic mass : Cr = 52; Zn = 65.4; Fe = 56; Cu = 63.5; Ni = 28; H = 1]

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 15


g-equivalents of NH3 produced = (100 × 1 - 80 × 0.25)×10-3
= 80 × 10-3
= 8 × 10-2
g-equivalents of ‘M’ = g-equivalents of NH3 = 
M = 65.4
So, ‘M’ is zinc (Zn)

So, mass of H2 gas produced = 20 × 2 = 40g.

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper I) - Question 16
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2.27 g of mercuric iodide is added into 100 mL, 0.2 M aqueous solution of KI. If KI is 90% dissociated and potassium tetraiodidomercurate(II) is 80% dissociated, then the osmotic pressure of this solution at 300 K is found to be ‘x’ atm. Calculate the value of ‘100x’ if R = 0.08 L atm mol-1 K-1. [Atomic mass : Hg = 200, I = 127]
[Assume volume of solution remains constant and formation constant of K2 [HgI4] is very large].

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 16




Now, π = MmixRT
0.32 × 0.08 × 300 = 7.68 atm = x
So, 100x = 768

*Answer can only contain numeric values
JEE Advanced Mock Test - 2 (Paper I) - Question 17
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Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 17

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JEE Advanced Mock Test - 2 (Paper I) - Question 18
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If 2 sin α sin β + 3 cos β + 5 cos α sin β = √38  ∀ α, β ∈ R, then |adj (adj A)| is equal to where  is

Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 18


JEE Advanced Mock Test - 2 (Paper I) - Question 19
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Monoatomic gas at pressure P0, volume V0 and temperature T0 is taken in a piston-cylinder such that piston is free to move inside the cylinder. The gas is compressed to volume V0/2 through various process. List-I describes four processes and List-II gives pressure of gas in final state C.


The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 19



JEE Advanced Mock Test - 2 (Paper I) - Question 20
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List-I describes four process and List-II gives loss of energy (in μJ) in each process. (C0 is capacitance of capacitor free from dielectric)
 

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 20




JEE Advanced Mock Test - 2 (Paper I) - Question 21
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In List-I solid sphere of mass m is placed on rough horizontal surface and an impulse J is applied on it and List-II gives velocity of sphere when pure rolling starts.

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 21

Using angular impulse momentum theorem

VC = J/M
(Q) Using angular impulse momentum theorem


(R) Using angular Impulse momentum theorem

(S) Using angular impulse momentum theorem

JEE Advanced Mock Test - 2 (Paper I) - Question 22
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List-I gives the situations in which a rod can oscillate. Length of rod is h, cross section area is A, density ρ, h >> √A, liquid density is σ and spring constant k = Aσg. List-II contains time period of oscillations.
 

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 22



JEE Advanced Mock Test - 2 (Paper I) - Question 23
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Some reactions are giv en in List-I. The product of each reaction (in List-I) containing the underlined element is treated with excess of PCl5 to form one or more product(s) given in List-II. Now match List-I with List-II and choose the correct option from the codes given below:

Codes:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 23

JEE Advanced Mock Test - 2 (Paper I) - Question 24
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Match the complex formed as a result of reactions giv en in List – I with their characteristics giv en in List II and choose the correct option, from the codes given below:

Codes:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 24

JEE Advanced Mock Test - 2 (Paper I) - Question 25
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Match the major organic product of reactions given in List I with their characteristics given in List II and choose the correct option from the codes given below:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 25


JEE Advanced Mock Test - 2 (Paper I) - Question 26
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Match the reactions giv en in List I with characteristics of their products giv en in List II and choose the correct option from the codes given below:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 26

JEE Advanced Mock Test - 2 (Paper I) - Question 27
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Let y = x and y = 0 are tangents to the parabola at A(4, 4) and B(2, 0) respectively, then

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 27

Let S be the focus.
So, equation of AS is 

JEE Advanced Mock Test - 2 (Paper I) - Question 28
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Match the following List-I with List-II
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 28

(P) α1 = α2 = α3 = 2, AM of α1, α2, α3 is HM of α1, α2, α3
(Q) Range of -x2 + x + a (-∞, -1], a = 1
(R) g(g(3)) = g(9) = 1 ⇒ g(g(g(3))) = g(1) = 2

(S) 

JEE Advanced Mock Test - 2 (Paper I) - Question 29
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Match the following List-I with List-II

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 29

(P) 

(Q) 

Given L.C.M. of a, b, c is 23 × 32 × 51
H.C.F of a, b, c is 2 × 3 × 5
So, for α1, α2, α3 there are 12 cases and β1, β2, β3 there are 6 cases
So, total number of triplets is 12 × 6 = 72

(R) {x/10} is discontinuous at 10, 20, 30, 40, 50
[x/2] is discontinuous at 2, 4, 6, ....., 50
So, number of points of discontinuous 25 – 5 = 20

(S) f(x) – g(x) = a(x – 1)2(x – 2)2  

JEE Advanced Mock Test - 2 (Paper I) - Question 30
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Match the following List-I with List-II

The correct option is:
Detailed Solution for JEE Advanced Mock Test - 2 (Paper I) - Question 30

(P) x = 1
(Q) 

(R) y2 = e2x – 2e2x ln y
So, given expression is 4

(S) 

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