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


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

JEE Advanced Mock Test - 3 (Paper I) for JEE 2025 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Advanced Mock Test - 3 (Paper I) questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Mock Test - 3 (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 - 3 (Paper I) below.
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JEE Advanced Mock Test - 3 (Paper I) - Question 1
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Match the following:

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

RMS value of alternating current =

x = x0sinωt cosωt

x = x0 / 22sinωt cosωt

x = x0 / 2sin2ωt

RMS value,

x = x0sinωt + x0cosωt

RMS value,

JEE Advanced Mock Test - 3 (Paper I) - Question 2
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A body is thrown with a velocity of 10 m s−1 at an angle of 45° to the horizontal. The radius of curvature of its trajectory in t = 1/√2 s after the body began to move is,

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

For projectile, we can write,

ux = 10cos45° = vx (ax = 0)

From the equation of motion at time t = 1√2 sec in y-direction,

Thus, the particle is at the highest point.

Velocity of the particle at the given time is

Let the radius of curvature be R.

Then, from centripetal acceleration,

or

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JEE Advanced Mock Test - 3 (Paper I) - Question 3
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The figure below shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 2 Ω/m. The position of the conducting rod at t = 0 is shown. A time-dependent magnetic field B = 4t T is switched on at t = 0.

The current in the loop at t = 0 due to induced emf is

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

Induced current,

=

= 0.21 A

Using Lenz's law, the upper end of the rod is negative which makes the current flow clockwise.

JEE Advanced Mock Test - 3 (Paper I) - Question 4
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A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 4

At equilibrium, let the extension of string be x. Then

Spring Force + up thrust = weight of the block

When the block is further displaced by “y” in downward direction and then released , resultant force acting in upward direction is

Hence,

Comparing with a = ω2y

JEE Advanced Mock Test - 3 (Paper I) - Question 5
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Answer the following by appropriately matching the lists based on the information given:

List I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as λf. Match each system with statements given in List II describing the nature and wavelength of the standing waves.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 5


Longitudinal wave

Longitudinal wave

JEE Advanced Mock Test - 3 (Paper I) - Question 6
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Calculate the heat supplied, work done and the change in the internal energy in the isobaric process.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 6

Process CA is an isobaric process:
As gas is monatomic,

Heat supplied in the process CA:

Change in the internal energy during the process CA:


Using first law of thermodynamics, we have:
∆Q = W + ∆U
Hence, work done, W = ∆Q - ∆U.
WCA = ∆QCA - ∆UCA
WCA = (-500R) - (-300R) = -200R

JEE Advanced Mock Test - 3 (Paper I) - Question 7
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Calculate the heat supplied, work done and the change in the internal energy in the isochoric process.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 7

Process AB is an isochoric process.
Volume remains constant during the isochoric process, so WAB = 0.
Hence,
∆QAB = ∆UAB
Change in the internal energy during the process AB:

Heat supplied in the process AB:
∆QAB = ∆UAB = 300R

JEE Advanced Mock Test - 3 (Paper I) - Question 8
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Calculate the heat supplied, work done and the change in the internal energy in the isothermal process.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 8

Process CA is an isothermal process.
UCA = 0
As AB is an isochoric process, therefore

Work done during isothermal process

Hence,
∆QBC = WBC = 250R

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 9
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Two identical objects A and B are at temperatures TA and TB respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature T (TA > T > TB). The objects A and B attain the temperature T eventually. Select the correct statements from the following–
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 9
Heat transfer from a body with a high temperature to a body with a lower temperature, when bodies are not in direct physical contact with each other or when they are separated in space, is called heat radiation.

Every object emits and absorbs the radiations simultaneously, if energy emitted is more than energy absorbed, temperature falls and vice versa.

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 10
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In a photoelectric effect experiment, the maximum kinetic energy of the ejected photoelectrons is measured for various wavelengths of the incident light. Diagram shows a graph of this maximum kinetic energy Kmax as a function of the wavelength λ of the light falling on the surface of the metal. Which of the following statement/s is/are correct?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 10

Cut-off wavelength, λ0 = 250nm

Threshold frequency,

Work function of the metal,

Maximum kinetic energy of photoelectrons corresponding to light of wavelength 100 nm is

= 7.432 eV ≈ 7.4 eV

Photoelectric effect takes place only for light of wavelength less than 250 nm, whereas

λred ≈ 700 nm

*Multiple options can be correct
JEE Advanced Mock Test - 3 (Paper I) - Question 11
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Three points are located at the vertices of an equilateral triangle whose side equals a. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 11
Since, the magnitude of velocity remains constant. So, instantaneous tangential acceleration is zero for each particle. So, only centripetal force is present on each particle. The vector sum of forces acting on the particles is zero. So, at every instant the momentum of system remains constant. If the mass of each particle are same. Then the momentum of the system is zero at every instant. So, the centre of mass (centroid) of system remains in rest.

Hence, collision takes place at the centre of mass of system (centroid).

Component of velocity towards O is V cos30

The time of converging = AO / Vcos30

Where

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 12
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A steady current I goes through a wire loop PQR having shape of a right-angled triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is  , find the value of k.

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

The magnetic field (B) due to elements PR and PQ is 0 as the point P is on the line of the conductor.
The magnetic induction due to the current element RQ at P:

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 13
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A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s-2.)


The magnitude (in Newton) of the normal reaction that acts on the block at the point Q is N. Find the value of 2N.

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

N - mg cos 60° = mv2 / R
N = 5 + 5/2 = 7.5 Newton
2N = 15

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 14
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A sample of air is initially at 533 K and 700 N/m2, and occupies a volume of 0.028 m3. The air is expanded at constant pressure to three times. A polytropic process with n = 1.5 is then carried out followed by an isothermal process which completes the cycle.

The efficiency of the cycle (in percentage in nearest integer) is ___.

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 14

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 15
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A point object is placed in front of a thin biconvex lens, of focal length 20 cm. When placed in air, the refractive index of material of lens is 1.5 The further surface of the lens is silvered and is having radius of curvature of 25 cm. The position of final image of object is at 25/x cm from lens. Determine the value of x?

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 15
We know that a thin silvered lens is equivalent to a combination of two lenses and a mirror. This equation system is working as a mirror whose

focal length is given by

where, fL = 20 cm and fM = −25 / 2cm

Let v is the image distance from lens, then

⇒ v = −12.5 cm

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 16
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Illuminate the surface of a certain metal alternately with wave length λ1 = 0.35 μm and λ2 = 0.54 μm. It is found that the corresponding maximum velocity is of photo electron having a ratio n = 2. Find the work function of that metal (in eV).

(Round off up to 2 decimal places)

Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 16

Using equation of two wave lengths,

- W

As, v1 = 2v2

3W = 4

3W =

= 5.64 eV

W =

= 1.88 eV

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 17
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A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls t without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is

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

Let the friction between surface and sphere be fs.

Let the friction between stick and sphere be fa.

Let N be the normal reaction between sphere and stick.

Now, N - fs = ma

Also N = 2N

fs = 1.4N

Writing torque about the centre, we get

(fs - fa)r = Iα

a = αr and I = mr2

On solving, fa = 0.8 = μN

μ = 0.4 = 4/10

So, P = 4

JEE Advanced Mock Test - 3 (Paper I) - Question 18
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Directions: The following question has four choices, out of which ONLY ONE is correct.

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 I) - Question 18

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 (d) is correct.

JEE Advanced Mock Test - 3 (Paper I) - Question 19
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Given below are two cleavage reactions:

(i) (CH3)3COCH3 → CH3I + (CH3)3COH

(ii) (CH3)3COCH3 → CH3OH + (CH3)3CI
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 19

The low polarity of solvent ether favours the SN2 mechanism.

The O-CH3 bond breaks and methyl iodide and tertiary butyl alcohol are formed as products.

The presence of a solvent with high polarity such as water favours the SN1 mechanism.

The (CH3)3-O bond breaks and methyl alcohol is formed.

The mechanism is

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 20
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EDTA4– is ethylenediaminetetraacetate ion. What is the total number of N – Co – O bond angles in [Co(EDTA)]1– complex ion?

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

Total No. of N-Co - O bond angles is 8.

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 21
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The value of log10 K for a reaction A ⇌ B is (Given Δr = -54.07 kJ mol-1, Δr = 10JK-1mol-1 and R = 8.314JK-1mol-1; 2.303 x 8.314 x 298 = 5705)

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

ΔG = -2.303 RT log K
ΔG = ΔH - TΔS= -54,070 - 2,980
= -57,050 J
-57,050 = -2.303 × 8.314 × 298 log K
-57,050 = -5705 log K
log K = 10

*Answer can only contain numeric values
JEE Advanced Mock Test - 3 (Paper I) - Question 22
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The number of paired electrons in O2 molecule is

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


The total number of electrons in O2 molecule are 16.
There are two unpaired electrons in the anti-bonding pi-2p orbitals. The remaining 14 electrons are paired.

JEE Advanced Mock Test - 3 (Paper I) - Question 23
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Match the names in List - 1 with their reactions in List - 2, and choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 23

Dehydrohalogenation: It is an elimination reaction that eliminates (removes) a hydrogen halide from a substrate.
C5H11X  CH3CH2CH=CH-CH3
Wurtz reaction: Alkyl halide reacts with sodium in the presence of dry ether to form alkane.
R-X + 2Na + R-X  R-R + 2NaX
Fittig reaction: Aryl halide reacts with sodium to form biphenyl.
C6H5X + 2Na + C6H5X   C6H5-C6H5 + 2NaX
Dehalogenation: It is an elimination reaction that eliminates halogens from the substrate.
Br-CH2CH2-Br + Zn  CH2=CH2 + ZnBr2

JEE Advanced Mock Test - 3 (Paper I) - Question 24
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Match the ions of 3d series of d-block elements in the modern periodic table in List - 1 with their values of magnetic moment in List - 2, and choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 24

Magnetic moment = , where n = no. of unpaired electrons
Ti3+ (Z = 22), electronic configuration: 3d1; no. of unpaired electrons: 1

Cr3+ (Z = 24), electronic configuration: 3d3; no. of unpaired electrons: 3

Ni2+ (Z = 28), electronic configuration: 3d8; no. of unpaired electrons: 2

Mn2+ (Z = 25), electronic configuration: 3d5; no. of unpaired electrons: 5

JEE Advanced Mock Test - 3 (Paper I) - Question 25
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Match the coordination compounds in List - 1 with their shapes/geometries and magnetic characters in List - 2, and choose the correct option.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 25

[Ni(CI)4]2-

Hybridisation: sp3
Tetrahedral, paramagnetic

[NiCN)4]2-

Hybridisation: dsp2
Square planar, diamagnetic

[Cu(NH3)4]+2

Hybridisation: dsp2
Square planar, paramagnetic

Ni(CO)4

Hybridisation: sp3
Tetrahedral, diamagnetic

JEE Advanced Mock Test - 3 (Paper I) - Question 26
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Answer by appropriately matching the information given in the three columns of the following table:

From the three columns, select a combination for the metal whose oxide and hydrated chloride form the components of Sorel's cement.
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 26

Sorel's cement is 2.5 to 3.5 parts by weight of MgO and 1 part of MgCl2.6H2O.
Mg reacts with hot water to form Mg(OH)2, which is basic in nature.
Epsom salt: MgSO4.7H2O

JEE Advanced Mock Test - 3 (Paper I) - Question 27
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Directions: Match the integrals in Column I with the values in Column II.

Which of the following is the correct option?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 27

For (i)

Put x = tanθ
⇒ dx = sec2θdθ


So, (i)  (D)
For (ii)


Put x = sinθ
⇒ dx = cosθdθ


So, (ii) - (E)
For (iii)


So, (iii) - (A)
For (iv)


So, (iv) -(C)​
Hence, option 3 is correct.

JEE Advanced Mock Test - 3 (Paper I) - Question 28
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Let (x, y) be such that sin-1(ax) + cos-1(y) + cos-1(bxy) = π/2
Which of the following options correctly matches the statements in Column I with the statements in Column II?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 28




Hence (A) → (p), (B)→  (q) , (C) → (p), (D)→  (s) is correct.

JEE Advanced Mock Test - 3 (Paper I) - Question 29
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Consider the lines given by:
L1 : x + 3y - 5 = 0, L2 : 3x - ky - 1 = 0, L3 : 5x + 2y - 12 = 0

Which of the following options correctly matches the statements in Column I with the statements in Column II?
Detailed Solution for JEE Advanced Mock Test - 3 (Paper I) - Question 29

Given equation of lines are L1 : x + 3y - 5 = 0, L2 : 3x - ky - 1 = 0, L3 : 5x + 2y - 12 = 0
(A) Solving equation L1 and L3,

∴ x = 2, y = 1
L1, L2, L3 are concurrent, if point (2, 1) lies on L2.
∴ 6 - k - 1 = 0
⇒ k = 5
(B) Either L1 is parallel to L2, or L3 is parallel to L2, then

(C) L1, L2, L3 will form a triangle, if they are not concurrent or not parallel.

(D) L1, L2, L3 do not form a triangle, if k = 5, -9,- 6/5
Hence, (A) → (r), (B) → (p), (C) → (q), (D) → (t)

JEE Advanced Mock Test - 3 (Paper I) - Question 30
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Directions: Answer the following question by appropriately matching the information given in the three columns in the following table.

Here,
Column I contains equation of first line
Column II contains equation of second line
Column III contains angle between the first and the second line

Which of the following options is an incorrect combination?
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