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JEE Advanced Practice Test- 3 Free Online Test 2026


MCQ Practice Test & Solutions: JEE Advanced Practice Test- 3 (54 Questions)

You can prepare effectively for JEE Mock Tests for JEE Main and Advanced 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "JEE Advanced Practice Test- 3". These 54 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Number of Questions: 54

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JEE Advanced Practice Test- 3 - Question 1

An engine whistling at a constant frequency n0 and moving with a constant velocity goes past a stationary observer. As the engine crosses him, the frequency of the sound heard by him changes by a factor f. The actual difference in the frequency of the sound heard by him before and after the engine across him is

 

Detailed Solution: Question 1

Simply use the above relation

JEE Advanced Practice Test- 3 - Question 2

A ball of density ρ0 falls from rest from point P onto the surface of a liquid of density ρ in time T. It enters the liquid stops moves up and returns to P in a total time 3T. Neglect viscosity, surface tension and splashing. The ratio p/p0 is equal to

Detailed Solution: Question 2

JEE Advanced Practice Test- 3 - Question 3

When an object is placed in front of a concave mirror of focal length f, a virtual image is produced with a magnification of 2. To obtain a real image with a magnification of 2. The object has to be moved by a distance equal to

Detailed Solution: Question 3

Calculate the initial and final distance of object from the concave mirror.

JEE Advanced Practice Test- 3 - Question 4

A solid cube is placed on a horizontal surface. The coefficient of friction between them is μ, where μ < 1/2 . A variable horizontal force is applied on the cube’s upper face, perpendicular to one edge and passing through the mid-point of that edge. The maximum acceleration with which it can move without toppling is

Detailed Solution: Question 4

To avoid t toppling

JEE Advanced Practice Test- 3 - Question 5

A spherical body of mass m, radius r and moment of inertia I about its centre moves along the x-axis. Its centre of mass moves with velocity = v0 and it rotates about its centre of mass with angular velocity ω0.LO = lω0 + mvor . The angular momentum of the body about the origin O is

Detailed Solution: Question 5

As we know
L0 = L1 + L2
L1 = due to rotational motion
L2 = due to linear motion.

*Multiple options can be correct
JEE Advanced Practice Test- 3 - Question 6

 DIRECTION ;- This section contains 2 multiple choice questions from 8 to 9. Each question has 4 choices (A), (B), (C) and (D) for its answer, out which ONE OR MORE is/are correct

The temperature of a solid object is observed to be constant during a period. In this period

Detailed Solution: Question 6

If temperature of a body is constant that means amount of hear energy received and amount of hear energy emitted are equal.

*Multiple options can be correct
JEE Advanced Practice Test- 3 - Question 7

The electric potential decreases uniformly from 120 V to 80 V as one moves on the axis from x = –1 cm to x = 1 cm. The electric field at the origin

Detailed Solution: Question 7

If direction of field is along the x-axis then its magnitude will be 20 V/cm other wise greatest then hat

JEE Advanced Practice Test- 3 - Question 8

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C) and D) out of WHICH ONLY ONE IS CORRECT.

paragraph for  question no.12 and 13

The base of a hollow right cone of semi vertical angle 30o is fixed to a horizontal plane. Two particle each of mass m are connected by a light inextensible string which passes through a small hole in the vertex of the cone. One particle A hangs at rest inside the cone. The other particle B moves on the outer smooth surface of the cone at a distance ℓ from vertex in a horizontal circle with centre at A. Neglecting friction, now answer the following.

The tention in the string is

Detailed Solution: Question 8

Now apply newton’s law of motion for the particle B.

JEE Advanced Practice Test- 3 - Question 9

The base of a hollow right cone of semi vertical angle 30o is fixed to a horizontal plane. Two particle each of mass m are connected by a light inextensible string which passes through a small hole in the vertex of the cone. One particle A hangs at rest inside the cone. The other particle B moves on the outer smooth surface of the cone at a distance ℓ from vertex in a horizontal circle with centre at A. Neglecting friction, now answer the following.

The angular velocity of B is

Detailed Solution: Question 9

Now apply newton’s law of motion for the particle B.

JEE Advanced Practice Test- 3 - Question 10

Paragraph For Question No.14 and 16

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C= 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

Detailed Solution: Question 10

Initially current will pass through the resistance only.

JEE Advanced Practice Test- 3 - Question 11

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C2 = 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

The charge on capacitance C1 and C2 are respectively

Detailed Solution: Question 11

Initially current will pass through the resistance only.

JEE Advanced Practice Test- 3 - Question 12

The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C1 = 3μF, C2 = 6μF, R1 = 4 kΩ and R2 = 7.0 kΩ. The power dissipated in R2 is 2.8 W.

The power dissipated to the resistor R1 is

Long time after switch is opened the charge on C1 is

Detailed Solution: Question 12

Option C is correct.

Given P_{R2} = 2.8 W and R_2 = 7.0 kΩ. Use P = I^2 R to find the steady current: I = √(P_{R2}/R_2) = 0.02 A (20 mA).

Power dissipated in R1: P_{R1} = I^2 R_1 = (0.02)^2 imes 4000 = 1.6 W.

Total resistance in the series path is R_1+R_2 = 11.0 kΩ, so the battery voltage is E = I(R_1+R_2) = 0.02 imes 11000 = 220 V.

When the switch is opened and a long time passes, the bottom plate of C1 is connected to the lower rail (battery negative) through R_2 while the top plate of C1 remains fixed at the battery positive potential. After transients die out, C1 charges to the full battery voltage E = 220 V.

Thus the final charge on C1 is Q = C_1 E = 3 × 10^{-6} F imes 220 V = 660 µC.

Therefore the correct option is 660 µC (Option C).


JEE Advanced Practice Test- 3 - Question 13

Column – I gives certain situation involving two thin conducting shells connected by a conducting wire via a key K. In all situation one sphere has net charge +q and other sphere has no net charge. After the key K is pressed

column – II gives some resulting effect. Match the figures in column with the statements in column – II.

 

Detailed Solution: Question 13

In such type of situation, we have to use these basic facts.
(a) charge will flow between the two body if there is potential difference between the two.
(b) during this process energy will lose due to sparking

JEE Advanced Practice Test- 3 - Question 14

A particle of mass 2 kg is moving on straight line under the action of force F = (8 – 2x)N. The
particle is released at rest from x = 6m. For the sub-sequent motion match the following (All the
values in the right column are in there SI units).

 

Detailed Solution: Question 14

At equilibrium Fnet will be zero and at the extreme point, particle will be in state of rest.

*Answer can only contain numeric values
JEE Advanced Practice Test- 3 - Question 15

This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

In the arrangement shown in figure, pulley are light and frictionless, threads are inextensible and mass of blocks A, B and C are m1 = 5 kg, m2 = 4 kg and m3 = 2.5 kg respectively and co-efficient of friction for both the planes is μ = 0.50. Calculate the acceleration of block A (in m/sec), when the system is released from rest. (g = 10 m/sec)


Detailed Solution: Question 15

In this situation block B will not move at all.

Now apply newton's laws of motion. 

*Answer can only contain numeric values
JEE Advanced Practice Test- 3 - Question 16

The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is


Detailed Solution: Question 16


Where R is radius of curvature.

*Answer can only contain numeric values
JEE Advanced Practice Test- 3 - Question 17

A disc ‘A’ of mass M is placed at rest on the smooth inclined surface of inclination θ. A ball B of mass m is suspended vertically from the centre of the disc A by a light inextensible string of light ℓ as shown in the figure. If the acceleration of the disc B immediately after the system is released from rest

 


Detailed Solution: Question 17

Correct option: A.

Let the acceleration of the disc along the plane (down the plane) be a. The vertical acceleration of the point of suspension (the disc centre) is then a sinθ downward.

For the ball (mass m), taking downward as positive, Newton's second law gives m a sinθ = m g - T, where T is the tension in the string.

Hence T = m (g - a sinθ).

For the disc (mass M) along the plane, the component of its weight down the plane is M g sinθ and the component of the tension along the plane (opposing downward motion) is T sinθ. Thus Newton's second law along the plane is M a = M g sinθ - T sinθ.

Substitute T = m (g - a sinθ) into the disc equation to get

M a = M g sinθ - sinθ · m (g - a sinθ).

Simplify: M a = (M - m) g sinθ + m a sin^2 θ.

Collecting terms in a gives a (M - m sin^2 θ) = (M - m) g sinθ.

Therefore the required acceleration is a = (M - m) g sinθ / (M - m sin^2 θ), which matches option A.

*Answer can only contain numeric values
JEE Advanced Practice Test- 3 - Question 18

A uniform rod AB of mass M and length R √2 is moving in a vertical plane inside a hollow sphere of radius R. The sphere is rolling on a fixed horizontal surface without slipping with velocity of its centre of mass 2v, when the end B is at the lowest position, its speed is found to be v as shown in the figure. If the kinetic energy of rod at this instant is 4/k mv2. Find k.


Detailed Solution: Question 18

length of rod is R √2

Velocity of point P
2v - ωR = v
KE of rod = rotational KE t translatory KE

JEE Advanced Practice Test- 3 - Question 19

Sulphur reacts with chlorine in 1 : 2 ratio and forms X. Hydrolysis of X gives a sulphur compoundY. What is the hybridization state of central atom in the compound Y?

Detailed Solution: Question 19


Hybridisation of S in H2SO3 = 1/2 (6+2+0) = 4 = sp3 (3σ + 1π)

JEE Advanced Practice Test- 3 - Question 20

Find out the correct representation of trans - decaline

Detailed Solution: Question 20

JEE Advanced Practice Test- 3 - Question 21

50 ml of a solution containing 10-3 mol of Ag+ is mixed with 50 ml of a 0.1 M HCl solution. Howmuch [Ag+] remains in solution? Given: Ksp of AgCl = 10-10

Detailed Solution: Question 21





 

JEE Advanced Practice Test- 3 - Question 22

Where X is a compound that forms azodye with benzene diazonium chloride in faintly basic medium.
Hence the products P, X and Y are respectively.

Detailed Solution: Question 22

*Multiple options can be correct
JEE Advanced Practice Test- 3 - Question 23

Which of the following solutions show lowering of vopour pressure on mixing?

Detailed Solution: Question 23

In B and D compound after mixing undergoes H – bonding and hence boiling point will be raised and hence vapour pressure lowered.

*Multiple options can be correct
JEE Advanced Practice Test- 3 - Question 24

Which of the following will produce aromatic compound as a product?

Detailed Solution: Question 24


Hence, option(A), (B) is correct.

JEE Advanced Practice Test- 3 - Question 25

This section contains 2 paragraphs. Based upon the first paragraph 2 multiple choice questions and based upon the second paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C) and D) out of WHICH ONLY ONE IS CORRECT.

Paragraph For Question No.12 and 13

A white compound (A) on strong heating decomposes to produce two products (B) and (C). (B) on reaction with white phosphorus produces (D), which is a strong dehydrating agent. (D) on reaction with perchloric acid converts it to its anhydride.

The compound (A) is

Detailed Solution: Question 25

JEE Advanced Practice Test- 3 - Question 26

Paragraph For Question No.12 and 13

A white compound (A) on strong heating decomposes to produce two products (B) and (C). (B) on reaction with white phosphorus produces (D), which is a strong dehydrating agent. (D) on reaction with perchloric acid converts it to its anhydride.
The product/s obtained on hydrolysis of (D) is

 

 

Detailed Solution: Question 26

JEE Advanced Practice Test- 3 - Question 27

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The percent distribution of C in the product mixture B, C and D at any time is equal to
 

 

Detailed Solution: Question 27

If k1, k2 and k3 be the rate constants of the reaction along path I, II and III respectively, then overall rate constant of consumption of A will be k1 + k2 + k3. So,

JEE Advanced Practice Test- 3 - Question 28

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The initial rate of consumption of A and the sum of the initial rate of formation of B, C and D arerespectively, taking [A] = 0.25 M, equal to

Detailed Solution: Question 28




JEE Advanced Practice Test- 3 - Question 29

A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometry

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min respectively. The energies of activation of the reaction along path I, path II and path III are 40, 60 and 80 kJ/mol respectively.

The overall energy of activation of A along all the three parallel path is equal to

 

Detailed Solution: Question 29

JEE Advanced Practice Test- 3 - Question 30

Match the following ;

Detailed Solution: Question 30

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