JEE Main Practice Test 2026

About the JEE Main exam

Also searched as: JEE Mains mock test. It's the same exam — JEE Main is the official name.

The Joint Entrance Examination - Main (JEE Main) is a engineering examination administered in India. It is designed to assess candidates' knowledge and proficiency across 3 key subject areas including Physics, Chemistry, Mathematics. This exam is widely recognized as an essential benchmark for aspiring engineers.

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8 free JEE Main practice questions (2026)

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1. A physical quantity X is represented by X = (A^2 * B^3) / (C * sqrt(D)). If the percentage errors in the measurements of A, B, C and D are 1%, 2%, 3% and 4% respectively, what is the maximum percentage error in the measurement of X?

   • 13%
   • 10%
   • 14%
   • 11%
   Show answer & explanation

   Answer: A. 13%

   Using error propagation, the relative error in X is given by dX/X = 2*(dA/A) + 3*(dB/B) + (dC/C) + 0.5*(dD/D). Substituting the given percentage errors: 2*(1%) + 3*(2%) + 3% + 0.5*(4%) = 2% + 6% + 3% + 2% = 13%.

   Topic: Physics · medium

2. A particle moves along the x-axis such that its coordinate x as a function of time t is given by x(t) = t^3 - 6t^2 + 9t + 5, where x is in meters and t is in seconds. Find the acceleration of the particle at the moments when it momentarily comes to rest.

   • -6 m/s^2
   • 6 m/s^2
   • either +6 m/s^2 or -6 m/s^2
   • Zero
   Show answer & explanation

   Answer: C. either +6 m/s^2 or -6 m/s^2

   Velocity v = dx/dt = 3t^2 - 12t + 9. Setting v = 0 gives t^2 - 4t + 3 = 0, which yields t = 1 s and t = 3 s. The acceleration is given by a = dv/dt = 6t - 12. At t = 1 s, a = -6 m/s^2. At t = 3 s, a = +6 m/s^2.

   Topic: Physics · medium

3. A projectile is fired at an angle of 45 degrees with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

   • tan^-1(1/2)
   • 45 degrees
   • tan^-1(sqrt(3)/2)
   • tan^-1(2)
   Show answer & explanation

   Answer: A. tan^-1(1/2)

   At the highest point, the vertical displacement is H and the horizontal displacement is R/2. The elevation angle theta satisfies tan(theta) = H / (R/2) = 2H/R. Since H = (u^2 * sin^2(alpha)) / (2g) and R = (u^2 * sin(2*alpha)) / g, we have H/R = 0.25 * tan(alpha). For alpha = 45 degrees, H/R = 1/4. Thus, tan(theta) = 2 * (1/4) = 1/2, meaning theta = tan^-1(1/2).

   Topic: Physics · medium

4. A block of mass 2 kg is resting on a rough horizontal surface with coefficient of static friction mu_s = 0.5. A horizontal force of 6 N is applied on the block. The frictional force acting on the block is (take g = 10 m/s^2):

   • 10 N
   • 6 N
   • 4 N
   • Zero
   Show answer & explanation

   Answer: B. 6 N

   The maximum static frictional force possible is f_max = mu_s * m * g = 0.5 * 2 * 10 = 10 N. Since the applied horizontal force (6 N) is less than the maximum limiting static friction, the block does not move, and the static frictional force adjusting itself is equal to the applied force, which is 6 N.

   Topic: Physics · easy

5. A body of mass 0.5 kg experiences a force F = (3x^2 + 4) N directed along the x-axis. If the body starts from rest at x = 0, its velocity at x = 2 m is:

   • 4 m/s
   • 8 m/s
   • 16 m/s
   • 2*sqrt(2) m/s
   Show answer & explanation

   Answer: B. 8 m/s

   The work done by the variable force is W = integral(F dx) from 0 to 2 = integral((3x^2 + 4) dx) = [x^3 + 4x] from 0 to 2 = 8 + 8 = 16 J. By the Work-Energy Theorem, W = Delta K = 0.5 * m * v^2. Thus, 16 = 0.5 * 0.5 * v^2, which leads to v^2 = 64, so v = 8 m/s.

   Topic: Physics · medium

6. A ball of mass m moving with velocity v_0 collides elastically with another stationary ball of identical mass. After the collision, the first ball moves at an angle of 30 degrees with its initial direction of motion. The speed of the second ball after the collision is:

   • v_0 / 2
   • sqrt(3) * v_0 / 2
   • v_0 / sqrt(2)
   • v_0 / 4
   Show answer & explanation

   Answer: A. v_0 / 2

   For an elastic collision between two identical masses where one is initially at rest, the two masses move perpendicular to each other after the collision. Since the first ball deflects by 30 degrees, the second ball must move at 60 degrees to the initial direction. By conservation of kinetic energy and momentum, the velocity vectors form a right-angled triangle with the hypotenuse equal to the initial velocity v_0. Therefore, the speed of the second ball is v_2 = v_0 * sin(30) = v_0 / 2.

   Topic: Physics · hard

7. A solid cylinder and a solid sphere, both of the same mass M and radius R, roll down the same inclined plane without slipping from rest. The ratio of their translational kinetic energies at the bottom of the incline, K_cyl / K_sph, is:

   • 14/15
   • 15/14
   • 5/6
   • 6/5
   Show answer & explanation

   Answer: A. 14/15

   The total kinetic energy is conserved and equals Mgh. The translational kinetic energy is K_t = 0.5 * M * v^2 = Mgh / (1 + k^2/R^2), where k is the radius of gyration. For a solid cylinder, k^2/R^2 = 1/2, so K_cyl = 2/3 * Mgh. For a solid sphere, k^2/R^2 = 2/5, so K_sph = 5/7 * Mgh. The ratio K_cyl / K_sph = (2/3) / (5/7) = 14/15.

   Topic: Physics · medium

8. At what depth d below the surface of the Earth (of radius R) is the acceleration due to gravity equal to its value at an altitude h = R/2 above the surface of the Earth?

   • 5R/9
   • 4R/9
   • R/2
   • 2R/3
   Show answer & explanation

   Answer: A. 5R/9

   Acceleration due to gravity at altitude h is g_h = g * R^2 / (R+h)^2. For h = R/2, g_h = g * R^2 / (1.5R)^2 = 4/9 * g. Acceleration due to gravity at depth d is g_d = g * (1 - d/R). Equating the two: 1 - d/R = 4/9, which gives d/R = 5/9, and thus d = 5R/9.

   Topic: Physics · medium

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