JEE Advanced Practice Test 2026

About the JEE Advanced exam

The Joint Entrance Examination - Advanced (JEE Advanced) 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 Advanced practice questions (2026)

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1. A uniform thin rod of mass M and length L is initially held horizontally and pivoted smoothly at one end. The rod is released from rest. When it reaches the vertical position, the pivot suddenly breaks. Immediately after the pivot breaks, the radius of curvature of the trajectory of the rod's center of mass is:

   • 3L/4
   • 2L/3
   • L/2
   • 4L/3
   Show answer & explanation

   Answer: A. 3L/4

   By conservation of mechanical energy, when the rod reaches the vertical position, Mg(L/2) = (1/2)I\omega^2 = (1/2)(ML^2/3)\omega^2, giving \omega = \sqrt{3g/L}. The velocity of the center of mass at this instant is purely horizontal and given by v_cm = \omega(L/2) = (1/2)\sqrt{3gL}. Immediately after the hinge breaks, the only force acting on the rod is gravity, so the acceleration of the center of mass is purely vertical, a_cm = g. Since the velocity is perpendicular to the acceleration, the normal acceleration is a_n = g. The radius of curvature is \rho = v_cm^2 / a_n = (3gL/4)/g = 3L/4.

   Topic: Physics · hard

2. A satellite moves in an elliptical orbit around Earth. The ratio of its maximum speed to its minimum speed is 3. What is the ratio of the semi-major axis of the orbit to the radius of curvature of the trajectory at the perigee?

   • 4/3
   • 3/4
   • 3/2
   • 9/8
   Show answer & explanation

   Answer: A. 4/3

   By angular momentum conservation, v_max * r_p = v_min * r_a, so r_a / r_p = v_max / v_min = 3. Since r_a/r_p = (1 + e)/(1 - e) = 3, we find eccentricity e = 1/2. The radius of curvature at the vertex (perigee) of an ellipse is \rho_p = b^2 / a = a(1 - e^2). Substituting e = 1/2 gives \rho_p = a(1 - 1/4) = 3a/4. Therefore, the ratio of the semi-major axis to the radius of curvature at perigee is a / \rho_p = 4/3.

   Topic: Physics · hard

3. One mole of a monoatomic ideal gas (\gamma = 5/3) undergoes a cyclic process consisting of three steps: 1 -> 2: isochoric heating from temperature T_0 to 2T_0; 2 -> 3: adiabatic expansion until its temperature returns to T_0; 3 -> 1: isothermal compression back to the initial state. The thermal efficiency of this cycle is:

   • 1 - ln 2
   • 1 - (2/3)ln 2
   • (2/3)(1 - ln 2)
   • 1 - (3/2)ln 2
   Show answer & explanation

   Answer: A. 1 - ln 2

   Heat is absorbed only during the isochoric heating step 1 -> 2: Q_in = C_v(2T_0 - T_0) = (3/2)RT_0. In the adiabatic step 2 -> 3, T_2 V_1^(\gamma-1) = T_3 V_3^(\gamma-1), so (2T_0) V_1^(2/3) = T_0 V_3^(2/3), giving V_3/V_1 = 2^(3/2). The work done in step 2 -> 3 is W_23 = -\Delta U = (3/2)RT_0. In the isothermal compression 3 -> 1 at temperature T_0, W_31 = -RT_0 ln(V_3/V_1) = -RT_0 ln(2^(3/2)) = -(3/2)RT_0 ln 2. The net work done is W_net = W_23 + W_31 = (3/2)RT_0(1 - ln 2). The efficiency is \eta = W_net / Q_in = 1 - ln 2.

   Topic: Physics · hard

4. A U-tube of uniform cross-sectional area A contains an incompressible, non-viscous liquid column of total length L. One arm of the tube is vertical, while the other arm is inclined at an angle of 30 degrees to the horizontal. If the liquid in the vertical arm is depressed slightly by a distance x and released, the angular frequency of small oscillations of the liquid column is:

   • \sqrt{3g / (2L)}
   • \sqrt{g / (2L)}
   • \sqrt{2g / L}
   • \sqrt{g / L}
   Show answer & explanation

   Answer: A. \sqrt{3g / (2L)}

   When the liquid is depressed by x in the vertical arm, the level in the inclined arm rises by x along the incline. The vertical level in the inclined arm rises by x * sin(30 deg) = x/2. The total height difference created between the two columns is \Delta h = x + x/2 = (3/2)x. The restoring force is F = -\rho A g \Delta h = -(3/2)\rho A g x. The total mass of the oscillating liquid is M = \rho A L. The equation of motion is M x'' + (3/2)\rho A g x = 0, which yields \omega = \sqrt{(3/2)g / L} = \sqrt{3g / (2L)}.

   Topic: Physics · hard

5. A spherical conductor of radius R_1 is concentric with a conducting spherical shell of inner radius R_2 and outer radius R_3 (R_1 < R_2 < R_3). The space between R_1 and R_2 is filled with a dielectric of relative permittivity \epsilon_r. The inner sphere carries charge Q, and the outer shell is initially isolated and neutral. If the outer shell is now connected to ground, the magnitude of the change in the electric potential of the inner conductor is:

   • Q / (4\pi\epsilon_0 R_3)
   • Q / (4\pi\epsilon_0 \epsilon_r R_1)
   • Q / (4\pi\epsilon_0) * (1/R_2 - 1/R_3)
   • Q / (4\pi\epsilon_0 \epsilon_r) * (1/R_1 - 1/R_2)
   Show answer & explanation

   Answer: A. Q / (4\pi\epsilon_0 R_3)

   Before grounding, the outer shell has an induced charge of -Q on its inner surface and +Q on its outer surface. The potential of the inner conductor is V_initial = (Q / 4\pi\epsilon_0) * [1/(\epsilon_r R_1) - 1/(\epsilon_r R_2) + 1/R_3]. After grounding, the charge on the outer surface of the shell flows to the earth, making the outer surface potential zero, so V_final = (Q / 4\pi\epsilon_0) * [1/(\epsilon_r R_1) - 1/(\epsilon_r R_2)]. The magnitude of the change in potential is |V_initial - V_final| = Q / (4\pi\epsilon_0 R_3).

   Topic: Physics · hard

6. An LC circuit contains an inductor L = 2.0 H and a capacitor C = 2.0 \mu F. The capacitor is initially charged to a potential difference of V_0 = 100 V, while the current through the inductor is zero. At t = 0, the switch is closed. The first non-zero time t_1 at which the energy is equally divided between the electric field of the capacitor and the magnetic field of the inductor, and the magnitude of the current at that moment, are:

   • t_1 = \pi / 2000 s, I = 1 / (10\sqrt{2}) A
   • t_1 = \pi / 1000 s, I = 1 / 10 A
   • t_1 = \pi / 4000 s, I = 1 / (5\sqrt{2}) A
   • t_1 = \pi / 2000 s, I = 1 / 5 A
   Show answer & explanation

   Answer: A. t_1 = \pi / 2000 s, I = 1 / (10\sqrt{2}) A

   The angular frequency is \omega = 1/\sqrt{LC} = 1/\sqrt{2 * 2 * 10^-6} = 500 rad/s. Charge varies as q(t) = Q_0 cos(\omega t). Energy is equally divided when q^2 / (2C) = (1/2)(Q_0^2 / 2C), which means cos^2(\omega t) = 1/2, so \omega t_1 = \pi/4. Hence, t_1 = \pi / (4 * 500) = \pi / 2000 s. The current is i(t) = -Q_0 \omega sin(\omega t). At \omega t = \pi/4, |I| = C V_0 \omega (1/\sqrt{2}) = (2 * 10^-6 * 100) * 500 / \sqrt{2} = 0.1 / \sqrt{2} = 1 / (10\sqrt{2}) A.

   Topic: Physics · hard

7. A particle of mass m and positive charge q is projected from the origin with initial velocity v_0 along the positive x-axis in a region containing a uniform electric field E_0 along the positive y-axis and a uniform magnetic field B_0 along the positive z-axis. The minimum positive value of v_0 for which the particle momentarily comes to rest at some point during its subsequent motion is:

   • 2 E_0 / B_0
   • E_0 / B_0
   • E_0 / (2 B_0)
   • 4 E_0 / B_0
   Show answer & explanation

   Answer: A. 2 E_0 / B_0

   Using cycloidal motion equations with drift velocity u_d = E_0/B_0 and cyclotron frequency \omega = qB_0/m: the velocity components are v_x(t) = u_d - (u_d - v_0) cos(\omega t) and v_y(t) = (u_d - v_0) sin(\omega t). For the particle to come to rest, both v_x = 0 and v_y = 0 must hold simultaneously. v_y = 0 requires \omega t = n\pi. For n = 1 (\omega t = \pi), cos(\omega t) = -1, which gives v_x = u_d + (u_d - v_0) = 2u_d - v_0. Setting v_x = 0 yields v_0 = 2u_d = 2E_0/B_0. (For even n, cos(\omega t) = 1 gives v_x = v_0, which only vanishes if v_0 = 0). Thus, the minimum positive value is 2E_0/B_0.

   Topic: Physics · hard

8. A conducting rod of mass m and length L slides vertically downward under gravity without friction along two vertical, parallel, perfectly conducting rails separated by distance L. The rails are connected at the top by a capacitor of capacitance C. A uniform horizontal magnetic field B is applied perpendicular to the plane of the rails. The downward acceleration of the rod is:

   • mg / (m + C B^2 L^2)
   • mg / (m - C B^2 L^2)
   • g * (1 - C B^2 L^2 / m)
   • C B^2 L^2 g / (m + C B^2 L^2)
   Show answer & explanation

   Answer: A. mg / (m + C B^2 L^2)

   At downward speed v, the induced EMF across the rod is \mathcal{E} = BLv. The charge on the capacitor is q = C\mathcal{E} = CBLv. The induced current is i = dq/dt = CBL(dv/dt) = CBLa, where a is the downward acceleration. The upward magnetic force acting on the rod is F_B = iLB = (CBLa)LB = C B^2 L^2 a. Applying Newton's second law: mg - F_B = ma => mg - C B^2 L^2 a = ma => a = mg / (m + C B^2 L^2).

   Topic: Physics · hard

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