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JEE MAIN 2026
22-01-2026 S1
Question
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is_________m/s.
(g = acceleration due to gravity)
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Select the correct option:
A
$\frac{2 \mathrm{mgR}}{\mathrm{B}^2 l^2}$
B
$\frac{8 \mathrm{mgR}}{\mathrm{B}^2 l^2}$
C
$\frac{2 m g R}{B^2 L^2}$
D
$\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Main
Physics
Hard
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