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JEE MAIN 2019
11-01-2019 S2
Question
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:
Select the correct option:
A
$\frac{2}{\sqrt{3}} \mathrm{~s}$
B
$\frac{3}{2} \mathrm{~s}$
C
$\frac{\sqrt{3}}{2} \mathrm{~s}$
D
$2 \sqrt{3} s$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$T = 2\pi\sqrt{\frac{\ell}{g}} \implies \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}}$$$$T_2 = 2 \sqrt{\frac{g_1}{g_2}}$$$$g_2 = \frac{3GM}{(3R)^2} = \frac{g}{3}$$$$T_2 = 2\sqrt{3} \text{ s}$$
Question Tags
JEE Main
Physics
Hard
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