JEE PYQ: Gravitation Question 46
Question 46 - 2019 (11 Jan Shift 2)
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 2 s. The period of oscillation of the same pendulum on the planet would be:
(1) $\frac{\sqrt{3}}{2}$ s
(2) $\frac{2}{\sqrt{3}}$ s
(3) $\frac{3}{2}$ s
(4) $2\sqrt{3}$ s
Show Answer
Answer: (4)
Solution
$g_p = \frac{GM_p}{R_p^2} = \frac{3GM_e}{(3R_e)^2} = \frac{g}{3}$. $T_p = T_e\sqrt{\frac{g_e}{g_p}} = 2\sqrt{3}$ s.
BETA
