JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Gravitation

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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:

(A) $\frac{3}{2} s$
(B) $\frac{2}{3} s$
(C) $\frac{3}{2} s$
(D) $23 s$ ✓ Correct

Answer: (D)

Solution

Surface gravity is

g = \frac{GM}{R ^{2}}

. The planet has three times Earth's mass and three times its diameter, hence three times its radius, so

\frac{g _{p}}{g _{e}} = 3 \times (\frac{1}{3})^{2} = \frac{1}{3}

. A pendulum's period varies as

T \propto \frac{1}{g}

, so

\frac{T _{p}}{T _{e}} = \frac{g _{e}}{g _{p}} = 3

. With

T_{e} = 2

T_{p} = 23

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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