If the time period of simple pendulum on the surface of earth is T. If the same pendulum is taken at the planet whose mass and radius both are three times as compare to the earth. The time periods of oscillation is –

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\frac{1}{\sqrt{3}}

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\sqrt{3}

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\frac{T}{3}

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Solution

The correct option is C $\sqrt{3}$ T
Time period of a simple pendulum = T = 2 $π \sqrt{\frac{l}{g}}$ The relation between g and G is given by g = $\frac{G M}{R^{2}}$

From this relations the time period of pendulum will become $\sqrt{3} T$ .

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If the time period of simple pendulum on the surface of earth is T. If the same pendulum is taken at the planet whose mass and radius both are three times as compare to the earth. The time periods of oscillation is –

The correct option is C √3 TTime period of a simple pendulum = T = 2π√lg The relation between g and G is given by g = GMR2 From this relations the time period of pendulum will become √3T.

The correct option is C $\sqrt{3}$ T
Time period of a simple pendulum = T = 2 $π \sqrt{\frac{l}{g}}$ The relation between g and G is given by g = $\frac{G M}{R^{2}}$

From this relations the time period of pendulum will become $\sqrt{3} T$ .