The length of a second's pendulum at the surface of earth is $1 m$ . The length of second's pendulum at the surface of moon where $g$ is $1 / 6 t h$ that at earth's surface is

1 / 6 m

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6 m

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1 / 36 m

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36 m

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Solution

The correct option is A $1 / 6 m$
For T to be same, $\sqrt{l_{e} / g_{e}} = \sqrt{l_{m} / g_{m}}$
or, $1 / g = l_{m} / (g / 6)$ , implies $l_{m} = 1 / 6 m$

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The length of a second's pendulum at the surface of earth is 1 m. The length of second's pendulum at the surface of moon where g is 1/6th that at earth's surface is

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The correct option is A $1 / 6 m$
For T to be same, $\sqrt{l_{e} / g_{e}} = \sqrt{l_{m} / g_{m}}$
or, $1 / g = l_{m} / (g / 6)$ , implies $l_{m} = 1 / 6 m$