Question

ii) The length of second's pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 of that on the earth's surface]
$6m$
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Answer 1

We know that the time period of a second pendulum is 2s .

At earth ,

$T = 2\pi \sqrt{ \frac{le}{ge} } \\ 2 = 2\pi \sqrt{ \frac{le}{ge} }$

Or,

$ge = \pi²1e$

At moon,

$T = 2\pi \sqrt{ \frac{1m}{gm} } \\ 2 = 2\pi \: \sqrt{ \frac{1m(ge)}{6} }$

Given gm =ge/6

Or,

$2 = 2\pi \sqrt{ \frac{61m}{ {\pi}^{2} 1e} }$

putting the value of gm from eq1

or

1m = le/6

Now, given le = 1m

Hence, Im = 1/6m