Question

Choose the correct option:

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` th of that on the earth’s surface]

Options

• `1/6`m

• 6 m

• `1/36` m

• `1/sqrt6`m

Answer 1

`1/6`m

Explanation:

Express the relation for the time period of a simple pendulum.

`T = 2pi sqrt(l/g)`

Here, T is the time period, l is the length and g is the acceleration due to the gravity of the pendulum.

If the time period is constant,

`l  oo  g`

Or

The length of the pendulum is proportional to the acceleration due to gravity.

Let `g_e` and `g_m` be the acceleration due to gravity of the earth and moon respectively. Also, let `l_e` and `l_m` be the lengths of the earth and moon respectively.

Hence, `l_e/l_m = g_e/g_m`

`l_m = g_m/g_e  l_e`

Substitute 1 m for `l_e` and `1/6  g_e` for `g_m` to find the value of `l_m`.

`l_m xx (1g_e)/(6g_e) xx 1 m` 

`l_m = 1/6 m`

Hence, the length of the second’s pendulum on the surface of the earth is `1/6 m`.