Question

iv) The length of the seconds pendulum on the surface of the earth
is 0.95m. What will be its length on the surface of the moon?
[ Smoon = (1 / 6 ) × g EARTH] (g: acceleration due to gravity)​

Answer 1

Answer:0.95/6m

Explanation:

The length of the seconds pendulum on the earth=0.95m

Its length on the moon=?

[Smoon=(1/6)*g]

g: acceleration due to gravity

TE=TM

$2\pi \sqrt \frac{0.95}{g} = 2\pi \sqrt \frac{1a}{ \frac{g}{6} }$

(let, length of the seconds pendulum on the surface of the moon=1a)

$1a = \frac{0.95}{6} m$