Question

1. A seconds pendulum is taken to a place where
acceleration due to gravity falls to one-fourth. How
is the time period of the pendulum affected, if at
all ? Give reason. What will be its new time period ?​

Answer 1

Answer:

The time period increases and it is given by ,

$T'= 4\pi \sqrt{\frac{l}{g}}$

Explanation:

time period of a pendulum is given by,

$T=2 \pi \sqrt{\frac{l}{g} }$  

where,  l = length of pendulum and

 g = acceleration due to gravity.

now, coming to the question.

acceleration due to gravity falls to one-fourth.

(i.e.,) g' = (1/4)g

let the new time period be T'

T' = $2\pi \sqrt{\frac{l}{(1/4)g}}$

$T' = 2\pi \sqrt{\frac{4l}{g}}$

$T'= 4\pi \sqrt{\frac{l}{g}}$

Thus you got the answer !