Question

Find the time period of a simple pendulum of length 90cm at a place where acceleration due to gravity is 10 metre pe second square

Answer 1

I hope this helps .......

Answer 2

Answer:

Time period will be doubled

Time period with acceleration due to gravity g

${T_1}$

$t \: = 2π \sqrt{ \frac{l}{g} }$

for a second's pendulum time period is 2s

when acceleration is one -fourth

${T_2}$

$t \: = 2π \sqrt{ \frac{ \frac{l}{g} }{4} } \\ \\ t \: = 2 \times 2π \sqrt{ \frac{l}{g}}$

2 × ${T_1}$

=4×2=4s