Question

A seconds pendulum is taken to a place where accelaration due to gravity falls to 4/25.find the time period

Answer 1

Explanation:

time period of a seconds pendulum is 2 seconds

considering g= 9.8 m/s²

$= >t = 2\pi \sqrt{ \frac{l}{g} } \\ = > 2 = 2\pi \sqrt{ \frac{l}{9.8} } \\ = > 4 = 4 {\pi}^{2} \frac{l}{9.8} \\ = > l = \frac{9.8}{ {\pi}^{2} } \\ \\ now \: g = \frac{4}{25} \\ t = 2\pi \sqrt{ \frac{ \frac{9.8}{ {\pi}^{2} } }{ \frac{4}{25} } } \\ = 2\pi \times \sqrt{ \frac{9.8 \times 25}{ {\pi}^{2} \times 4 } } \\ = 2\pi \times \sqrt{9.8} \times \frac{5}{4\pi} \\ = 7.82 \: s$