Question

A pendulum clock is accurate at a place where $g=9.8m/s^2$. Find the value of $g$ at another place where clock becomes slow by $24$ seconds in a day $\left(24hrs\right)$.

Answer 1

we know that,

$T=2\pi \sqrt{\frac{l}{g}}$

$T=2\pi \sqrt{\frac{l}{9.8}}$

24sec loss in 24hrs

24sec loss in $24\times 60\times 6sec$

$\frac{24}{24\times 60\times 6}$loss in 1sec

$\frac{24T}{24\times 60\times 6}$ loss in T sec

$T_{new}=2\pi \sqrt{\frac{1}{g_{new}}}$

$T-\frac{24T}{24\times 60\times 60}=2\pi \sqrt{\frac{1}{g_{new}}}$

$2\pi \sqrt{\frac{l}{9.8}(1-\frac{24}{24\times 60\times 60})}=2\pi \sqrt{\frac{l}{g_{new}}}$

$\sqrt{\frac{l}{9.8}}(1-\frac{24}{24\times 60\times 60})=\sqrt{\frac{l}{g_{new}}}$

$\sqrt{\frac{l}{9.8}}(1-2.77\times {10}^{-4})=\sqrt{\frac{l}{g_{new}}}$

$\sqrt{\frac{l}{9.8}}\left(0.999\right)=\sqrt{\frac{l}{g_{new}}}$

$g_{new}=\frac{9.8}{0.998}=9.81m/s^2$ Answer