Question

A pendulum clock shows accurate time. If the length increases by 0.1 %, deduce the error in time per day.

Answer 1

$L^{\prime }=1.001L$
$T=2\pi \sqrt{\frac{L}{g}}=ls$
$T^{\prime }=2\pi \sqrt{\frac{L^{\prime }}{g}}=2\pi \sqrt{\frac{1.001L}{g}}=T\times \sqrt{1.001}=1.0005\times T$
$T^{\prime }=1.0005s$
Error in time per day=$1.0005\times 60\times 60\times 24-1\times 60\times 60\times 24$
Error = 86443.2-86400
Error = 43.2 s