If the length of a simple pendulum is decreased by 3%, the percentage change in its period T is :
[Note : $T=2\pi \sqrt{\frac{L}{g}}$ where L is the length of pendulum and g is constant'
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.5% decrease
$T=2\pi \sqrt{\frac{L}{g}}$
$ΔT=ΔL\frac{dT}{dL}$
$ΔT=\frac{3}{100}L×\frac{1}{2}×\frac{2π}{\sqrt{Lg}}$
$ΔT=\frac{1.5}{100}×2π\sqrt{\frac{L}{g}}$
so $\frac{ΔT}{T}=\frac{1.5}{100}$
so $\frac{ΔT}{T}×100=1.5\%$