Question

A simple pendulum is taken to $ \, 64 \, km$ above the earth's surface. It's time period will :

• A. Increase by $1\%$
• B. Decrease by $1\%$
• C. Increase by $2\%$
• D. Decrease by $2\%$

Answer 1

Answer: (A)

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{l \left(R + h\right)^{2}}{G M}}$
$T \propto \left(R + h\right)$
or $\frac{T^{'}}{T}=\frac{R + h}{R}=1+\frac{h}{R}=1+\frac{1}{1 0 0}$
$\therefore \frac{T^{'}}{T}-1=\frac{1}{1 0 0}$
$\therefore \left(\frac{T^{'} - T}{T}\right)\times 100=1$
$\therefore $ : Time period will increase by 1%