Question

A simple pendulum is taken to $64km$ 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)

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{l{\left(R+h\right)}^2}{GM}}$
$T\propto \left(R+h\right)$
or $\frac{T^{{}^{\prime }}}{T}=\frac{R+h}{R}=1+\frac{h}{R}=1+\frac{1}{100}$
$\therefore \frac{T^{{}^{\prime }}}{T}-1=\frac{1}{100}$
$\therefore \left(\frac{T^{{}^{\prime }}-T}{T}\right)\times 100=1$
$\therefore$ : Time period will increase by 1%