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

The correct option is A increase by $1$%

$T=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{l(R+h{)}^2}{GM}}$
$T\propto (R+h)$

$\frac{T^{\prime }}{T}=\frac{R+h}{R}=1+\frac{h}{R}=1+\frac{1}{100}\therefore \frac{T^{\prime }}{T}-1=\frac{1}{100}$

$\therefore \left(\frac{T^{\prime }-T}{T}\right)\times 100=1$

$\therefore$ Time period will increase by 1%