Question

A simple pendulum has length $1m$. Find its time period at a height of $9R$ from Earth surface.(Consider $g=10m/s^2$)

• A. $\sqrt{10}\pi$
• B. $\sqrt{20}\pi$
• C. $\sqrt{40}\pi$
• D. $\sqrt{50}\pi$

Answer 1

The correct option is C $\sqrt{40}\pi$

$T=2\pi \sqrt{\frac{L}{g}}$

$T=2\pi \sqrt{\frac{1}{g^{\prime }}}$ ($l=1m$)

$g^{\prime }$ is acceleration due to gravity at height $\left(9R\right)$

$\frac{g^{\prime }}{g}={\left(\frac{R}{R+h}\right)}^2$

$g^{\prime }=g{\left(\frac{R}{R+9R}\right)}^2$
$g^{\prime }=g{\left(\frac{R}{10R}\right)}^2$

$g^{\prime }=\frac{g}{100}=\frac{10}{100}=\frac{1}{10}$

Now,

$T=2\pi \sqrt{\frac{1}{\frac{1}{10}}}$

$T=2\pi \sqrt{10}$

$T=\sqrt{40}\pi$