Question

If a simple pendulum with a time period of 2 seconds is made to oscillate at a height R equal to the Earth's radius, then its time period at this height will be

• A. 1/2 s
• B. 1/4 s
• C. 2 s
• D. 4 s

Answer 1

The correct option is D 4 s

$\begin{array}{c}\text{Given}\to \text{Time period of pendulum on the surface}\\ \text{of earth}=2\text{s.}\\ \text{The time period upto a Height}R\text{equal to}\\ \text{the surface Radius of earth is to be measured}\end{array}$

$\text{Formula used}[T=2\pi \cdot \sqrt{\frac{l}{g}}]$

$\begin{array}{c}\text{for the surface}\\ \begin{array}{cc}T=2S\Rightarrow & 2=2\pi \cdot \sqrt{\frac{l}{g}}\\ & \Rightarrow [l=\frac{g}{{\pi }^2}]\end{array}\end{array}$

$\begin{array}{c}\text{for a very Large Height the acceleration due to gravity}\\ \text{significantly changes}\to \end{array}$

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

$\begin{array}{c}\text{for}(h=R)\\ \text{we get}{g}^{\prime }=\frac{g}{4}\end{array}$

$\Rightarrow T^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}}$

$\begin{array}{cc}\Rightarrow T^{\prime }=2\pi & \sqrt{\frac{g}{{\pi }^2}\cdot \frac{4}{g}}\\ [T^{\prime }=4\sec ]& \end{array}$