Question

A simple pendulum has a time period $T_1$ when on the Earth's surface, and $T_2$ when taken to a height $R$ above the Earth's surface, where ${}^{\prime }{R}^{\prime }$ is the radius of the Earth. The value of $\frac{T_2}{T_1}$ is

• A. $1$
• B. $\sqrt{2}$
• C. $0.5$
• D. $2$

Answer 1

The correct option is D $2$

$\begin{array}{c}\text{Given- Time period on the surface of earth}=T_1\\ \text{Time period on a height}R\text{above earth surface}=T_2\end{array}$

$\text{Time period of a pendulum formula}[T=2\pi \cdot \sqrt{\frac{l}{g}}]$

$\text{we have}\to T_1=2\pi \cdot \sqrt{\frac{l}{g}}$

$\begin{array}{c}\text{At a Height above the surface of earth we have}\\ g^{\prime }=g\cdot \left[\frac{R^2}{(R+h{)}^2}\right]\begin{array}{c}h=\text{height}\\ R=\text{Radius of earth}\end{array}\end{array}$

$\begin{array}{cc}\text{for this case}(h=R)& \\ \text{we have}{g}^{\prime }=\frac{g}{4}& \end{array}$

$\begin{array}{cc}\text{Hence}{T}_2& =2\pi \cdot \sqrt{\frac{4l}{g}}=2\cdot \{2\pi \cdot \sqrt{\frac{l}{g}}\}\\ & T_2=2T_1\\ \Rightarrow \frac{T_2}{T_1}=& 2\end{array}$