Question

$T$ is the time period of simple pendulum on the earth's surface Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface Then, the value of $x$ will be:

• A. $\frac{1}{4}$
• B. 4
• C. $\frac{1}{2}$
• D. 2

Answer 1

The Correct Option is D

At surface of earth time period
$T=2\pi \sqrt{\frac{\ell }{g}}$
At height $h=R$
$g^{\prime }=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}=\frac{g}{4}$
$\therefore xT=2\pi \sqrt{\frac{\ell }{(g/4)}}$
$\Rightarrow xT=2\times 2\pi \sqrt{\frac{\ell }{g}}$
$\Rightarrow xT=2T\Rightarrow x=2$

Answer 2

1. The time period of a pendulum is given by: \[ T \propto \frac{1}{\sqrt{g}}. \]
2. At height \(R\) above the Earth's surface, the acceleration due to gravity is: \[ g' = \frac{g}{(1+h/R)^2}, \quad h = R \implies g' = \frac{g}{(1+1)^2} = \frac{g}{4}. \]
3. The new time period \(T'\) becomes: \[ T' = T \times \sqrt{\frac{g}{g'}} = T \times \sqrt{\frac{g}{g/4}} = T \times \sqrt{4} = 2T. \]
Thus, \(x = 2\). The time period increases as the effective gravity decreases with height. At a height equal to Earth's radius, the time period doubles.