Question

Time period of a simple pendulum on earth's surface is ' $T$ '. It time period becomes ' $xT$ ' when taken to a height ' $2R$ ' above earth's surface. The value of $x$ will be ($R = \text{radius of earth}$)

• A. 2
• B. 4
• C. 1
• D. 3

Hint:

If gravity becomes \(g/n^2\), the pendulum time period becomes \(n\) times.

Answer 1

The Correct Option is D

Concept:
Time period of a pendulum is: \[ T=2\pi\sqrt{\frac{l}{g}} \] So, \[ T\propto \frac{1}{\sqrt{g}} \] Also, acceleration due to gravity at height \(h\) is: \[ g_h=g\left(\frac{R}{R+h}\right)^2 \] ip

Step 1: Find gravity at height \(2R\).
\[ h=2R \] So, \[ g_h=g\left(\frac{R}{R+2R}\right)^2 =g\left(\frac{R}{3R}\right)^2 =\frac{g}{9} \] ip

Step 2: Find the new time period.
Since \[ T_h\propto \frac{1}{\sqrt{g_h}} \] \[ T_h=T\sqrt{\frac{g}{g_h}} =T\sqrt{\frac{g}{g/9}} =T\sqrt{9}=3T \] Thus, \[ x=3 \] ip Hence, the correct answer is:
\[ \boxed{(D)\ 3} \]