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

Step 1: Understanding the Concept:
The time period of a simple pendulum is inversely proportional to the square root of the local acceleration due to gravity ($g$). Acceleration due to gravity decreases with altitude from the Earth's surface.
Step 2: Key Formula or Approach:
1. Time Period: $T = 2\pi \sqrt{\frac{l}{g}} \implies T \propto \frac{1}{\sqrt{g}}$
2. Variation of $g$: $g' = g \left( \frac{R}{R+h} \right)^2$
Step 3: Detailed Explanation:
1. Calculate the gravity at height $h = 2R$:
\[ g' = g \left( \frac{R}{R+2R} \right)^2 = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9} \]
2. Calculate the new time period $T'$:
Since $T \propto \frac{1}{\sqrt{g}}$, we have $\frac{T'}{T} = \sqrt{\frac{g}{g'}}$.
\[ \frac{T'}{T} = \sqrt{\frac{g}{g/9}} = \sqrt{9} = 3 \]
\[ T' = 3T \]
Comparing this with $T' = xT$, we get $x = 3$.
Step 4: Final Answer:
The value of $x$ is 3.