Question

Time period of a simple pendulum is \(T_1\) when on the earth's surface and \(T_2\) when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio \(T_1:T_2\) is

• A. \(1:2\)
• B. \(1:3\)
• C. \(1:4\)
• D. \(1:5\)

Hint:

Period is inversely proportional to the square root of g, and g falls with the square of distance from the centre.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
\(T = 2\pi\sqrt{\frac lg}\), so \(T \propto \frac{1}{\sqrt g}\). The value of \(g\) changes with distance \(r\) from the centre of the earth as \(g \propto \frac{1}{r^2}\).

Step 2: Key Formula or Approach:
At the surface, \(r = R\). At a height \(2R\) above the surface, \(r = 3R\).

Step 3: Detailed Explanation:
\(g_2 = g_1\left(\frac{R}{3R}\right)^2 = \frac{g_1}{9}\).
\[ \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac19} = \frac13 \]
So \(T_1 : T_2 = 1 : 3\). The ratio \(1:2\) would come from taking the distance as \(2R\) instead of \(3R\), forgetting that the height is measured from the surface.

Final Answer:
\(T_1 : T_2 = 1 : 3\), option (B). \[ \boxed{1:3} \]