Question

The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. The value of x is

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

Hint:

The period varies inversely with the square root of g, and g falls to a quarter at distance 2R.

Answer 1

The Correct Option is C

Step 1: g at height R
\(g' = g\left(\frac{R}{R+h}\right)^2 = g\left(\frac R{2R}\right)^2 = \frac g4\).

Step 2: Period
\(T = 2\pi\sqrt{\frac l g}\), so \(T' = 2\pi\sqrt{\frac{l}{g/4}} = 2T\).

Step 3: Result
\(x = 2\). Option (C).

Final Answer:
x = 2. \[ \boxed{\text{(C)}\ 2} \]