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: Distance from the centre:
Height $2R$ above the surface means $3R$ from the centre.

Step 2: Period:
$T_2 = 2\pi\sqrt{\frac{l}{g/9}} = 3 \times 2\pi\sqrt{\frac lg} = 3T_1$.

Final Answer:
The ratio is $1:3$, option (B). \[ \boxed{1:3} \]