A simple pendulum doing small oscillations at a place R height above earth surface has time period of $$T_1 = 4 \text{ s}$$. $$T_2$$ would be its time period if it is brought to a point which is at a height $$2R$$ from earth surface. Choose the correct relation $$[R = \text{radius of earth}]$$ :

A simple pendulum at height $$R$$ above Earth's surface has time period $$T_1 = 4$$ s. We need $$T_2$$ at height $$2R$$.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{l}{g'}}$$ where $$g'$$ is the effective gravitational acceleration at height $$h$$:

$$ g' = \frac{g}{(1 + h/R)^2} $$

At height $$h = R$$:

$$ g_1 = \frac{g}{(1+1)^2} = \frac{g}{4} $$ $$ T_1 = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}} $$

At height $$h = 2R$$:

$$ g_2 = \frac{g}{(1+2)^2} = \frac{g}{9} $$ $$ T_2 = 2\pi\sqrt{\frac{l}{g/9}} = 2\pi\sqrt{\frac{9l}{g}} $$

Dividing to find the relation between $$T_2$$ and $$T_1$$:

$$ \frac{T_2}{T_1} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} $$

Therefore,

$$ T_2 = \frac{3}{2}T_1 $$

Multiplying both sides by 2:

$$ 2T_2 = 3T_1 $$

This corresponds to $$3T_1 = 2T_2$$.

The correct answer is Option (4): $$3T_1 = 2T_2$$.