A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1=4 \mathrm{~s}$. $\mathrm{T}_2$ would be it's time period if it is brought to a point which is at a height $2 \mathrm{R}$ from earth surface. Choose the correct relation [$\mathrm{R}=$ radius of earth] :

<p>The time period of a simple pendulum is given by the formula:</p> <p>$T = 2\pi \sqrt{\frac{l}{g}}$</p> <p>where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.</p> <p>The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:</p> <p>$g' = g \left(\frac{R}{R + h}\right)^2$</p> <p>where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.</p> <p>Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.</p> <p>For the initial case at height $R$:</p> <p>$$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$$</p> <p>For the new case at height $2R$:</p> <p>$$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$$</p> <p>The time period is proportional to the square root of the inverse of $g$, so:</p> <p>$$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$</p> <p>Therefore:</p> <p>$T_1 = \frac{2}{3}T_2$</p> <p>Rearranging this equation:</p> <p>$3T_1 = 2T_2$</p> <p>This corresponds to Option A.</p>