JEE Main 2024 — Gravitation Question with Solution

To find the length of the second's pendulum at a height $h=2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $$T=2\pi \sqrt{\frac{L}{g}}$$ Since we are talking about a second's pendulum, the period, $T$, is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T=2\text{ seconds}$.

Now let's find the gravitational acceleration at height $h=2R$ where $R$ is the radius of the earth. The general formula for gravitational acceleration at a height $h$ above the surface is: $$g_h=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$$ Plugging $h=2R$ into the formula, we get:

$$g_h=\frac{g}{{(1+\frac{2R}{R})}^2}$$

$$g_h=\frac{g}{{(1+2)}^2}$$

$$g_h=\frac{g}{3^2}=\frac{g}{9}$$

So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth. Given that $g={\pi }^2{\text{ m/s}}^2$, we get: $$g_h=\frac{{\pi }^2}{9}{\text{ m/s}}^2$$

Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$:

$$2=2\pi \sqrt{\frac{L_h}{g_h}}$$

$$1=\pi \sqrt{\frac{L_h}{g_h}}$$

$$\frac{1}{\pi }=\sqrt{\frac{L_h}{g_h}}$$

Squaring both sides, we get:

$$\frac{1}{{\pi }^2}=\frac{L_h}{g_h}$$

Multiplying both sides by $g_h$ gives us the length $L_h$:

$$L_h=\frac{g_h}{{\pi }^2}$$

Substituting $g_h$ into the equation yields:

$$L_h=\frac{{\pi }^2}{9{\pi }^2}$$

$$L_h=\frac{1}{9}\text{ m}$$

Therefore, the length of the second's pendulum at a height $h=2R$ from the surface of the Earth is $\frac{1}{9}$ meters. The correct answer is Option A.