QJEEM24Phdb34: JEE Main 2024 Physics - Class 11 JEEM24Phdb34 Solution: If is the radius of the earth and the acceleration due to…

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.