If $\mathrm{R}$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $\mathrm{h}=2 R$ from the surface of earth will be, :

<p>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 $T = 2 \text{ seconds}$. </p> <p>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: <p>$g_h = \frac{g}{{(1 + \frac{2R}{R})}^2}$</p> <p>$g_h = \frac{g}{{(1 + 2)}^2}$</p> <p>$g_h = \frac{g}{3^2} = \frac{g}{9}$ </p></p> <p>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: $g_h = \frac{\pi^2}{9} \text{ m/s}^2$ </p> <p>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 $L_h$: <p>$2 = 2\pi\sqrt{\frac{L_h}{g_h}}$ <br/><br/>$1 = \pi\sqrt{\frac{L_h}{g_h}}$ <br/><br/>$\frac{1}{\pi} = \sqrt{\frac{L_h}{g_h}}$ <br/><br/> Squaring both sides, we get:</p> <p>$\frac{1}{\pi^2} = \frac{L_h}{g_h}$ <br/><br/> Multiplying both sides by $g_h$ gives us the length $L_h$:</p> <p>$L_h = \frac{g_h}{\pi^2}$ </p></p> <p>Substituting $g_h$ into the equation yields: <p>$L_h = \frac{\pi^2}{9\pi^2}$</p> <p>$L_h = \frac{1}{9} \text{ m}$ </p></p> <p>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.</p>