1. Time period of a seconds pendulum

A second's pendulum has time period $T=2\,\text{s}$

For a simple pendulum, $T=2\pi\sqrt{\frac{L}{g'}}$ where $g'$ is the acceleration due to gravity at the place where the pendulum is located.

2. Gravity at height $h=2R$ above earth's surface

Distance from the center of the earth becomes $r=R+h=R+2R=3R$

Acceleration due to gravity at this height is $g'=g\left(\frac{R}{R+h}\right)^2=g\left(\frac{R}{3R}\right)^2=\frac{g}{9}$

Given $g=\pi^2\,\text{m/s}^2$ so $g'=\frac{\pi^2}{9}\,\text{m/s}^2$

3. Use pendulum formula

Using $T=2\pi\sqrt{\frac{L}{g'}}$ with $T=2$ s:

$2=2\pi\sqrt{\frac{L}{g'}}$

Divide by 2: $1=\pi\sqrt{\frac{L}{g'}}$

So, $\sqrt{\frac{L}{g'}}=\frac{1}{\pi}$

Squaring both sides, $\frac{L}{g'}=\frac{1}{\pi^2}$

Hence, $L=\frac{g'}{\pi^2}$

Substitute $g'=\dfrac{\pi^2}{9}$: $L=\frac{\pi^2/9}{\pi^2}=\frac{1}{9}\,\text{m}$

4. Option check

• A: $\frac{1}{9}$ m ✅
• B: $\frac{8}{9}$ m ❌
• C: $\frac{2}{9}$ m ❌
• D: $\frac{4}{9}$ m ❌

Therefore, the correct answer is A.