Question

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

• A. $\frac{2}{9} \, \text{m}$
• B. $\frac{1}{9} \, \text{m}$
• C. $\frac{4}{9} \, \text{m}$
• D. $\frac{8}{9} \, \text{m}$

Answer 1

Answer: (B)

$\frac{1}{9} \, \text{m}$

Answer 2

The time period $T$ of a simple pendulum is related to the acceleration due to gravity $g$ by the formula:

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

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Step 1: Gravitational acceleration at height $h$

At a height $h = 2R$ from the surface of the Earth, the acceleration due to gravity $g'$ is given by:

$g' = g \left( \frac{R}{R + h} \right)^2$

Substituting $h = 2R$:

$g' = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9}$

Therefore, the new value of gravitational acceleration at height $h = 2R$ is $\frac{g}{9}$.

Step 2: Time period of the pendulum

The time period $T$ of a pendulum is related to the length $L$ and the acceleration due to gravity $g$ by:

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

At height $h = 2R$, the new time period $T'$ will be:

$T' = 2\pi \sqrt{\frac{L'}{g'}}$

Since the time period remains 2 seconds, we equate the time periods:

$2 = 2\pi \sqrt{\frac{L'}{g/9}}$

Squaring both sides:

$1 = \pi^2 \frac{L'}{g}$

Solving for $L'$:

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

Substitute $g = \pi^2 \, \text{m/s}^2$ into the equation:

$L' = \frac{\pi^2}{9\pi^2} = \frac{1}{9} \, \text{m}$

Thus, the length of the second's pendulum at a height $h = 2R$ is $\frac{1}{9} \, \text{m}$.