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

The Correct Option is B

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 \(\frac{1}{9} \, \text{m}\).

Answer 2

We are given: Earth's radius \(R\), gravity at surface \(g = \pi^2 \ \text{m/s}^2\). We must find the length of a seconds pendulum at height \(h = 2R\) above the surface. A seconds pendulum has a period \(T = 2 \ \text{s}\).

Concept Used:

The period of a simple pendulum is \(T = 2\pi \sqrt{\frac{L}{g'}}\), where \(g'\) is the acceleration due to gravity at that location. For a seconds pendulum, \(T = 2 \ \text{s}\).

Gravity varies with height as \(g' = g \left( \frac{R}{R + h} \right)^2\).

Step-by-Step Solution:

Step 1: Find length \(L_0\) at Earth's surface.

At surface: \(T = 2\pi \sqrt{\frac{L_0}{g}} = 2 \ \text{s}\). Given \(g = \pi^2\):

\[ 2\pi \sqrt{\frac{L_0}{\pi^2}} = 2 \] \[ 2\pi \cdot \frac{\sqrt{L_0}}{\pi} = 2 \] \[ 2\sqrt{L_0} = 2 \] \[ \sqrt{L_0} = 1 \] \[ L_0 = 1 \ \text{m} \]

Step 2: Find \(g'\) at height \(h = 2R\).

At height \(h\), \(g' = g \left( \frac{R}{R + h} \right)^2 = g \left( \frac{R}{R + 2R} \right)^2 = g \left( \frac{R}{3R} \right)^2 = \frac{g}{9}\).

Given \(g = \pi^2\), so \(g' = \frac{\pi^2}{9}\).

Step 3: Find length \(L\) at height \(h\) for seconds pendulum.

At height \(h\), \(T = 2\pi \sqrt{\frac{L}{g'}} = 2\):

\[ 2\pi \sqrt{\frac{L}{\pi^2/9}} = 2 \] \[ 2\pi \cdot \frac{\sqrt{L}}{\pi/3} = 2 \] \[ 2\pi \cdot \frac{3\sqrt{L}}{\pi} = 2 \] \[ 6\sqrt{L} = 2 \] \[ \sqrt{L} = \frac{1}{3} \] \[ L = \frac{1}{9} \ \text{m} \]

Hence, the length of the second's pendulum at height \(2R\) is \(\frac{1}{9} \ \text{m}\).