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