Question

The lengths of seconds pendulums on the surface of the earth and at an altitude '\(h\)' from the surface of the earth are \(l_s\) and \(l_h\) respectively. The radius of the earth is

• A. \(\frac{h\sqrt{l_h}}{\sqrt{l_s}-\sqrt{l_h}}\)
• B. \(\frac{h\sqrt{l_h}}{\sqrt{l_h}-\sqrt{l_s}}\)
• C. \(\frac{\sqrt{l_h}}{h(\sqrt{l_s}-\sqrt{l_h})}\)
• D. \(\frac{\sqrt{l_s}}{h(\sqrt{l_h}-\sqrt{l_s})}\)

Hint:

A seconds pendulum has period 2 s, so its length is proportional to g.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The period is \(T = 2\pi\sqrt{\frac{l}{g}}\). For a seconds pendulum \(T = 2\) s at both places, so \(\frac{l}{g}\) is the same, which means \(l \propto g\).

Step 2: Key Formula or Approach:
At height \(h\), \(g_h = g\left(\frac{R}{R + h}\right)^2\). So \(\frac{l_h}{l_s} = \frac{g_h}{g_s} = \frac{R^2}{(R + h)^2}\).

Step 3: Detailed Explanation:
Take the square root: \(\frac{\sqrt{l_h}}{\sqrt{l_s}} = \frac{R}{R + h}\).
Cross multiply: \(R\sqrt{l_s} = (R + h)\sqrt{l_h}\).
So \(R(\sqrt{l_s} - \sqrt{l_h}) = h\sqrt{l_h}\).
\[ R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}} \]
Since \(l_s > l_h\), the denominator is positive, which fits option A. Option B has a negative denominator.

Final Answer:
The radius of the earth is \(\frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}\), option (A). \[ \boxed{\frac{h\sqrt{l_h}}{\sqrt{l_s}-\sqrt{l_h}}} \]