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: Use the ratio of lengths:
$\sqrt{\frac{l_s}{l_h}} = \frac{R + h}{R} = 1 + \frac hR$.

Step 2: Isolate R:
$\frac hR = \sqrt{\frac{l_s}{l_h}} - 1 = \frac{\sqrt{l_s} - \sqrt{l_h}}{\sqrt{l_h}}$, so $R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$.

Final Answer:
$R = \frac{h\sqrt{l_h}}{\sqrt{l_s} - \sqrt{l_h}}$, option (A). \[ \boxed{\frac{h\sqrt{l_h}}{\sqrt{l_s}-\sqrt{l_h}}} \]