Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is - (consider radius of earth R_E=6400 ~km and g on earth 10 ~m / s^2 )

Simple Harmonic Motion PYQ — JEE Main 2022 · 28 Jul · Shift 2 · Q45 with Solution

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Correct answer: C

For a pendulum of length $l$ , $T = 2\pi \sqrt{\frac{l}{g}}$

Since the two clocks are identical, their lengths are the same. Hence, $T \propto \frac{1}{\sqrt{g}}$

So, $\frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}}$

Given:

Thus, $\frac{6}{4} = \sqrt{\frac{10}{g_2}}$

$\frac{3}{2} = \sqrt{\frac{10}{g_2}}$

Squaring both sides, $\frac{9}{4} = \frac{10}{g_2}$

So, $g_2 = 10 \cdot \frac{4}{9} = \frac{40}{9}\,\text{m/s}^2$

Acceleration due to gravity at height $h$ above earth is $g_h = g\left(\frac{R_E}{R_E+h}\right)^2$

So, $\frac{40}{9} = 10\left(\frac{R_E}{R_E+h}\right)^2$

Divide by $10$ : $\frac{4}{9} = \left(\frac{R_E}{R_E+h}\right)^2$

Taking square root, $\frac{2}{3} = \frac{R_E}{R_E+h}$

Hence, $R_E + h = \frac{3}{2}R_E$

Therefore, $h = \frac{1}{2}R_E$

$h = \frac{1}{2} \times 6400 = 3200\,\text{km}$

$\boxed{h = 3200\,\text{km}}$

So the correct option is C.