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 $R_{E}=6400 \mathrm{~km}$ and $\mathrm{g}$ on earth $10 \mathrm{~m} / \mathrm{s}^{2}$ )
Solution
<p>$T \propto \sqrt {1/g}$</p>
<p>$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}$$</p>
<p>${4 \over 6} = {R \over {R + h}}$</p>
<p>$\Rightarrow h = R/2$</p>
<p>$= 3200$ km</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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