JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 6)
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 $$10 \mathrm{~m} / \mathrm{s}^{2}$$ )
1200 km
1600 km
3200 km
4800 km
Explanation
$$T \propto \sqrt {1/g} $$
$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}$$
$${4 \over 6} = {R \over {R + h}}$$
$$ \Rightarrow h = R/2$$
$$ = 3200$$ km
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