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Q.

The height at which the weight of a body becomes $\frac{1}{16} th$ its weight on the surface of earth (radius R) is

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a

4R

b

15R

c

5R

d

3R

answer is D.

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Detailed Solution

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$g = \frac{G M}{R^{2}} g^{1} = \frac{G M}{{(R + h)}^{2}} = \frac{g R^{2}}{R^{2} {(1 + \frac{h}{R})}^{2}}$

$\frac{g}{16} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow (1 + \frac{h}{R}) = 4 \Rightarrow h = 3 R$

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