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Q.

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

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a

4R

b

15R

c

5R

d

3R

answer is A.

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Detailed Solution

${mg}_{h} = \frac{mg}{{(1 + \frac{h}{R})}^{2}} \Rightarrow \frac{mg}{16} = \frac{mg}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 3 R$

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The height at which the weight of a body become 116th, its weight on the surface of earth (radius R) is :