NEET Physics Question - Gravitation - Acceleration Due to Gravity and its variation

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

(2012 Pre)

$5R$

$15R$

$3R$

$4R$

Solution:

Weight at height $h$ is given by $W_h = \frac{W}{(1 + \frac{h}{R})^2}$. Given $W_h = \frac{W}{16}$, we equate: $\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2}$. Taking the square root gives $$ 1 + \frac{h}{R} = 4 \Rightarrow \frac{h}{R} = 3 \Rightarrow h = 3R$$.

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