Gravitation - Result Question 16
16. The height at which the weight of a body becomes $1 / 16$ th, its weight on the surface of earth (radius $R$ ), is :
[2012]
(a) $5 R$
(b) $15 R$
(c) $3 R$
(d) $4 R$
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Answer:
Correct Answer: 16. (c)
Solution:
(c) $\frac{G M_m}{(R+h)^{2}}=\frac{1 G M_m}{16 R^{2}}$
$\Rightarrow \frac{1}{(R+h)^{2}}=\frac{1}{16 R^{2}}$
$\Rightarrow \frac{R}{(R+h)}=\frac{1}{4}$
$\Rightarrow \quad \frac{R+h}{R}=4$
$\Rightarrow h=3 R$
