Question

If ${}^{\prime }{R}^{\prime }$ is the radius of the earth, then the height at which the weight of a body becomes ${\frac{1}{4}}^{th}$ of its weight on the surface of the earth is:

• A. $2R$
• B. $R$
• C. $\frac{3R}{8}$
• D. $\frac{R}{4}$

Answer 1

Answer: (C)

we know that acceleration due to gravity varies height h and the surface of the earth

$g^{\prime }=g(1-\frac{2h}{R})$

hence the weight of the body

$m{g}^{\prime }=mg(1-\frac{2h}{R})$

$w^{\prime }=w(1-\frac{2h}{R})$

$w^{\prime }=w(1-\frac{2h}{R})$

$\frac{1}{4}=(1-\frac{2h}{R})$

$\frac{2h}{R}=1-\frac{1}{4}=\frac{3}{4}$

$h=\frac{3R}{8}$