Question

The ratio of the weight of a body at a height of $\frac{R}{10}$ from the surface of the earth to that at a depth of $\frac{R}{10}$ is ($R$ is radius of earth)

• A. $4 : 5$
• B. $1 : 1$
• C. $9 : 8$
• D. $2 : 3$
• E. $8 : 9$

Answer 1

Answer: (E)

$\omega_{h}=\frac{m g}{\left(1+\frac{h}{R}\right)^{2}}$
Given $h=\frac{R}{10}$
$=\frac{100}{121} m g$
$\omega_{d}=m g\left(1-\frac{d}{R}\right)$
$=m g\left(1-\frac{1}{10}\right) $
$\left(\therefore d=\frac{R}{10}\right)$
$=\frac{9}{10} m g$
$\therefore \frac{\omega_{h}}{\omega_{d}}=\frac{100\, m g}{121} \times \frac{10}{9} m g$
$=\frac{1000}{121 \times 9} \approx \frac{8}{9}$