Weight of a body of mass m decreases by $1\%$ when it is raised to height h above the Earth's surface. If the body is taken to a depth h in a mine, then its weight will:

Hint :- Gravitational acceleration g varies decreasingly with height as well as depth.

Step 1: Calculate height h

g = $\dfrac{GM}{r^{2}}$

Then g$_{h}$ = $\dfrac{GM}{(r + h)^{2}}$

$\dfrac{g}{g_{h}}$ = $\dfrac{100}{99}$ = ($\dfrac{r + h}{r}$)$^{2}$

$\dfrac{r + h}{r}$ = 1.005

r + h = 1.005r

h= 0.005r

Step 2: Calculate % variation in g with given depth

g$_{d}$ = g(1 - $\dfrac{d}{r}$)

d =h= 0.005r

g$_{d}$ = 1 - 0.005 = 0.995

Thus the weight will be decreased by 0.5%.