Question

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

• A. $3R$
• B. $4R$
• C. $5R$
• D. $15R$

Answer 1

The correct option is A $3R$

weight w = mg

m = mass (constent)

$\frac{w_1}{w_2}=\frac{g_1}{g_2};g_1=\frac{GM}{R^2}$

$9_2=\frac{GM}{(R+h{)}^2}$ ;h = height

So$\frac{w_1}{w_2}=\frac{(R+h{)}^2}{R^2}$

putting $\frac{w_1}{W_2}=16=\frac{(R+h{)}^2}{R^2}$

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