Answer is (A)
$g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$
When $g^{\prime}=\frac{g}{4}$ then.
$\begin{array}{l}
\frac{g}{4}=g \times\left(\frac{R}{R+h}\right)^{2} \Rightarrow \frac{1}{2}=\frac{R}{R+h} \\
\therefore 2 R=R+h \Rightarrow R=h
\end{array}$
The value of gravitational acceleration ‘ $\mathrm{g}$ ‘ at a height ‘h’ above the earth’s surface is $\frac{\mathrm{g}}{4}$ then $(\mathrm{R}=$ radius of earth)
A) $h=R$
B) $\mathrm{h}=\frac{\mathrm{R}}{2}$
C) $\mathrm{h}=\frac{\mathrm{R}}{3}$
D) $\mathrm{h}=\frac{\mathrm{R}}{4}$
A) $h=R$
B) $\mathrm{h}=\frac{\mathrm{R}}{2}$
C) $\mathrm{h}=\frac{\mathrm{R}}{3}$
D) $\mathrm{h}=\frac{\mathrm{R}}{4}$
The value of gravitational acceleration ‘ $\mathrm{g}$ ‘ at a height ‘h’ above the earth’s surface is $\frac{\mathrm{g}}{4}$ then $(\mathrm{R}=$ radius of earth)
A) $h=R$
B) $\mathrm{h}=\frac{\mathrm{R}}{2}$
C) $\mathrm{h}=\frac{\mathrm{R}}{3}$
D) $\mathrm{h}=\frac{\mathrm{R}}{4}$
A) $h=R$
B) $\mathrm{h}=\frac{\mathrm{R}}{2}$
C) $\mathrm{h}=\frac{\mathrm{R}}{3}$
D) $\mathrm{h}=\frac{\mathrm{R}}{4}$