question_answer

If R is the radius of Earth, the height at which the weight of a body becomes 1/4 its weight on the surface of the Earth is (a) \[2R\]                                             (b) \[R\] (c) \[\frac{R}{2}\]                                             (d) \[\frac{R}{4}\]

Answer:

Weight of a body at a height becomes \[\frac{1}{4}\]of its value on the surface of earth \[\Rightarrow \]g at the height becomes \[\frac{1}{4}\]of its value on surface of earth.

Surface	Earth
\[{{g}_{1}}=g\]	\[{{g}_{2}}=\frac{g}{4}\]
\[{{R}_{1}}=R\]	\[{{R}_{2}}=R+h\]

\[g=\frac{GM}{{{R}^{2}}}\] \[\Rightarrow \] \[g\propto \frac{1}{{{R}^{2}}}\]    (\[\because \]G, M are constant) \[\Rightarrow \]\[\frac{{{g}_{2}}}{{{g}_{1}}}=\frac{R_{1}^{2}}{R_{2}^{2}}\]\[\Rightarrow \]\[\frac{g}{4g}=\frac{{{(R)}^{2}}}{{{(R+h)}^{2}}}={{\left( \frac{R}{R+h} \right)}^{2}}\] \[\Rightarrow \]\[\frac{1}{2}=\frac{R}{R+h}\]\[\Rightarrow \]\[R+h=2R\]\[\Rightarrow \]\[h=R\]