Question

Q- ​How much distance Should a person travel inside the Earth from the surface of the Earth, So that its
weight becomes one ninth of its value on the surface of the Earth?
$$\begin{gathered} \mathrm{A})\mathrm{R} \\ \mathrm{B})\frac{\mathrm{R}}{9} \\ \mathrm{C})\frac{8\mathrm{R}}{9} \\ \mathrm{D})\frac{9\mathrm{R}}{8} \end{gathered}$$

Answer 1

Dear Student


$$\begin{gathered} Theaccelerationduetogravityinsideearth \\ g\text{'}=g\left(1-\frac{h}{R}\right) \\ Nowasgiven \\ Weightinsideearthsurfaceis1/9timesweightonearth \\ So \\ mg=9mg\text{'} \\ g=9g\text{'} \\ g\text{'}=9g\text{'}\left(1-\frac{h}{R}\right) \\ 1=9-9\frac{h}{R} \\ 9\frac{h}{R}=8 \\ h=\frac{8}{9}R \end{gathered}$$
Hence option c is right
Regards