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7.Gravitation
normal
At what height above the earth's surface is the value of $'g'$ is same as in a $200\, km$ deep mine ........ $km$
A
$50$
B
$100$
C
$150$
D
$400$
Solution
$\mathrm{g}\left(1-\frac{2 \mathrm{h}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{\mathrm{h}^{\prime}}{\mathrm{R}}\right)$
Above surface $\quad$ Below surface
$\Rightarrow \frac{2 \mathrm{h}}{\mathrm{R}}=\frac{\mathrm{h}^{\prime}}{\mathrm{R}} \Rightarrow \mathrm{h}=\frac{\mathrm{h}^{\prime}}{2}=\frac{200}{2}=100 \mathrm{km}$
Standard 11
Physics