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At a certain depth $\mathrm{"d "}$ below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $\mathrm{3R}$ above earth surface. Where $\mathrm{R}$ is Radius of earth (Take $\mathrm{R=6400 km}$ ). The depth $\mathrm{d}$ is equal to
Option: 1
$\mathrm{4800km}$

Option: 2
$\mathrm{2560km}$

Option: 3
$\mathrm{640km}$

Option: 4
$\mathrm{5260km}$

Answers (1)

best_answer

$\mathrm{Given}$

$\mathrm{g\left(1-\frac{d}{R}\right)=4 \frac{g}{\left(1+\frac{h}{R}\right)^2} }$

$\mathrm{ 1-\frac{d}{R}=\frac{4}{(1+3)^2}=\frac{1}{4} }$

$\mathrm{ \frac{d}{R}=\frac{3}{4} }$

$\mathrm{ d=\frac{3 R}{4}=\frac{3}{4} \times 6400 }$

$\mathrm{ d=4800 \mathrm{~km}}$

Posted by

Ajit Kumar Dubey

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