Discussion Forum



1)

The depth 'd' at which the value of acceleration  due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is 

(R= radius of earth)


A) $d=R\left(\frac{n}{n-1}\right)$

B) $d=R\left(\frac{n-1}{2n}\right)$

C) $d=R\left(\frac{n-1}{n}\right)$

D) $d=R^2\left(\frac{n-1}{n}\right)$

Answer:

Option C

Explanation:

 Acceleration due to gravity at depth d is given as, 

     g'= $g\left(1-\frac{d}{R}\right)$   ...........(i)

 Given,  $g'=\frac{g}{n}$

 Substituting the value of 'g' in Eq.(i) , we  get

$\frac{g}{n}=g\left(1-\frac{d}{R}\right)$

 $\Rightarrow $    $\frac{1}{n}=1-\frac{d}{R}$

$\therefore$  $ \frac{d}{R}=1-\frac{1}{n}=\frac{n-1}{n}$

 $d=R\left(\frac{n-1}{n}\right)$