The value of g (acceleration due to gravity) at earth's surface is 10 ms -2. Its value in ms -2 at the centre of the earth which is assumed to be a sphere of radius R metre and uniform mass density is

Question

The value of g (acceleration due to gravity) at earth's surface is 10 ms -2. Its value in ms -2 at the centre of the earth which is assumed to be a sphere of radius R metre and uniform mass density is (A) 5 (B) (10/R) (C) (10/2 R) (D) zero

Tardigrade Admin · Accepted Answer

Inside the earth g'=g[1-(h/R)] At centre, g'=g[1-(R/R)]=0

Q. The value of $g$ (acceleration due to gravity) at earth's surface is $10 m s^{- 2}$ . Its value in $m s^{- 2}$ at the centre of the earth which is assumed to be a sphere of radius $R$ metre and uniform mass density is

5

$\frac{10}{R}$

$\frac{10}{2 R}$

zero

Inside the earth $g^{'} = g [1 - \frac{h}{R}]$
At centre, $g^{'} = g [1 - \frac{R}{R}] = 0$

Q. The value of g (acceleration due to gravity) at earth's surface is 10ms−2. Its value in ms−2 at the centre of the earth which is assumed to be a sphere of radius R metre and uniform mass density is

5

R10​

2R10​

zero

Inside the earth g′=g[1−Rh​] At centre, g′=g[1−RR​]=0

Q. The value of $g$ (acceleration due to gravity) at earth's surface is $10 m s^{- 2}$ . Its value in $m s^{- 2}$ at the centre of the earth which is assumed to be a sphere of radius $R$ metre and uniform mass density is

$\frac{10}{R}$

$\frac{10}{2 R}$

Inside the earth $g^{'} = g [1 - \frac{h}{R}]$
At centre, $g^{'} = g [1 - \frac{R}{R}] = 0$