The weight of a body at earth's surface is W . At a depth halfway to the centre of the earth, it will be (assuming uniform density in the earth) -

Question

The weight of a body at earth's surface is W . At a depth halfway to the centre of the earth, it will be (assuming uniform density in the earth) - (A) W (B) (W/2) (C) (W/4) (D) (W/8)

Tardigrade Admin · Accepted Answer

Due to depth gâ²=g 1 - (d/Re) ∴ d=Re / 2 so gâ²=g 1 - (Re / 2/Re) =g / 2 Now gâ² is half so weight is also half.

Q. The weight of a body at earth's surface is $W$ . At a depth halfway to the centre of the earth, it will be (assuming uniform density in the earth) -

$W$

$\frac{W}{2}$

$\frac{W}{4}$

$\frac{W}{8}$

Due to depth $g^{'} = g {1 - \frac{d}{R _{e}}} ∴ d = R_{e} /2$
so $g^{'} = g {1 - \frac{R _{e} /2}{R _{e}}} = g /2$
Now $g^{'}$ is half so weight is also half.

Q. The weight of a body at earth's surface is W . At a depth halfway to the centre of the earth, it will be (assuming uniform density in the earth) -

W

2W​

4W​

8W​

Due to depth g′=g{1−Re​d​}∴d=Re​/2 so g′=g{1−Re​Re​/2​}=g/2 Now g′ is half so weight is also half.

Q. The weight of a body at earth's surface is $W$ . At a depth halfway to the centre of the earth, it will be (assuming uniform density in the earth) -

$W$

$\frac{W}{2}$

$\frac{W}{4}$

$\frac{W}{8}$

Due to depth $g^{'} = g {1 - \frac{d}{R _{e}}} ∴ d = R_{e} /2$
so $g^{'} = g {1 - \frac{R _{e} /2}{R _{e}}} = g /2$
Now $g^{'}$ is half so weight is also half.