Assuming the earth to be a sphere of uniform density how much could a body weigh at a depth one third of radius of earth. from the surface of earth, when the body weight 300 N on the surface of earth.?

200

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300

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400

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500

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Solution

The correct option is A 200
In depth, we know $g^{'} = g (1 - \frac{d}{R})$ .

Here $d = \frac{R}{3}$

And, weight at surface $= m g = 300 N$

Weight at depth $W^{'} = m g^{'} = m g (1 - \frac{1}{3})$

$⟹ W^{'} = 200 N$

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Assuming the earth to be a sphere of uniform density how much could a body weigh at a depth one third of radius of earth. from the surface of earth, when the body weight 300 N on the surface of earth.?

The correct option is A 200In depth, we know g′=g(1−dR).Here d=R3 And, weight at surface=mg=300NWeight at depth W′=mg′=mg(1−13)⟹W′=200N

The correct option is A 200
In depth, we know $g^{'} = g (1 - \frac{d}{R})$ .

Here $d = \frac{R}{3}$

And, weight at surface $= m g = 300 N$

Weight at depth $W^{'} = m g^{'} = m g (1 - \frac{1}{3})$

$⟹ W^{'} = 200 N$