Question

a body weighs 300 N on the surface of the earth surface what is its weight at a depth of 1/3 of the radius of earth

Answer 1

Answer:-

200 N

Explanation:-

Let the depth be 'd' km from the surface of earth and let the radius of earth be 'r' km.

Hence, according to the question :-

=> d = 1/3 × r

=> d = r/3 ---(1)

Now, we know that if value of acceleration due to gravity is g and at a depth d from the earth surface is g', then :-

g' = g(1 - d/r)

Weight at a depth of r/3 km :-

=> mg' = mg(1 - d/r)

=> W' = W(1 - d/r)

Here :- {W = 300 N, r = r km, d = r/3 km}

=> W' = 300(1 - r/3/r) [∵ d = r/3]

=> W' = 300(1 - 1/3)

=> W' = 300(2/3)

=> W' = 100(2)

=> W' = 200 N

Thus, weight of the body at a depth of 1/3 of the radius of the earth is 200N .