Question

The distance from the centre of earth, where the weight of a body is zero and one-fourth of that of the weight of the body on the surface of earth are (assume R is the radius of the earth).
A.) $\left( {0,\dfrac{R}{4}} \right)$
B.) $\left( {0,\dfrac{3}{4}R} \right)$
C.) $\left( {\dfrac{R}{4},0} \right)$
D.) $\left( {\dfrac{3}{4}R,0} \right)$

Answer 1

Hint: As we travel through the depth of the earth then the value of G begins to decrease and even the gravitational acceleration will enable us to solve this problem.

Step By Step Answer:

Formula used: -

${g_a}$=$g\left( {1 - \dfrac{d}{R}} \right)$

Gravitational acceleration is the free fall acceleration without any drag of an object in the vacuum. It is the steady gain in speed resulting solely from the force of gravitational attraction.

We know gravitational acceleration at depth $d$ is given by

${g_a}$=$g\left( {1 - \dfrac{d}{R}} \right)$
Depth at which weight is $\left( {\dfrac{{mg}}{4}} \right)$
$m{g_{depth}}$=$\dfrac{{mg}}{4}$=$mg\left( {1 - \dfrac{d}{R}} \right)$
$\dfrac{d}{R} = \left( {1 - \dfrac{1}{4}} \right) = \dfrac{3}{4}R$
Finally, the distance will come $\left( {R - \dfrac{3}{4}R} \right) = \dfrac{R}{4}$

Therefore, distance of this point from centre = $\dfrac{R}{4}$

We also know that the weight of the body at the centre of earth is 0.

Hence our answer is option A i.e., $\left( {0,\dfrac{R}{4}} \right)$ is the distance from the centre of earth, where the weight of a body is zero and one-fourth of that of the weight of the body on the surface of earth.

Note: For solving these types of questions we must remember the important definition, properties and formulae related to gravitational acceleration at depth $d$. It will be very helpful for us if we remember these properties.