A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

The correct option is 27 N.

Let's understand the step-by-step physics and mathematical derivations to arrive at this answer.

1. Understanding the Gravitational Force at the Earth's Surface:
The gravitational force experienced by a body of mass m on the surface of the earth is its weight W. This force is given by Newton's law of universal gravitation:

$F=W=\frac{G·M·m}{R^2}$

where:
- $G$ is the universal gravitational constant,
- $M$ is the mass of the earth,
- $m$ is the mass of the body, and
- $R$ is the radius of the earth.
We are given that the weight on the surface is 48 N:

$\frac{G·M·m}{R^2}=48\text{N}$

2. Finding the Gravitational Force at a Height h:
When the body is at a height h from the surface of the earth, its distance from the center of the earth becomes:

$r=R+h$

The gravitational force F' at this height is:

$F^{\text{'}}=\frac{G·M·m}{{\left(R+h\right)}^2}$

3. Substituting the Given Value of Height:
The height is given as one-third of the radius of the earth:

$h=\frac{R}{3}$

Now, let's find the new distance r from the center of the earth:

$r=R+\frac{R}{3}=\frac{4R}{3}$

4. Calculating the New Gravitational Force:
Substitute r into the formula for F':

$F^{\text{'}}=\frac{G·M·m}{{\left(\frac{4R}{3}\right)}^2}$

Simplify the denominator:

$F^{\text{'}}=\frac{G·M·m}{\frac{16}{9}{R}^2}=\frac{9}{16}·\left(\frac{G·M·m}{R^2}\right)$

Since we know that $\frac{G·M·m}{R^2}=48\text{N}$, we substitute this value back into the equation:

$F^{\text{'}}=\frac{9}{16}·48$

$F^{\text{'}}=9·3=27\text{N}$

Thus, the gravitational force experienced by the body at the given height is 27 N.