To solve this problem, we need to find out how deep inside the earth a man should go so that his weight becomes one-fourth of his weight on the Earth's surface.
Let's start with the basic concept of gravitational force:
The gravitational force on an object of mass \( m \) at a distance \( r \) from the center of the Earth is given by:
\(F = \frac{G \cdot M \cdot m}{r^2}\)
Where:
The weight of an object on the Earth's surface is:
\(W_0 = \frac{G \cdot M \cdot m}{R^2}\)
Where \(R\) is the radius of the Earth.
Now, when the man goes inside the Earth to a depth \(d\), his distance from the center becomes \(R-d\). The effective mass of the Earth that contributes to the gravitational force at this point is proportional to the volume of the sphere of radius \(R-d\). Thus, the new mass is given by:
\(M' = M \cdot \left(\frac{R-d}{R}\right)^3\)
The new gravitational force, or weight \(W\)\) at this depth is:
\(W = \frac{G \cdot M' \cdot m}{(R-d)^2} = \frac{G \cdot M \cdot m}{R^2} \cdot \left(\frac{R-d}{R}\right)\)
We want the weight \(W\) to be one-fourth of the original weight:
\(\frac{W_0}{4} = \frac{G \cdot M \cdot m}{4 \cdot R^2}\)
Equating it to the new weight, we have:
\(\frac{G \cdot M \cdot m}{R^2} \cdot \left(\frac{R-d}{R}\right) = \frac{G \cdot M \cdot m}{4 \cdot R^2}\)
On simplifying, we get:
\(R-d = \frac{R}{4}\)
This implies:
\(d = R - \frac{R}{4} = \frac{3R}{4}\)
Therefore, the man must go to a depth of \(\frac{3R}{4}\) from the Earth's surface for his weight to become one-fourth of the weight on the surface.
Thus, the correct answer is:
\( \frac{3R}{4} \)