The weight of an object measured on the surface of earth is 300 N. What will be the weight of the same object at depth R/4 inside the earth? (Given R = Radius of earth)
To solve the problem of finding the weight of an object at a depth \( R/4 \) inside the earth, where \( R \) is the radius of the earth, we need to understand how gravity changes with depth.
The weight of an object at the surface of the earth is given by:
W = mg,
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
As given, the weight at the surface (W_surface) is 300 N.
Inside the Earth, the gravitational force decreases linearly with depth. The formula for weight at a depth \( d \) inside the earth is:
W_d = \left(1 - \frac{d}{R}\right) W_{surface}.
Here, \( d = R/4 \). So, substituting the values, we get:
Therefore, the correct answer is that the weight of the object at a depth \( R/4 \) inside the earth will be 225 N.
This matches the given correct answer option. The other options can be ruled out based on this calculation, as they do not satisfy the equation derived for gravitational force at a depth.
