JEE Physics 2025 Question with Solution

Given: A point mass $m$ is at distance $2R$ from the center of a uniform solid sphere of mass $M$ and radius $R$. A smaller sphere of radius $R/3$ is removed along the radius toward the point mass.

Find: The ratio $F_1 : F_2$.

For a point outside a uniform sphere, the whole mass acts as if concentrated at its center. Hence

$$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$$

Now use superposition for the remaining body:

• force due to original full sphere
• minus force due to the removed small sphere.

The removed sphere has the same density as the original sphere, so its mass is

$$M_{\text{removed}} = M\left(\frac{R/3}{R}\right)^3 = \frac{M}{27}$$

Its center is at distance

$$R - \frac{R}{3} = \frac{2R}{3}$$

from $O$ toward the point mass. Therefore the distance from $m$ to the center of the removed sphere is

$$d = 2R - \frac{2R}{3} = \frac{4R}{3}$$

So

$$F_2 = \frac{GMm}{(2R)^2} - \frac{Gm(M/27)}{(4R/3)^2}$$

Now,

$$(4R/3)^2 = \frac{16R^2}{9}$$

Therefore

$$\frac{Gm(M/27)}{(4R/3)^2} = \frac{GMm}{48R^2}$$

Thus

$$F_2 = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$$ $$F_2 = \frac{GMm}{R^2}\left(\frac{1}{4} - \frac{1}{48}\right) = \frac{11GMm}{48R^2}$$

Also,

$$F_1 = \frac{GMm}{4R^2} = \frac{12GMm}{48R^2}$$

Hence

$$F_1 : F_2 = 12 : 11$$

Therefore, the correct option is C.

Given: The external point mass is at distance $2R$ from the center of the large sphere. A spherical cavity of radius $R/3$ is cut from the side nearer to the point mass.

Find: The ratio of the original force to the new force after removal.

The key idea is that the field of the remaining mass equals the field of the full sphere plus the field of a sphere of negative mass placed where the cavity is located.

1. Force due to the full sphere:

$$F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$$

2. The removed sphere has radius $R/3$, so its volume is $1/27$ of the original sphere. Hence its mass is

$$\frac{M}{27}$$

3. Since the cavity touches the outer surface along the radius toward the point mass, its center is at distance

$$\frac{2R}{3}$$

from the center of the big sphere. 4. The external mass is at distance $2R$ from the big sphere's center, so the distance to the cavity center is

$$2R - \frac{2R}{3} = \frac{4R}{3}$$

5. Force due to the removed part:

$$F_{\text{removed}} = \frac{Gm(M/27)}{(4R/3)^2} = \frac{GMm}{48R^2}$$

6. Net force after removal:

$$F_2 = F_1 - F_{\text{removed}} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2} = \frac{11GMm}{48R^2}$$

7. Ratio:

$$F_1 : F_2 = \frac{12GMm}{48R^2} : \frac{11GMm}{48R^2} = 12 : 11$$

Therefore, the ratio is $12 : 11$.