A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius ( R \over 2 ) is made in the sphere (as shown in figure) and the force becomes F2. The value of F1 : F2 is

Gravitation PYQ — JEE Main 2021 · 25 Feb · Shift 1 · Q47 with Solution

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Correct answer: C

Force due to the original solid sphere

Since the particle is at a distance $3R$ from the centre of the sphere, and $3R > R$ , the sphere acts like a point mass placed at its centre.

Let the mass of the full sphere be $M$ .

So,

F_1 = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}

where $m$ is the mass of the particle.

After making the cavity

A cavity of radius $\dfrac{R}{2}$ is removed. From the figure (standard configuration), the cavity is tangent to the outer surface and lies along the line joining the centre of the big sphere to the particle.

So the centre of the cavity is at a distance

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

from the centre of the big sphere.

The mass removed is proportional to volume:

M_c = M\left(\frac{R/2}{R}\right)^3 = M\cdot \frac{1}{8} = \frac{M}{8}

Now use superposition:

Force due to full sphere = $F_1$
Force due to removed smaller sphere must be subtracted.

Force due to the removed sphere

The particle is at distance $3R$ from the big sphere's centre. The cavity centre is at $\dfrac{R}{2}$ from the big centre toward the particle, so distance of particle from cavity centre is

3R - \frac{R}{2} = \frac{5R}{2}

Hence force due to the removed mass would be

F_c = \frac{G\left(\frac{M}{8}\right)m}{\left(\frac{5R}{2}\right)^2} = \frac{GMm}{8} \cdot \frac{4}{25R^2} = \frac{GMm}{50R^2}

Net force after cavity is made

F_2 = F_1 - F_c = \frac{GMm}{9R^2} - \frac{GMm}{50R^2}

Taking LCM $450R^2$ ,

F_2 = GMm\left(\frac{50-9}{450R^2}\right) = \frac{41GMm}{450R^2}

Also,

F_1 = \frac{GMm}{9R^2} = \frac{50GMm}{450R^2}

Therefore,

F_1 : F_2 = 50 : 41

Checking options

A: $36:25$ ❌
B: $41:50$ ❌
C: $50:41$ ✅
D: $25:36$ ❌

So the correct option is C.

More from Gravitation

View written solutionFree

Correct answer: C

Force due to the original solid sphere

Since the particle is at a distance $3R$ from the centre of the sphere, and $3R > R$ , the sphere acts like a point mass placed at its centre.

Let the mass of the full sphere be $M$ .

So,

F_1 = \frac{GMm}{(3R)^2} = \frac{GMm}{9R^2}

where $m$ is the mass of the particle.

After making the cavity

So the centre of the cavity is at a distance

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

from the centre of the big sphere.

The mass removed is proportional to volume:

M_c = M\left(\frac{R/2}{R}\right)^3 = M\cdot \frac{1}{8} = \frac{M}{8}

Now use superposition:

Force due to full sphere = $F_1$
Force due to removed smaller sphere must be subtracted.

Force due to the removed sphere

The particle is at distance $3R$ from the big sphere's centre. The cavity centre is at $\dfrac{R}{2}$ from the big centre toward the particle, so distance of particle from cavity centre is

3R - \frac{R}{2} = \frac{5R}{2}

Hence force due to the removed mass would be

F_c = \frac{G\left(\frac{M}{8}\right)m}{\left(\frac{5R}{2}\right)^2} = \frac{GMm}{8} \cdot \frac{4}{25R^2} = \frac{GMm}{50R^2}

Net force after cavity is made

F_2 = F_1 - F_c = \frac{GMm}{9R^2} - \frac{GMm}{50R^2}

Taking LCM $450R^2$ ,

F_2 = GMm\left(\frac{50-9}{450R^2}\right) = \frac{41GMm}{450R^2}

Also,

F_1 = \frac{GMm}{9R^2} = \frac{50GMm}{450R^2}

Therefore,

F_1 : F_2 = 50 : 41

Checking options

A: $36:25$ ❌
B: $41:50$ ❌
C: $50:41$ ✅
D: $25:36$ ❌

So the correct option is C.

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