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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/2 is made in the sphere as shown in the figure, and the force becomes F₂. The value of F₁:F₂ is

Asked in JEE Main 25th Feb 1st Shift 2021 · Field inside a cavity

Figure: Field inside a cavity
Answer: (4) 50:41

Step-by-step solution

The intact sphere acts like a point mass at its centre, so F₁=(GMm)/((3R)²)=(GMm)/(9R²).

The scooped ball has radius R/2, so its mass is M/8, and in the figure its centre lies at R/2 from the centre on the side facing the particle, that is at 3R-R/2=(5R)/2 from it.

By superposition F₂=F₁-(G(M/8)m)/(((5R)/2)²)=(GMm)/(9R²)-(GMm)/(50R²)=(41GMm)/(450R²).

So F₁:F₂=(50)/(450):(41)/(450)=50:41.

Why the other options are wrong

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