JEE Main 2020 · 8 January · evening · Single correct · +4 / −1
As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e, on the surface of the cavity. R can be detemined by the equation :
- (a) (R² + R – 1) (2 – R) = 1
- (b) (R² – R – 1) (2 – R) = 1
- (c) (R² – R + 1) (2 – R) = 1
- (d) (R² + R + 1) (2 – R) = 1
Answer and solution
Answer (d) (R² + R + 1) (2 – R) = 1
Solution
Physics sir · 4 steps
Treat the cavity as a negative mass, then set the shaded part's centre of mass exactly at G.
- Read the geometry. The cavity touches the big sphere from inside, so CO = R - 1. G is on the cavity's surface on the far side from O, so CG = 1 - (R - 1) = 2 - R.
- Negative mass. Take C as the origin. The full sphere has mass proportional to R³ with its centre at C. The cavity is a mass proportional to 1³ = 1 at distance R - 1, and we subtract it. The remainder has mass proportional to R³ - 1, and its centre lies on the side opposite O.
CG = (1 × (R-1))/(R³ - 1) = (R-1)/(R-1)(R²+R+1) = 1/(R²+R+1)
- Match with geometry. This distance must equal 2 - R. So 1/(R²+R+1) = 2 - R.
- So. R satisfies (R² + R + 1)(2 - R) = 1, which is option (d).
The idea to keep. For a body with a hole, treat the hole as negative mass placed at its own centre. Then get one more equation for distances from the figure, and remember that R³ - 1 = (R-1)(R²+R+1).
Answer from the NTA final answer key. Solution by PrepWiser.