Question:
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 determined by the equation :
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 determined by the equation :
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Treat the cavity as a negative mass. The cavity is tangent to the sphere surface, so \(CO=R-1\); G lies on the cavity surface at distance \(2-R\) from C on the side away from O.
Updated On: Nov 5, 2026
- \((R^2+R-1)(2-R)=1\)
- \((R^2-R-1)(2-R)=1\)
- \((R^2-R+1)(2-R)=1\)
- \((R^2+R+1)(2-R)=1\)
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The Correct Option is D
Solution and Explanation
Take C as origin, positive towards O. The cavity just touches the sphere surface, so \(CO=R-1\).
Mass is proportional to volume: full sphere \(\propto R^3\), cavity \(\propto 1\), remaining \(\propto R^3-1\).
\(x_G=\frac{R^3\cdot 0-1\cdot(R-1)}{R^3-1}=-\frac{R-1}{R^3-1}=-\frac{1}{R^2+R+1}\).
G is on the cavity surface, a distance 1 from O on the side towards C, so \(x_G=(R-1)-1=R-2\).
\(R-2=-\frac{1}{R^2+R+1}\Rightarrow (2-R)(R^2+R+1)=1\).
Answer: (4) \((R^2+R+1)(2-R)=1\)
Mass is proportional to volume: full sphere \(\propto R^3\), cavity \(\propto 1\), remaining \(\propto R^3-1\).
\(x_G=\frac{R^3\cdot 0-1\cdot(R-1)}{R^3-1}=-\frac{R-1}{R^3-1}=-\frac{1}{R^2+R+1}\).
G is on the cavity surface, a distance 1 from O on the side towards C, so \(x_G=(R-1)-1=R-2\).
\(R-2=-\frac{1}{R^2+R+1}\Rightarrow (2-R)(R^2+R+1)=1\).
Answer: (4) \((R^2+R+1)(2-R)=1\)
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