As shown in figure, when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred 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) \left(\mathrm{R}^{2}-\mathrm{R}+1\right)(2-\mathrm{R})=1

(B) \left(\mathrm{R}^{2}+\mathrm{R}-1\right)(2-\mathrm{R})=1

(D) \left(\mathrm{R}^{2}-\mathrm{R}-1\right)(2-\mathrm{R})=1

JEE Mains · Physics · STD 11 - System of particles and rotational motion

As shown in figure, when a spherical cavity (centred at \(\mathrm{O})\) of radius \(1\) is cut out of a uniform sphere of radius \(\mathrm{R} \text { (centred at } \mathrm{C}),\) the centre of mass of remaining (shaded) part of sphere is at \(G\), i.e, on the surface of the cavity. \(\mathrm{R}\) can be detemined by the equation

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\left(\mathrm{R}^{2}+\mathrm{R}+1\right)(2-\mathrm{R})=1\)

Detailed explanation

By concept of \(COM\)

\(\mathrm{m}_{1} \mathrm{R}_{1}=\mathrm{m}_{2} \mathrm{R}_{2}\)

Remaining mass \(\times(2-\mathrm{R})=\) cavity mass \(\times(\mathrm{R}-1)\)

\(\left(\frac{4}{3} \pi \mathrm{R}^{3} \rho-\frac{4}{3} \pi l^{3} \rho\right)(2-\mathrm{R})=\frac{4}{3} \pi l^{3} \rho \times(\mathrm{R}-1)\)

\(\left(\mathrm{R}^{3}-1\right)(2-\mathrm{R})=\mathrm{R}-1\)

\(\left(\mathrm{R}^{2}+\mathrm{R}+1\right)(2-\mathrm{R})=1\)

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List \(I\)	List \(II\)
\((A)\) Young's Modulus \((Y)\)	\((I)\) \(\left[ M L ^{-1} T ^{-1}\right]\)
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\((C)\) Planck's Constant \((h)\)	\((III)\) \(\left[ M L ^{-1} T ^{-2}\right]\)
\((D)\) Work Function \((\phi)\)	\((IV)\) \(\left[ M L ^2 T ^{-2}\right]\)

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