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  4. As shown in fig. 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 determined by the equation :

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Q. As shown in fig. 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 determined by the equation :Physics Question Image

JEE MainJEE Main 2020System of Particles and Rotational Motion

Solution:

By concept of $COM$
$m_{1}R_{1}=m_{2}R_{2}$
Remaining mass $× \left(2-R\right) =$ cavity mass $× \left(R-1\right)$
$\left(\frac{4}{3}\pi R^{3}\,\rho-\frac{4}{3}\pi l^{3}\,\rho\right)\left(2-R\right)=\frac{4}{3}\pi l^{3}\,\rho\times\left(R-1\right)$
$\left(R^{3}-1\right)\left(2-R\right)=R-1$
$\left(R^{2}+R+1\right)\left(2-R\right)=1$

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