jee-main 2020 Q4

jee-main · India · session1_08jan_shift2 Not Maths
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 determined by the equation:
(1) $\left(R^{2} + R + 1\right)(2 - R) = 1$
(2) $\left(R^{2} - R - 1\right)(2 - R) = 1$
(3) $\left(R^{2} - R + 1\right)(2 - R) = 1$
(4) $\left(R^{2} + R - 1\right)(2 - R) = 1$
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 determined by the equation:\\
(1) $\left(R^{2} + R + 1\right)(2 - R) = 1$\\
(2) $\left(R^{2} - R - 1\right)(2 - R) = 1$\\
(3) $\left(R^{2} - R + 1\right)(2 - R) = 1$\\
(4) $\left(R^{2} + R - 1\right)(2 - R) = 1$