Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

57 of 130

Question 57 of 130

As shown in figure. When a spherical cavity (centred at OO ) of radius 11 is cut out of a uniform sphere of radius RR (centred at CC ), the centre of mass of remaining (shaded part of sphere is at G,G, i.e., on the surface of the cavity. RR can be determined by the equation:

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Answer: (A) (R2+R+1)(2−R)=1\left({R}^{2}+R+1\right)\left(2-R\right)=1

M1=43πR3ρ{M}_{1}=\dfrac{4}{3}\pi {R}^{3}\rho
M2=43π(1)3(−ρ){M}_{2}=\dfrac{4}{3}\pi {\left(1\right)}^{3}\left(-\rho \right)
Xcom=M1X1+M2X2M1+M2{X}_{com}=\dfrac{{M}_{1}{X}_{1}+{M}_{2}{X}_{2}}{{M}_{1}+{M}_{2}}
⇒[43πR3ρ]0+[43π(1)3(−ρ)][R−1]43πR3ρ+43π(1)3(−ρ)−(2−R)\Rightarrow \dfrac{\left[\dfrac{4}{3}\pi {R}^{3}\rho \right]0+\left[\dfrac{4}{3}\pi {\left(1\right)}^{3}\left(-\rho \right)\right]\left[R-1\right]}{\dfrac{4}{3}\pi {R}^{3}\rho +\dfrac{4}{3}\pi {\left(1\right)}^{3}\left(-\rho \right)}-\left(2-R\right)
⇒(R−1)(R3−1)=(2−R)(R≠1)\Rightarrow \dfrac{\left(R-1\right)}{{(R}^{3}-1)}=\left(2-R\right)\left(R\neq 1\right)
(R−1)(R−1)(R2+R+1)=2−1\dfrac{\left(R-1\right)}{\left(R-1\right)\left({R}^{2}+R+1\right)}=2-1
(R2+R+1)(2−R)=1\left({R}^{2}+R+1\right)\left(2-R\right)=1
Alternative:
Mremaining(2−R)=Mcavity(1−R){M}_{remaining}\left(2-R\right)={M}_{cavity}\left(1-R\right)
⇒(R3−13)(2−R)=13[R−1]\Rightarrow \left({R}^{3}-{1}^{3}\right)\left(2-R\right)={1}^{3}\left[R-1\right]
⇒(R2+R+1)(2−R)=1\Rightarrow \left({R}^{2}+R+1\right)\left(2-R\right)=1

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