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JEE Main 20208 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2020 · 8 January · Evening Shift · Physics · Center of Mass, Momentum and Collision

As shown in figure. When a spherical cavity (centred at O ) of...

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:
As shown in figure. When a spherical cavity (centred at O ) of radius 1 is cut out of a uniform sphere of radius R

Options

  1. A(R2+R+1)(2−R)=1\left({R}^{2} + R + 1\right)\left(2 - R\right) = 1
  2. B(R2−R−1)(2−R)=1\left({R}^{2} - R - 1\right)\left(2 - R\right) = 1
  3. C(R2−R+1)(2−R)=1\left({R}^{2} - R + 1\right)\left(2 - R\right) = 1
  4. D(R2+R−1)(2−R)=1\left({R}^{2} + R - 1\right)\left(2 - R\right) = 1

Correct answer

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

Step-by-step solution

M1=43πR3ρ{M}_{1} = \dfrac{4}{3}\pi{R}^{3}ρ
M2=43π(1)3(−ρ){M}_{2} = \dfrac{4}{3}\pi{\left(1\right)}^{3}\left(- ρ\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)⇒ \dfrac{\left[\dfrac{4}{3}\pi{R}^{3}ρ\right]0 + \left[\dfrac{4}{3}\pi{\left(1\right)}^{3}\left(- ρ\right)\right]\left[R - 1\right]}{\dfrac{4}{3}\pi{R}^{3}ρ + \dfrac{4}{3}\pi{\left(1\right)}^{3}\left(- ρ\right)} - \left(2 - R\right)
⇒(R−1)(R3−1)=(2−R)(R≠1)⇒ \dfrac{\left(R - 1\right)}{{( R}^{3} - 1 )} = \left(2 - R\right)\left(R \ne 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
\left({R}^{2} + R + 1\right)\left(2 - R\right) = 1<br/>Alternative:<br/><br />Alternative:<br />{M}_{remaining}\left(2 - R\right) = {M}_{cavity}\left(1 - R\right)<br/><br />⇒ \left({R}^{3} - {1}^{3}\right)\left(2 - R\right) = {1}^{3}\left[R - 1\right]<br/><br />⇒ \left({R}^{2} + R + 1\right)\left(2 - R\right) = 1

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