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JEE Main 2020 (8 Jan, Shift 2) · Composite & Cut-Out Bodies

Composite & Cut-Out Bodies question from JEE Main 2020 (8 Jan, Shift 2)

Q8JEE Main 2020 · 8 JAN · SHIFT 2Difficult

Composite & Cut-Out Bodies

As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered 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:

Figure

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Answer: (d)

  1. Take C as origin. The cavity touches the sphere surface, so CO=R−1CO = R - 1; mass ∝\propto volume ∝r3\propto r^3.
  2. xG=−13(R−1)R3−1=−R−1(R−1)(R2+R+1)=−1R2+R+1x_G = \frac{-1^3(R-1)}{R^3 - 1} = -\frac{R-1}{(R-1)(R^2+R+1)} = -\frac{1}{R^2+R+1}
  3. G lies on the cavity surface on the far side: xG=(R−1)−1=−(2−R)x_G = (R-1) - 1 = -(2-R).
  4. 1R2+R+1=2−R⇒(R2+R+1)(2−R)=1\frac{1}{R^2+R+1} = 2 - R \Rightarrow (R^2+R+1)(2-R) = 1
  5. So the answer is (d).
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