JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
As shown in figure, when a spherical cavity (centred at \(\mathrm{O})\) of radius \(1\) is cut out of a uniform sphere of radius \(\mathrm{R} \text { (centred at } \mathrm{C}),\) the centre of mass of remaining (shaded) part of sphere is at \(G\), i.e, on the surface of the cavity. \(\mathrm{R}\) can be detemined by the equation
- A \(\left(\mathrm{R}^{2}-\mathrm{R}+1\right)(2-\mathrm{R})=1\)
- B \(\left(\mathrm{R}^{2}+\mathrm{R}-1\right)(2-\mathrm{R})=1\)
- C \(\left(\mathrm{R}^{2}+\mathrm{R}+1\right)(2-\mathrm{R})=1\)
- D \(\left(\mathrm{R}^{2}-\mathrm{R}-1\right)(2-\mathrm{R})=1\)
Answer & Solution
Correct Answer
(C) \(\left(\mathrm{R}^{2}+\mathrm{R}+1\right)(2-\mathrm{R})=1\)
Step-by-step Solution
Detailed explanation
By concept of \(COM\) \(\mathrm{m}_{1} \mathrm{R}_{1}=\mathrm{m}_{2} \mathrm{R}_{2}\) Remaining mass \(\times(2-\mathrm{R})=\) cavity mass \(\times(\mathrm{R}-1)\)…
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