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AP EAMCET · PHYSICS · Center of Mass Momentum and Collision

A circular plate \(A\) of radius \(1.5 \mathrm{r}\) is removed from one edge of a uniform circular plate \(B\) of radius \(2 r\). The distance of centre of mass of the remaining portion from the centre of the plate \(B\) is

  1. A \(\frac{5 \mathrm{r}}{12}\)
  2. B \(\frac{9 \mathrm{r}}{14}\)
  3. C \(\frac{3 r}{4}\)
  4. D \(\frac{7 r}{8}\)
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Answer & Solution

Correct Answer

(B) \(\frac{9 \mathrm{r}}{14}\)

Step-by-step Solution

Detailed explanation

Radius of removed circular plate, \(\mathrm{A}=1.5 \mathrm{r}\) Radius of circular plate, \(\mathrm{B}=2 \mathrm{r}\) Let mass of circular plate \(A, M_1=M\) So surface mass density, \(\sigma=\frac{M}{\pi(2 r)^2}=\frac{M}{4 \pi r^2}\) Then mass of removed circular plate, A…
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