A uniform square plate has a side of length $2 \mathrm{R}$. A circular piece of maximum possible area is cut…

A uniform square plate has a side of length $2 \mathrm{R}$. A circular piece of maximum possible area is cut and removed from one of the quadrants of the plate as shown in the figure. Shift in the centre of mass of the plate is.
  1. $\frac{\pi R}{\sqrt{2}(16-\pi)}$
  2. $\frac{\mathrm{R}}{(16-\pi)}$
  3. $\frac{\mathrm{R}}{\pi(16-\pi)}$
  4. $\frac{\mathrm{R} \pi}{(16-\pi)}$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2017 (25 Apr Shift 2)

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