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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

There is a thin uniform disc of radius R and mass per unit area σ ,...

There is a thin uniform disc of radius RR and mass per unit area σ\sigma, in which a hole of radius R/2R / 2 has been cut out as shown in the figure. Inside the hole, a square plate of same mass per unit area σ\sigma is inserted so that its corners touch the periphery of the hole. The distance of the centre of mass of the system from the origin is

Options

  1. A\dfrac{\text{R(}2 - \pi )}{2\left(3\pi + 2\right)}
  2. B\dfrac{\text{R(}1 - \pi )}{2\left(2\pi + 1\right)}
  3. C\dfrac{2 \text{R} \pi}{2 \left(3 \pi + 2\right)}
  4. D\dfrac{3 \text{R} \pi}{2 \left(2 \pi + 1\right)}

Correct answer

A. \dfrac{\text{R(}2 - \pi )}{2\left(3\pi + 2\right)}

Step-by-step solution

Side of square =Rcos⁡45∘=R2= R \cos 4 {5}^{∘} = \dfrac{\text{R}}{\sqrt{2}}

Area of square =R22= \dfrac{R^{2}}{2}

X(COM)=(π×R2×σ×0+π×R24(−σ)×R2+R22×σ×R2)(π×R2×σ+π×R24(−σ)+R22×σ)<br/><br/>=R(2−π)2(3π+2)<br/><br/>\text{X}\left(\text{COM}\right) = \left(\pi \times R^{2} \times \sigma \times 0 + \pi \times \dfrac{\text{R}2}{4}\left(- \sigma\right) \times \dfrac{R}{2} + \dfrac{\text{R}2}{2} \times \sigma \times \dfrac{R}{2}\right)\left(\pi \times R^{2} \times \sigma + \pi \times \dfrac{\text{R}2}{4} \left(- \sigma\right) + \dfrac{\text{R}2}{2} \times \sigma\right)<br /><br />=\dfrac{\text{R(}2 - \pi )}{2 ( 3\pi + 2 )}<br /><br />∴ThecentreofmassofthesystemisatadistanceofR(2−π)2(3π+2)fromthecentre and The centre of mass of the system is at a distance of \dfrac{\text{R(}2 - \pi )}{2 ( 3\pi + 2 )} from the centre \text{ and } Otowardstheplateasshowninthefiguretowards the plate as shown in the figure
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