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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

A hemispherical cavity of radius R is created in a solid sphere of...

A hemispherical cavity of radius RR is created in a solid sphere of radius 2R2R as shown in the figure. The yy-coordinate of the centre of mass of the remaining sphere is
A hemispherical cavity of radius R is created in a solid sphere of radius 2R as shown in the figure. The y -coordinate

Options

  1. AyCM=−R15{y}_{CM} = - \dfrac{R}{15}
  2. ByCM=−R40{y}_{CM} = - \dfrac{\text{R}}{4 0}
  3. CyCM=−R30{y}_{CM} = - \dfrac{R}{30}
  4. DyCM=−R20{y}_{CM} = - \dfrac{R}{20}

Correct answer

B. yCM=−R40{y}_{CM} = - \dfrac{\text{R}}{4 0}

Step-by-step solution

xCM=43π(2R)30−23πR3R43π(2R)3−23πR3=−R15{x}_{CM} = \dfrac{\dfrac{4}{3} \pi {\left(2 \text{R}\right)}^{3} 0 - \dfrac{2}{3} \pi R^{3} \text{R}}{\dfrac{4}{3} \pi {\left(2 \text{R}\right)}^{3} - \dfrac{2}{3} \pi R^{3}} = - \dfrac{\text{R}}{1 5}

yCM=43π(2R)30−23πR33R843π(2R)3−23πR3=−R40{y}_{CM} = \dfrac{\dfrac{4}{3} \pi {\left(2 \text{R}\right)}^{3} 0 - \dfrac{2}{3} \pi R^{3} \dfrac{3 \text{R}}{8}}{\dfrac{4}{3} \pi {\left(2 \text{R}\right)}^{3} - \dfrac{2}{3} \pi R^{3}} = - \dfrac{\text{R}}{4 0}

zCM=0{z}_{CM} = 0
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