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JEE Main 202124 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2021 · 24 February · Evening Shift · Physics · Center of Mass, Momentum and Collision

A circular hole of radius ((a)/(2)) is cut out of a circular disc...

A circular hole of radius (a2)\left(\dfrac{a}{2}\right) is cut out of a circular disc of radius aa as shown in figure. The centroid of the remaining circular portion with respect to point OO will be:

A circular hole of radius ((a)/(2)) is cut out of a circular disc of radius a as shown in figure. The centroid of the

Options

  1. A23a\dfrac{2}{3}a
  2. B56a\dfrac{5}{6}a
  3. C16a\dfrac{1}{6}a
  4. D1011a\dfrac{10}{11}a

Correct answer

B. 56a\dfrac{5}{6}a

Step-by-step solution

A circular hole of radius ((a)/(2)) is cut out of a circular disc of radius a as shown in figure. The centroid of the

Let σ\sigma be the uniform mass density of disc then

xCOM=(σπa2)a−σπ(a24)×3a2σπa2−σπa24{x}_{COM} = \dfrac{\left(\sigma\pi{a}^{2}\right)a - \sigma\pi\left(\dfrac{{a}^{2}}{4}\right) \times \dfrac{3a}{2}}{\sigma\pi{a}^{2} - \dfrac{\sigma\pi{a}^{2}}{4}}

=a−3a81−14=5a6= \dfrac{a - \dfrac{3a}{8}}{1 - \dfrac{1}{4}} = \dfrac{5a}{6}

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