JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A circular hole of radius \(\left(\frac{ a }{2}\right)\) is cut out of a circular disc of radius \('a'\) as shown in figure. The centroid of the remaining circular portion with respect to point \('O'\) will be :
- A \(\frac{1}{6} a\)
- B \(\frac{10}{11} a\)
- C \(\frac{5}{6} a\)
- D \(\frac{2}{3} a\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{6} a\)
Step-by-step Solution
Detailed explanation
Let \(\sigma\) be the uniform mass density of disc then \(x _{ COM }=\frac{\left(\sigma \pi a ^{2}\right) a -\sigma \pi\left(\frac{ a ^{2}}{4}\right) \times \frac{3 a }{2}}{\sigma \pi a ^{2}-\frac{\sigma \pi a ^{2}}{4}}\) \(=\frac{a-\frac{3 a}{8}}{1-\frac{1}{4}}=\frac{5 a}{6}\)
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