JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The disc of mass \(M\) with uniform surface mass density \(\sigma\) is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position \(\frac{x}{3} \frac{a}{\pi}, \frac{x}{3} \frac{a}{\pi}\) where \(x\) is ....... . (Round off to the Nearest Integer) \([ a\) is an area as shown in the figure \(]\)
- A \(4\)
- B \(2\)
- C \(1\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(C.O.M.\) of quarter disc is at \(\frac{4 a}{3 \pi}, \frac{4 a}{3 \pi}\) \(=4\)
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