JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Three identical spheres each of mass \(M\) are placed at the corners of a right angled triangle with mutually perpendicular sides equal to \(3\,m\) each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be \(\sqrt{x} m\). The value of \(x\) is
- A \(2\)
- B \(3\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ r }_{\text {com }}=\frac{ M (0 \hat{ i }+0 \hat{ j })+ M (3 \hat{ i })+ M (3 \hat{ j })}{3 M }\) \(\overrightarrow{ r }_{\text {coma }}=\hat{ i }+\hat{ j }\) \(\left|\overrightarrow{ r }_{\text {com }}\right|=\sqrt{2}=\sqrt{ x }\) \(x =2\)
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