JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A square shaped hole of side \(l=\frac{a}{2}\) is carved out at a distance \(d =\frac{ a }{2}\) from the centre \('O'\) of a uniform circular disk of radius \(a\). If the distance of the centre of mass of the remaining portion from \(O\) is \(-\frac{a}{X},\) value of \(X\) (to the nearest integer) is.......
- A \(15\)
- B \(45\)
- C \(30\)
- D \(23\)
Answer & Solution
Correct Answer
(D) \(23\)
Step-by-step Solution
Detailed explanation
\(X _{ com }=\frac{ m _{1} x _{1}- m _{2} x _{2}}{ m _{1}- m _{2}}\) where : \(m _{1}=\) mass of complete disc \(m _{2}=\) removed mass Let \(\sigma=\) surface mass density of disc material wrt…
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