JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A child of mass \(5\,kg\) is going round a merry-goround that makes \(1\) rotation in \(3.14\,s\). The radius of the merry-go-round is \(2\,m\). The centrifugal force on the child will be \(.......\,N\)
- A \(80\) \(N\)
- B \(50\) \(N\)
- C \(100\) \(N\)
- D \(40\) \(N\)
Answer & Solution
Correct Answer
(D) \(40\) \(N\)
Step-by-step Solution
Detailed explanation
\(\omega=\frac{2 \pi}{3.14}=2\,rad / s\) \(\left|\bar{f}_{\text {centrifiugal }}\right|=\left|-m \bar{a}_{\operatorname{Re} f .}\right|\) \(=M \omega^2 R\) \(=40\,N\)
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