MHT CET · Physics · Motion In Two Dimensions
A particle performing U.C.M. of radius \(\frac{\pi}{2} \mathrm{~m}\) makes \({ }_x\) revolutions in time
\(t\). Its tangential velocity is
- A \(\frac{\pi x}{t}\)
- B \(\frac{\pi^2 x}{t}\)
- C \(\frac{\pi^2 x^2}{t}\)
- D \(\frac{2 \pi x}{t}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi^2 x}{t}\)
Step-by-step Solution
Detailed explanation
Particle is under uniform circular motion,
So angular velocity is \(\omega=\frac{2 \pi x}{t}\)
The tangential velocity is \(v=\omega R=\left(\frac{2 \pi x}{t}\right) R\)
Given, \(R=\frac{\pi}{2}\), therefore,
\(v=\frac{\pi^2 x}{t}\)
So angular velocity is \(\omega=\frac{2 \pi x}{t}\)
The tangential velocity is \(v=\omega R=\left(\frac{2 \pi x}{t}\right) R\)
Given, \(R=\frac{\pi}{2}\), therefore,
\(v=\frac{\pi^2 x}{t}\)
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