MHT CET · Physics · Motion In Two Dimensions
A particle performing uniform circular motion of radius \(\frac{\pi}{2} \mathrm{~m}\) makes ' \(\mathrm{x}\) ' revolutions in time ' \(\mathrm{t}\) '. Its tangential velocity is
- A \(\frac{\pi \mathrm{X}}{\mathrm{t}}\)
- B \(\frac{\pi \mathrm{x}^2}{\mathrm{t}}\)
- C \(\frac{\pi^2 x^2}{t}\)
- D \(\frac{\pi^2 x}{t}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi^2 x}{t}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Circumference of the circle }=2 \pi \mathrm{r} \\ & \therefore \quad \text { Tangential velocity }=\frac{\text { Distance Travelled }}{\text { Time }} \\ & \qquad \frac{\pi^2 \times x}{\mathrm{t}} \\ & =\frac{\pi^2 x}{\mathrm{t}}\end{aligned}\)
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