AP EAMCET · PHYSICS · Laws of Motion
A body of mass \(0.6 \mathrm{~kg}\) is moving along a circular path of radius \(1 \mathrm{~m}\). If the body moves with \(\frac{900}{\pi}\) revolutions per minute, its kinetic energy
- A \(120 \mathrm{~J}\)
- B \(270 \mathrm{~J}\)
- C \(360 \mathrm{~J}\)
- D \(240 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(270 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
We have \[ \begin{aligned} & \mathrm{V}=\omega \mathrm{r} \\ & =2 \pi \mathrm{fr} \\ & =2 \pi \times \frac{15}{\pi} \times 1 \\ & =30 \mathrm{~m} / \mathrm{s} \end{aligned} \] So, Kinetic energy \(=\frac{1}{2} \times 0.6 \times 900\) \[ =270 \mathrm{~J} \]
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