MHT CET · Physics · Motion In Two Dimensions
A particle is performing U.C.M. along the circumference of circle of diameter \(50 \mathrm{~cm}\) with frequency \(2 \mathrm{~Hz}\). The acceleration of the particle in \(\mathrm{m} / \mathrm{s}^2\) is
- A \(2 \pi^2\)
- B \(4 \pi^2\)
- C \(8 \pi^2\)
- D \(\pi^2\)
Answer & Solution
Correct Answer
(B) \(4 \pi^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{d}=50 \mathrm{~cm} \\ & \therefore \mathrm{r}=25 \times 10^{-2} \mathrm{~m}, \mathrm{f}=2 \mathrm{~Hz} \\ & \mathrm{a}=\mathrm{r} \omega^2=4 \pi^2 \mathrm{f}^2 \mathrm{r}=4 \pi^2 \times 4 \times 25 \times 10^{-2}=4 \pi^2\end{aligned}\)
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