MCQ
A particle moves in a circle of radius $25\,cm$ at two revolutions per sec. The acceleration of the particle in $m/s^2$ is
- A$\pi ^2$
- B$8\pi ^2$
- ✓$4\pi ^2$
- D$2\pi ^2$
✓
Answer
Correct option: C.
$4\pi ^2$
c
$\mathrm{r}=25 \times 10^{-2} \mathrm{m}, \mathrm{f}=2 / \mathrm{sec}$
$\mathrm{r}=25 \times 10^{-2} \mathrm{m}, \mathrm{f}=2 / \mathrm{sec}$
$\therefore \quad \omega=2 \pi f=4 \pi \mathrm{rad} / \mathrm{sec}$
Acceleration $=\omega^{2} \mathrm{r}=(4 \pi)^{2} \times 25 \times 10^{-2}$
$=16 \times 25 \times 10^{-2} \pi^{2} \mathrm{m} / \mathrm{s}^{2}=4 \pi^{2} \mathrm{m} / \mathrm{s}^{2}$
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