Q108 / 150 · MHT CET 2023 (14 May Shift 1) · Physics
A particle moves around a circular path of radius ' $r$ ' with uniform speed ' $V$ '. After moving half the circle, the average acceleration of the particle is
- A$\frac{\mathrm{V}^2}{\mathrm{r}}$
- B$\frac{2 \mathrm{~V}^2}{\mathrm{r}}$
- Correct answerC$\frac{2 \mathrm{~V}^2}{\pi \mathrm{r}}$
- D$\frac{\mathrm{V}^2}{\pi \mathrm{r}}$
Solution
At end points of the half revolution magnitude of the velocity is same but it directs in opposite direction.
$$
\begin{array}{ll}
\therefore & \Delta V=V-(-V) \\
\therefore & \Delta V=2 V
\end{array}
$$
Time taken to complete the half revolution is $\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}$
Average acceleration is, $a=\frac{\Delta V}{t}=\frac{2 \mathrm{~V}}{\frac{\pi \mathrm{r}}{\mathrm{V}}}$
$$
\therefore \quad a=\frac{2 V^2}{\pi r}
$$
Hence, the correct answer is option: 'C'.