System of Particles and Rotational Motion - Result Question 12
12. A particle starting from rest, moves in a circle of radius ’ $r$ ‘. It attains a velocity of $V_0 m / s$ in the $n^{\text{th }}$ round. Its angular acceleration will be,
[NEET Odisha 2019]
(a) $\frac{V_0^{2}}{4 \pi n r} rad / s^{2}$
(b) $\frac{V_0}{n} rad / s^{2}$
(c) $\frac{V_0^{2}}{2 \pi n r^{2}} rad / s^{2}$
(d) $\frac{V_0^{2}}{4 \pi n r^{2}} rad / s^{2}$
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Answer:
Correct Answer: 12. (d)
Solution:
- (d) $V^{2}=u^{2}+2 a s$
$ V_0^{2}=0+2 r \alpha(2 \pi r) n \Rightarrow \alpha=\frac{V_0^{2}}{4 \pi n r^{2}} $
