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 OD 2019]

Correct Answer is : V024πnr2rad∕s2

{ {\V}_0}/{ {\n}} {\rad} \/ {\s}^2

\frac{ {\V}_0^2}{2 \pi n r^2} {\rad} \/ {\s}^2

\frac{V_0^2}{4 \pi n r^2} {\rad} \/ {\s}^2

\frac{ {\V}_0^2}{4 \pi n r} {\rad} \/ {\s}^2

NEET Odisha 2019 Physics and Chemistry

Question : 8 of 90

Marks: +1, -0

A particle starting from rest, moves in a circle of radius ' r '. It attains a velocity of V0m∕s in the n‌th ‌ round. Its angular acceleration will be :-

[NEET OD 2019]

‌
V0
n
‌rad∕s2
‌
V02
2πnr2
‌rad∕s2
‌
V02
4πnr2
‌rad∕s2
‌
V02
4πnr
‌rad∕s2

Validate

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