System of Particles and Rotational Motion - Result Question 82
89. A thin uniform circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclination plane will be
[1994]
(a) $\frac{g}{2}$
(b) $\frac{g}{3}$
(c) $\frac{g}{4}$
(d) $\frac{2 g}{3}$
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Answer:
Correct Answer: 89. (c)
Solution:
- (c)
$ a=\frac{g \sin \theta}{(1+K^{2} / R^{2})}=\frac{g \sin 30^{\circ}}{1+1}=\frac{g}{4} $
