A thin uniform circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linear acceleration along the inclined plane will be

The correct option is g/4.

Let's understand the physics of a thin uniform circular ring rolling down an inclined plane without slipping.

For a body of mass $M$ and radius $R$ rolling down an inclined plane of inclination $\theta$, the linear acceleration $a$ along the inclined plane is given by the standard formula:
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}$
where:
• $g$ is the acceleration due to gravity,
• $\theta$ is the angle of inclination of the plane, and
• $I$ is the moment of inertia of the rolling body about its central axis.

For a thin uniform circular ring, the moment of inertia $I$ about its central axis is:
$I=M{R}^2$

Substituting this moment of inertia into our acceleration formula, we get:
$a=\frac{g\sin \theta }{1+\frac{M{R}^2}{M{R}^2}}$
Simplifying the denominator:
$a=\frac{g\sin \theta }{1+1}=\frac{g\sin \theta }{2}$

Given that the angle of inclination of the plane is $\theta =30°$, we substitute $\sin (30°)=\frac{1}{2}$ into the equation:
$a=\frac{g\times \frac{1}{2}}{2}=\frac{g}{4}$

Thus, the linear acceleration of the ring along the inclined plane is indeed $\frac{g}{4}$.