A solid cylinder is placed gently over an incline plane of inclination 60°. The acceleration of cylinder when it start rolling without slipping is , g/√x where μ is coefficient of friction. (Take g = 10 m/s²)

The correct answer is 3.

Let us analyze the motion of a solid cylinder of mass $m$ and radius $R$ rolling down an inclined plane of inclination $\theta =60°$ without slipping.

The diagram representing this setup is shown below:

The forces acting on the cylinder are:
1. The gravitational force ($mg$) acting vertically downwards.
2. The normal force ($N$) acting perpendicular to the incline.
3. The static friction force ($f$) acting up the incline, opposing the tendency to slide down.

For translational motion down the incline:
$mg\sin \theta -f=ma$

For rotational motion about the center of mass:
$\tau =I\alpha$

Here, the torque is provided by the friction force $f$ acting at the surface of the cylinder of radius $R$:
$fR=I\alpha$

For a solid cylinder, the moment of inertia about its central axis is:
$I=\frac{1}{2}m{R}^2$

Since the cylinder rolls without slipping, the linear acceleration $a$ and angular acceleration $\alpha$ are related by:
$a=R\alpha$
or
$\alpha =\frac{a}{R}$

Substituting $I$ and $\alpha$ into the torque equation:
$fR=\left(\frac{1}{2}m{R}^2\right)\left(\frac{a}{R}\right)$
Simplifying this gives:
$f=\frac{1}{2}ma$

Now, substitute this expression for friction $f$ back into the force equation:
$mg\sin \theta -\frac{1}{2}ma=ma$

Adding $\frac{1}{2}ma$ to both sides:
$mg\sin \theta =\frac{3}{2}ma$

Dividing by $m$ and solving for $a$:
$a=\frac{2}{3}g\sin \theta$

Given the angle of inclination $\theta =60°$, we have:
$\sin 60°=\frac{\sqrt{3}}{2}$

Substitute this value back into the acceleration formula:
$a=\frac{2}{3}g\left(\frac{\sqrt{3}}{2}\right)$
Simplifying this gives:
$a=\frac{g}{\sqrt{3}}$

We are given that the acceleration of the cylinder is $\frac{g}{\sqrt{x}}$. By comparing the two expressions:
$\frac{g}{\sqrt{x}}=\frac{g}{\sqrt{3}}$
This yields:
$x=3$