A body is in limiting equilibrium on a rough inclined plane at angle 30° with horizontal. Calculate the acceleration with which the body will slide down when the inclination of the plane is changed to 60°. (Take g = 10 m/s²)

To solve this problem, we'll use the concepts of forces on an inclined plane and friction. Here's the step-by-step approach:

### Step 1: Understanding the Problem
- The body is in limiting equilibrium at an inclination of \[30^\circ\]. This means that the frictional force is exactly equal to the component of the gravitational force trying to pull the body down the plane.
- When the inclination is changed to \[60^\circ\], the friction will still act, but it won't be enough to stop the body from sliding down. We need to calculate the acceleration of the body under these conditions.

### Step 2: Determine the Coefficient of Friction
In limiting equilibrium at \[30^\circ\], the frictional force (\(f\)) equals the component of the gravitational force along the plane:

\[
f = \mu mg \cos(30^\circ)
\]
\[
mg \sin(30^\circ) = \mu mg \cos(30^\circ)
\]

Here, \(m\) is the mass of the body, \(g\) is the acceleration due to gravity, and \(\mu\) is the coefficient of friction. Simplifying, we get:

\[
\mu = \tan(30^\circ)
\]
\[
\mu = \frac{1}{\sqrt{3}}
\]

### Step 3: Calculate the Net Force when the Inclination is \[60^\circ\]
When the angle is changed to \[60^\circ\], the force components become:

- Gravitational force down the plane: \(mg \sin(60^\circ)\)
- Frictional force up the plane: \(\mu mg \cos(60^\circ)\)

The net force (\(F_{\text{net}}\)) down the plane is:

\[
F_{\text{net}} = mg \sin(60^\circ) - \mu mg \cos(60^\circ)
\]

### Step 4: Calculate the Acceleration
The acceleration \(a\) of the body is given by Newton's second law:

\[
a = \frac{F_{\text{net}}}{m}
\]

Substituting the expressions, we get:

\[
a = g \left(\sin(60^\circ) - \mu \cos(60^\circ)\right)
\]

Using \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), \(\cos(60^\circ) = \frac{1}{2}\), and \(\mu = \frac{1}{\sqrt{3}}\):

\[
a = 10 \left(\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{3}} \cdot \frac{1}{2}\right)
\]

Simplifying further:

\[
a = 10 \left(\frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}}\right)
\]
\[
a = 10 \times \frac{3 - 1}{2\sqrt{3}}
\]
\[
a = 10 \times \frac{2}{2\sqrt{3}}
\]
\[
a = \frac{10}{\sqrt{3}} \text{ m/s}^2
\]
\[
a \approx 5.77 \text{ m/s}^2
\]

### Final Answer:
The acceleration with which the body will slide down when the inclination of the plane is changed to \[60^\circ\] is approximately \(5.77 \, \text{m/s}^2\).