A body of 6 kg rests in limiting equilibrium on an inclined plane whose slope is `30^(@)`. If the plane is raised to a slope of `60^(@)`, the force in kg along the plane required to support it is

Options

A. 3 kg
B. `2sqrt(3)` kg
C. `sqrt(3)` kg
D. `3sqrt(3)` kg

Answer

To solve the problem, we need to determine the force required to support a 6 kg body on an inclined plane when the slope is raised from 30 degrees to 60 degrees. We will use the concepts of friction and the forces acting on the body.

### Step-by-Step Solution:

1. **Identify the Mass and Weight**:
- Given mass \( m = 6 \, \text{kg} \).
- Weight \( W = mg = 6 \times 10 = 60 \, \text{N} \).

2. **Determine the Coefficient of Friction**:
- The angle of repose at 30 degrees gives us the coefficient of friction \( \mu \).
- \( \mu = \tan(30^\circ) = \frac{1}{\sqrt{3}} \).

3. **Analyze Forces on the Inclined Plane**:
- When the plane is at 60 degrees, the forces acting on the body are:
- The weight component along the incline: \( W_{\parallel} = mg \sin(60^\circ) \).
- The normal force: \( N = mg \cos(60^\circ) \).
- The frictional force opposing the motion: \( F_f = \mu N \).

4. **Calculate the Weight Components**:
- Calculate \( W_{\parallel} \):
\[ W_{\parallel} = mg \sin(60^\circ) = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{N} \]
- Calculate the normal force \( N \):
\[ N = mg \cos(60^\circ) = 60 \times \frac{1}{2} = 30 \, \text{N} \]

5. **Calculate the Frictional Force**:
- Calculate the frictional force \( F_f \):
\[ F_f = \mu N = \frac{1}{\sqrt{3}} \times 30 = \frac{30}{\sqrt{3}} = 10\sqrt{3} \, \text{N} \]

6. **Set Up the Equation for Equilibrium**:
- For the body to be in equilibrium, the applied force \( F \) along the incline must balance the weight component down the incline minus the frictional force:
\[ F + F_f = W_{\parallel} \]
- Rearranging gives:
\[ F = W_{\parallel} - F_f = 30\sqrt{3} - 10\sqrt{3} = 20\sqrt{3} \, \text{N} \]

7. **Convert Force to kg**:
- Since 1 N = 0.1 kg, we convert the force:
\[ F = \frac{20\sqrt{3}}{10} = 2\sqrt{3} \, \text{kg} \]

### Final Answer:
The force required to support the body when the plane is raised to a slope of 60 degrees is \( 2\sqrt{3} \, \text{kg} \).