A cubical block rests on a plane of `mu= sqrt(3)` . The angle through which the plane be inclined to the horizontal so that the block just slides down will be -

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem of determining the angle at which a cubical block will just slide down an inclined plane with a coefficient of friction \( \mu = \sqrt{3} \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
- The block experiences two main forces: its weight (\( mg \)) acting downward and the normal force (\( N \)) acting perpendicular to the inclined plane.
- The weight can be resolved into two components: one parallel to the incline (\( mg \sin \theta \)) and one perpendicular to the incline (\( mg \cos \theta \)).

### Step 2: Write the Expression for the Forces
- The frictional force (\( F_f \)) acting against the motion of the block is given by:
\[ F_f = \mu N \]
- The normal force can be expressed as:
\[ N = mg \cos \theta \]
- Therefore, the frictional force becomes:
\[ F_f = \mu mg \cos \theta \]

### Step 3: Set Up the Inequality for Motion
- For the block to just start sliding down, the component of the gravitational force along the incline must equal the frictional force:
\[ mg \sin \theta = F_f \]
- Substituting the expression for the frictional force, we have:
\[ mg \sin \theta = \mu mg \cos \theta \]

### Step 4: Simplify the Equation
- We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

### Step 5: Substitute the Value of \( \mu \)
- Given \( \mu = \sqrt{3} \), we substitute this into the equation:
\[ \sin \theta = \sqrt{3} \cos \theta \]

### Step 6: Divide Both Sides by \( \cos \theta \)
- This gives:
\[ \tan \theta = \sqrt{3} \]

### Step 7: Find the Angle \( \theta \)
- The angle \( \theta \) for which \( \tan \theta = \sqrt{3} \) is:
\[ \theta = 60^\circ \]

### Conclusion
- Therefore, the angle through which the plane must be inclined for the block to just slide down is \( \theta = 60^\circ \).