A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased till the block just begins to slide down. The inclination of the plane is than 3 in 5. The coefficient of friction between the block and the plane is (Take, `g = 10m//s^(2)`)

Options

A. `(3)/(5)`
B. `(3)/(4)`
C. `(4)/(5)`
D. `(2)/(3)`

Explanation

To solve the problem, we need to determine the coefficient of friction (μ) between the block and the inclined plane when the block just begins to slide down. The inclination of the plane is given as 3 in 5, which we can interpret as a right triangle where the height is 3 units and the base is 4 units.

### Step-by-Step Solution:

1. **Understand the Geometry of the Incline:**
- The incline forms a right triangle where the opposite side (height) is 3, and the adjacent side (base) is 4.
- We can find the angle of inclination (θ) using the sine function:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5} \]

2. **Calculate the Angle θ:**
- From the sine value, we can also find the cosine value:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \]

3. **Set Up the Forces Acting on the Block:**
- The forces acting on the block include:
- Gravitational force (mg) acting downwards.
- Normal force (N) acting perpendicular to the inclined plane.
- Frictional force (F_f) acting up the incline when the block is about to slide down.

4. **Write the Equations for Forces:**
- The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\text{parallel}} = mg \sin \theta \)
- Perpendicular to the incline: \( F_{\text{perpendicular}} = mg \cos \theta \)

5. **Express the Normal Force:**
- The normal force (N) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

6. **Frictional Force at the Point of Sliding:**
- The frictional force at the point of sliding is given by:
\[ F_f = \mu N = \mu (mg \cos \theta) \]

7. **Set Up the Equation for Equilibrium:**
- At the point of sliding, the frictional force equals the component of gravitational force parallel to the incline:
\[ mg \sin \theta = \mu (mg \cos \theta) \]

8. **Cancel out mg from both sides:**
- Since mass (m) and gravitational acceleration (g) are common on both sides, we can simplify:
\[ \sin \theta = \mu \cos \theta \]

9. **Rearranging for μ:**
- Rearranging gives:
\[ \mu = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

10. **Calculate the Coefficient of Friction:**
- From the triangle, we know:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4} \]
- Therefore, the coefficient of friction (μ) is:
\[ \mu = \frac{3}{4} \]

### Final Answer:
The coefficient of friction between the block and the inclined plane is \( \frac{3}{4} \).