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`

Answer

To solve the problem, we need to find the coefficient of friction (μ) between the block and the inclined plane when the block just begins to slide down.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block (m) = 5 kg
- Inclination of the plane (θ) = 3 in 5
- Acceleration due to gravity (g) = 10 m/s²

2. **Understanding the Inclination:**
- The inclination of 3 in 5 can be interpreted as a right triangle where the opposite side (height) is 3 units and the hypotenuse is 5 units.
- Using Pythagoras theorem, we can find the adjacent side (base):
\[ \text{Base} = \sqrt{(5^2 - 3^2)} = \sqrt{25 - 9} = \sqrt{16} = 4 \]

3. **Calculate the Angle θ:**
- From the triangle, we can find:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \]

4. **Setting Up the Forces:**
- The forces acting on the block are:
- Weight (mg) acting downwards.
- Normal force (N) acting perpendicular to the inclined plane.
- Frictional force (f) acting up the incline.

5. **Resolving the Weight:**
- The weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos(\theta) \)
- Parallel to the incline: \( mg \sin(\theta) \)

6. **Applying Newton's Second Law:**
- At the point of impending motion (just before sliding), the frictional force equals the component of weight down the incline:
\[ f = mg \sin(\theta) \]
- The maximum static frictional force is given by:
\[ f = \mu N \]
- The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos(\theta) \]

7. **Equating the Forces:**
- At the point of sliding:
\[ \mu N = mg \sin(\theta) \]
- Substituting for N:
\[ \mu (mg \cos(\theta)) = mg \sin(\theta) \]
- Dividing both sides by \( mg \):
\[ \mu \cos(\theta) = \sin(\theta) \]
- Rearranging gives:
\[ \mu = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \]

8. **Calculating the Coefficient of Friction:**
- From our earlier calculation, we found:
\[ \tan(\theta) = \frac{3}{4} \]
- Thus, the coefficient of friction (μ) is:
\[ \mu = \frac{3}{4} \]

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