A partical is realeased from the top of two inclined rought surface of height `h` each. The angle of inclination of the two planes are `30^(@)` and `60^(@)` respectively. All other factors (e.g. coefficient of friction , mass of the block etc) are same in both the cases. Let `K_(1) and K_(2)` be the kinetic energy of the partical at the bottom of the plane in two cases. Then

Options

A. (a) `K_1=K_2`
B. (b) `K_1gtK_2`
C. (c) `K_1ltK_2`
D. (d) Data insufficient

Explanation

To solve the problem, we need to analyze the motion of a particle released from the top of two inclined planes with different angles of inclination (30° and 60°). We will derive the expressions for the kinetic energy at the bottom of each incline and compare them.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle:**
- When the particle is on the inclined plane, the gravitational force \( mg \) can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

2. **Determine the Normal Force:**
- The normal force \( N \) acting on the particle is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

3. **Calculate the Frictional Force:**
- The frictional force \( F_f \) acting against the motion of the particle is given by:
\[ F_f = \mu N = \mu mg \cos \theta \]

4. **Work Done by Gravity:**
- The work done by gravity as the particle moves down the incline is:
\[ W_g = mgh \]
- This is the same for both inclines since they have the same height \( h \).

5. **Work Done by Friction:**
- The distance \( s \) traveled along the incline can be related to the height \( h \) and the angle \( \theta \) using the relationship:
\[ s = \frac{h}{\sin \theta} \]
- Therefore, the work done by friction \( W_f \) is:
\[ W_f = -F_f \cdot s = -\mu mg \cos \theta \cdot \frac{h}{\sin \theta} \]

6. **Apply the Work-Energy Principle:**
- According to the work-energy principle, the change in kinetic energy \( K \) is given by:
\[ K = W_g + W_f \]
- Substituting the expressions for work done:
\[ K = mgh - \left(-\mu mg \cos \theta \cdot \frac{h}{\sin \theta}\right) \]
- Simplifying this gives:
\[ K = mgh - \mu mg \frac{h \cos \theta}{\sin \theta} \]
- Thus, we can express the kinetic energy at the bottom of the incline for both angles.

7. **Calculate Kinetic Energies for Each Incline:**
- For \( \theta = 30^\circ \):
\[ K_1 = mgh - \mu mg \frac{h \cos 30^\circ}{\sin 30^\circ} = mgh - \mu mg \frac{h \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = mgh - 2\mu mg h \cdot \frac{\sqrt{3}}{2} \]
- For \( \theta = 60^\circ \):
\[ K_2 = mgh - \mu mg \frac{h \cos 60^\circ}{\sin 60^\circ} = mgh - \mu mg \frac{h \cdot \frac{1}{2}}{\frac{\sqrt{3}}{2}} = mgh - \frac{\mu mg h}{\sqrt{3}} \]

8. **Compare Kinetic Energies:**
- Since both expressions for kinetic energy depend on the frictional work done, we can analyze the coefficients:
- The term \( 2\mu mg h \cdot \frac{\sqrt{3}}{2} \) for \( K_1 \) is greater than \( \frac{\mu mg h}{\sqrt{3}} \) for \( K_2 \).
- Therefore, it follows that:
\[ K_1 < K_2 \]

### Conclusion:
The kinetic energy at the bottom of the incline for the 30° angle is less than that for the 60° angle:
\[ K_1 < K_2 \]