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. `K_(1)= K_(2)`
B. `K_(1) gt K_(2)`
C. `K_(1) lt K_(2)`
D. Data insufficient

Answer

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. The key points to consider are the work done by gravity and the work done against friction.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Height of both inclined planes, \( h \)
- Angles of inclination: \( \theta_1 = 30^\circ \) and \( \theta_2 = 60^\circ \)
- Coefficient of friction, \( \mu \) (same for both planes)

2. **Calculate the Work Done by Gravity:**
The work done by gravity when the particle descends from height \( h \) is given by:
\[ W_{\text{gravity}} = mgh \]
This value is the same for both planes since the height \( h \) is the same.

3. **Determine the Distance Traveled Along the Inclined Planes:**
The distance \( X \) traveled along the inclined plane can be calculated using the relationship:
\[ X = \frac{h}{\sin \theta} \]
- For \( \theta_1 = 30^\circ \):
\[ X_1 = \frac{h}{\sin 30^\circ} = \frac{h}{\frac{1}{2}} = 2h \]
- For \( \theta_2 = 60^\circ \):
\[ X_2 = \frac{h}{\sin 60^\circ} = \frac{h}{\frac{\sqrt{3}}{2}} = \frac{2h}{\sqrt{3}} \]

4. **Calculate the Work Done Against Friction:**
The frictional force acting on the particle is given by:
\[ F_{\text{friction}} = \mu mg \cos \theta \]
The work done against friction is:
\[ W_{\text{friction}} = -F_{\text{friction}} \cdot X \]
- For \( \theta_1 = 30^\circ \):
\[ W_{\text{friction},1} = -\mu mg \cos 30^\circ \cdot 2h = -\mu mg \cdot \frac{\sqrt{3}}{2} \cdot 2h = -\mu mg \sqrt{3} h \]
- For \( \theta_2 = 60^\circ \):
\[ W_{\text{friction},2} = -\mu mg \cos 60^\circ \cdot \frac{2h}{\sqrt{3}} = -\mu mg \cdot \frac{1}{2} \cdot \frac{2h}{\sqrt{3}} = -\frac{\mu mg h}{\sqrt{3}} \]

5. **Calculate the Kinetic Energy at the Bottom of the Inclined Planes:**
Using the work-energy principle, the kinetic energy \( K \) at the bottom of each plane can be expressed as:
\[ K = W_{\text{gravity}} + W_{\text{friction}} \]
- For \( K_1 \):
\[ K_1 = mgh - \mu mg \sqrt{3} h = mg(h - \mu \sqrt{3} h) = mg(1 - \mu \sqrt{3})h \]
- For \( K_2 \):
\[ K_2 = mgh - \frac{\mu mg h}{\sqrt{3}} = mg(h - \frac{\mu h}{\sqrt{3}}) = mg(1 - \frac{\mu}{\sqrt{3}})h \]

6. **Compare \( K_1 \) and \( K_2 \):**
To determine which kinetic energy is greater, we compare the terms:
- Since \( \sqrt{3} > 1 \), it follows that:
\[ \mu \sqrt{3} > \mu \]
Thus,
\[ 1 - \mu \sqrt{3} < 1 - \frac{\mu}{\sqrt{3}} \]
Therefore, we conclude that:
\[ K_1 < K_2 \]

### Final Conclusion:
The kinetic energy of the particle at the bottom of the inclined plane with \( 30^\circ \) inclination is less than that at the bottom of the plane with \( 60^\circ \) inclination:
\[ K_1 < K_2 \]