A wooden block sliding down from the top of a smooth inclined plane starting from rest takes `t_(1)`, seconds to reach the bottom of the plane and attains velocity `V_(1)`. Another block of twice the mass falling freely from the same height takes `t_(2)` sec. to reach the bottom of the plane and attains `V_(2)`. If angle of inclination of the plane is `30^(@)`.

Options

A. `v_(1)=v_(2) & t_(1)=t_(2)`
B. `v_(1)=v_(2)& t_(1)=2t_(2)`
C. `v_(1)gt v_(2) & t_(1) gt t_(2)`
D. `v_(1)=v_(2)& t_(2)=2t_(1)`

Answer

To solve the problem, we need to analyze the motion of two blocks: one sliding down a smooth inclined plane and the other falling freely from the same height.

### Step-by-Step Solution:

1. **Define the Problem**:
- We have a wooden block sliding down a smooth inclined plane starting from rest. It takes time \( t_1 \) seconds to reach the bottom and attains velocity \( V_1 \).
- Another block of twice the mass falls freely from the same height and takes time \( t_2 \) seconds to reach the bottom, attaining velocity \( V_2 \).
- The angle of inclination of the plane is \( 30^\circ \).

2. **Analyze the Block on the Inclined Plane**:
- The height of the inclined plane is \( h \).
- The length of the inclined plane \( l \) can be related to the height by \( h = l \cos(30^\circ) \).
- The acceleration \( a \) of the block on the incline can be derived from the gravitational force component acting along the incline:
\[ a = g \sin(30^\circ) = g \cdot \frac{1}{2} = \frac{g}{2} \]

3. **Use Kinematic Equations**:
- Since the block starts from rest, we can use the kinematic equation:
\[ V_1^2 = u^2 + 2as \]
Here, \( u = 0 \), \( a = \frac{g}{2} \), and \( s = l \):
\[ V_1^2 = 0 + 2 \cdot \frac{g}{2} \cdot l = g \cdot l \]
- Thus, we have:
\[ V_1 = \sqrt{g \cdot l} \]

4. **Relate Length \( l \) to Height \( h \)**:
- From the relationship \( h = l \cos(30^\circ) \), we can express \( l \):
\[ l = \frac{h}{\cos(30^\circ)} = \frac{h}{\frac{\sqrt{3}}{2}} = \frac{2h}{\sqrt{3}} \]
- Substituting \( l \) back into the equation for \( V_1 \):
\[ V_1 = \sqrt{g \cdot \frac{2h}{\sqrt{3}}} = \sqrt{\frac{2gh}{\sqrt{3}}} \]

5. **Analyze the Free-Falling Block**:
- For the block falling freely from height \( h \):
\[ V_2^2 = u^2 + 2gh \]
Here, \( u = 0 \):
\[ V_2^2 = 0 + 2gh \implies V_2 = \sqrt{2gh} \]

6. **Determine Time Taken for Each Block**:
- For the inclined block:
\[ V_1 = u + at_1 \implies V_1 = 0 + \frac{g}{2} t_1 \implies t_1 = \frac{2V_1}{g} \]
- For the free-falling block:
\[ V_2 = u + gt_2 \implies V_2 = 0 + gt_2 \implies t_2 = \frac{V_2}{g} \]

7. **Relate \( t_1 \) and \( t_2 \)**:
- From the expressions for \( t_1 \) and \( t_2 \):
\[ t_1 = \frac{2V_1}{g} \quad \text{and} \quad t_2 = \frac{V_2}{g} \]
- Since \( V_1 = \sqrt{\frac{2gh}{\sqrt{3}}} \) and \( V_2 = \sqrt{2gh} \), we can find the relationship between \( t_1 \) and \( t_2 \).

8. **Final Relationships**:
- From the calculations, we find:
\[ V_1 = V_2 \quad \text{and} \quad t_1 = 2t_2 \]

### Conclusion:
- The correct relationships are \( V_1 = V_2 \) and \( t_1 = 2t_2 \).