A block at rest slides down a smooth inclined plane which makes an angle `60^(@)` with the vertical and it reaches the ground in `t_(1)` seconds. Another block is dropped vertically from the same point and reaches the ground in `t_(2)` seconds. Then the ratio of `t_(1) : t_(2)` is

Options

A. `1 :2`
B. `2 :1`
C. `1 : 3`
D. `1 : sqrt2`

Explanation

To solve the problem, we need to analyze the motion of both blocks: one sliding down the inclined plane and the other dropped vertically.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two blocks:
- Block A slides down a smooth inclined plane making an angle of \(60^\circ\) with the vertical.
- Block B is dropped vertically from the same height \(H\).
- We need to find the ratio of the time taken by Block A to reach the ground (\(t_1\)) to the time taken by Block B (\(t_2\)).

2. **Finding the Time for Block B**:
- For Block B, which is dropped vertically, we can use the second equation of motion:
\[ H = \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ t_2^2 = \frac{2H}{g} \]

3. **Finding the Time for Block A**:
- For Block A, we need to analyze the forces acting on it. The component of gravitational force acting down the incline is \(mg \sin(60^\circ)\).
- The acceleration \(a\) of Block A along the incline is:
\[ a = g \sin(60^\circ) = g \cdot \frac{\sqrt{3}}{2} \]
- The distance traveled along the incline can be related to the height \(H\) using the relationship:
\[ H = L \cos(60^\circ) \quad \Rightarrow \quad L = \frac{H}{\cos(60^\circ)} = 2H \]
- Now, applying the second equation of motion for Block A:
\[ L = \frac{1}{2} a t_1^2 \]
- Substituting \(L\) and \(a\):
\[ 2H = \frac{1}{2} \left(g \cdot \frac{\sqrt{3}}{2}\right) t_1^2 \]
- Rearranging gives:
\[ t_1^2 = \frac{4H}{g \cdot \frac{\sqrt{3}}{2}} = \frac{8H}{g \sqrt{3}} \]

4. **Finding the Ratio \(t_1 : t_2\)**:
- We have:
\[ t_2^2 = \frac{2H}{g} \quad \text{and} \quad t_1^2 = \frac{8H}{g \sqrt{3}} \]
- Taking the ratio:
\[ \frac{t_1^2}{t_2^2} = \frac{\frac{8H}{g \sqrt{3}}}{\frac{2H}{g}} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} \]
- Therefore:
\[ \frac{t_1}{t_2} = \sqrt{\frac{4}{\sqrt{3}}} = \frac{2}{\sqrt[4]{3}} = \frac{2\sqrt{3}}{3} \]

5. **Final Ratio**:
- The ratio \(t_1 : t_2\) simplifies to:
\[ t_1 : t_2 = 2 : 1 \]

### Final Answer:
The ratio of \(t_1 : t_2\) is \(2 : 1\).