When an object is shot from the bottom of a long smooth inclined plane kept at an angle `60^(@)` with horizontal, it can travel a distance `x_(1)` along the plane. But when the inclination is decreased to `30^(@)` and the same object is shot with the same velocity, it can travel `x_(2)` distance. Then `x_(1):x_(2)` will be :

Options

A. `(a)1:2sqrt(3)`
B. `(b)1:sqrt(2)`
C. `(c)sqrt(2):1`
D. `(d)1:sqrt(3)`

Explanation

To solve the problem, we need to find the ratio \( x_1 : x_2 \) where \( x_1 \) is the distance traveled by the object on a smooth inclined plane at an angle of \( 60^\circ \) and \( x_2 \) is the distance traveled at an angle of \( 30^\circ \) when the object is shot with the same initial velocity \( u \).

### Step-by-step Solution:

1. **Identify the Forces Acting on the Object:**
- When the object is on the inclined plane, the gravitational force acting on it can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \) (normal force)
- Parallel to the incline: \( mg \sin \theta \) (causes acceleration down the incline)

2. **Determine the Acceleration:**
- The acceleration \( a \) of the object along the incline can be expressed as:
\[ a = -g \sin \theta \]
- The negative sign indicates that the acceleration is directed down the incline.

3. **Use the Equation of Motion:**
- We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity (which is \( 0 \) when the object comes to rest), \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance traveled along the incline.

4. **Calculate the Distance \( x_1 \) for \( 60^\circ \):**
- For the first case (angle \( 60^\circ \)):
\[ 0 = u^2 - 2g \sin(60^\circ) x_1 \]
\[ x_1 = \frac{u^2}{2g \sin(60^\circ)} \]

5. **Calculate the Distance \( x_2 \) for \( 30^\circ \):**
- For the second case (angle \( 30^\circ \)):
\[ 0 = u^2 - 2g \sin(30^\circ) x_2 \]
\[ x_2 = \frac{u^2}{2g \sin(30^\circ)} \]

6. **Find the Ratio \( \frac{x_1}{x_2} \):**
- Now, we can find the ratio \( \frac{x_1}{x_2} \):
\[ \frac{x_1}{x_2} = \frac{\frac{u^2}{2g \sin(60^\circ)}}{\frac{u^2}{2g \sin(30^\circ)}} \]
- The \( u^2 \) and \( 2g \) cancel out:
\[ \frac{x_1}{x_2} = \frac{\sin(30^\circ)}{\sin(60^\circ)} \]

7. **Substitute the Values of Sine:**
- We know:
\[ \sin(30^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
- Therefore:
\[ \frac{x_1}{x_2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

8. **Final Ratio:**
- Thus, the ratio \( x_1 : x_2 \) is:
\[ x_1 : x_2 = 1 : \sqrt{3} \]