The direction of motion of a projectile at a certain instant is inclined at an angle `60^(@)` to the horizontal. After `sqrt(3)` seconds it is inclined an angle `30^(@)`. If the horizontal compound of velocity of projection is `3n` (in `m//sec)` then value of n is : `(take g = 10 m//s^(2))`

Explanation

To solve the problem, we need to analyze the projectile motion of the object and use the given angles and time to find the value of \( n \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The projectile's direction changes from \( 60^\circ \) to \( 30^\circ \) after \( \sqrt{3} \) seconds.
- The horizontal component of the velocity of projection is given as \( 3n \) m/s.

2. **Components of Initial Velocity**:
- Let the initial velocity of projection be \( u \).
- The horizontal component of the initial velocity \( u_x = u \cos(60^\circ) = \frac{u}{2} \).
- The vertical component of the initial velocity \( u_y = u \sin(60^\circ) = \frac{\sqrt{3}u}{2} \).

3. **Vertical Velocity After \( \sqrt{3} \) Seconds**:
- The vertical velocity \( v_y \) after \( \sqrt{3} \) seconds can be calculated using the formula:
\[ v_y = u_y - g t \]
- Substituting the values:
\[ v_y = \frac{\sqrt{3}u}{2} - 10 \cdot \sqrt{3} \]

4. **Using the Angle at \( \sqrt{3} \) Seconds**:
- At \( \sqrt{3} \) seconds, the projectile is inclined at \( 30^\circ \).
- The tangent of the angle gives us the relationship between the vertical and horizontal components of the velocity:
\[ \tan(30^\circ) = \frac{v_y}{u_x} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{\frac{\sqrt{3}u}{2} - 10\sqrt{3}}{\frac{u}{2}} \]

5. **Cross-Multiplying and Simplifying**:
- Cross-multiplying gives:
\[ u - 20 = \sqrt{3} \left( \frac{\sqrt{3}u}{2} - 10\sqrt{3} \right) \]
- Simplifying further:
\[ u - 20 = \frac{3u}{2} - 30 \]
- Rearranging the equation:
\[ 2u - 40 = 3u - 60 \]
- This simplifies to:
\[ 60 - 40 = 3u - 2u \implies u = 20 \text{ m/s} \]

6. **Finding the Horizontal Component**:
- The horizontal component of the initial velocity is:
\[ u_x = \frac{u}{2} = \frac{20}{2} = 10 \text{ m/s} \]

7. **Relating to \( 3n \)**:
- We know \( u_x = 3n \):
\[ 10 = 3n \]
- Solving for \( n \):
\[ n = \frac{10}{3} \approx 3.33 \]

### Final Answer:
The value of \( n \) is \( \frac{10}{3} \).