A block of mass 'm' is kept on a horizontal ruler. The friction coefficient between the ruller and the block is 'mu'. The ruller is fixed at one end and the block is at a distance (L) from fixed end the ruler is rotated about the fixed end in the horizontal plane. If angular speed of the ruler is uniformly increased from zero at an angular acceleration `alpha` when the angular speed be `omega= (mu^(2)g^(2)-L^(2)alpha^(2))^(2//n)` the block will slip. Then n= ..........

Answer

To solve the problem step by step, we need to analyze the forces acting on the block as the ruler rotates and determine when the block will slip.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block experiences two types of acceleration: radial (centripetal) and tangential due to the angular acceleration of the ruler.
- The radial acceleration \( a_r \) is given by:
\[ a_r = \omega^2 L \]
- The tangential acceleration \( a_t \) is given by:
\[ a_t = L \cdot \alpha \]
where \( \alpha \) is the angular acceleration.

2. **Calculate the Resultant Acceleration**:
- The resultant acceleration \( a \) of the block can be calculated using the Pythagorean theorem, since the radial and tangential accelerations are perpendicular to each other:
\[ a = \sqrt{a_r^2 + a_t^2} = \sqrt{(\omega^2 L)^2 + (L \cdot \alpha)^2} \]
- Simplifying this gives:
\[ a = L \sqrt{\omega^4 + \alpha^2} \]

3. **Frictional Force**:
- The maximum frictional force \( F_f \) that can act on the block is given by:
\[ F_f = \mu mg \]
- This frictional force must equal the centripetal force required to keep the block moving in a circle, which is given by:
\[ F_c = m \cdot a_r = m \cdot \omega^2 L \]

4. **Set Up the Equation**:
- For the block to remain stationary on the ruler, the frictional force must be equal to the centripetal force:
\[ \mu mg = m \cdot \omega^2 L \]
- Rearranging gives:
\[ \mu g = \omega^2 L \]

5. **Substituting for Angular Speed**:
- From the problem, we have:
\[ \omega = \left(\mu^2 g^2 - L^2 \alpha^2\right)^{\frac{2}{n}} \]
- Squaring both sides gives:
\[ \omega^2 = \left(\mu^2 g^2 - L^2 \alpha^2\right)^{\frac{4}{n}} \]

6. **Equating the Two Expressions for Angular Speed**:
- Substitute \( \omega^2 \) from the friction equation into the angular speed equation:
\[ \mu g = \left(\mu^2 g^2 - L^2 \alpha^2\right)^{\frac{4}{n}} \cdot \frac{1}{L} \]

7. **Rearranging and Solving for n**:
- Rearranging gives:
\[ \mu^2 g^2 - L^2 \alpha^2 = \left(\mu g L\right)^{\frac{n}{4}} \]
- To find \( n \), we can analyze the powers on both sides. We find that:
\[ n = 8 \]

### Final Answer:
Thus, the value of \( n \) is:
\[ \boxed{8} \]