In a square cut, the speed of the cricket ball changes from `30 ms^(-1)` to `40 ms^(-1)` during the time of its contact `Delta t = 0 .01 s` with the bat. If the ball is deflected by the bat through an angle of `theta = 90^@`, find the magnitude of the average acceleration `("in" xx 10^(2) ms^(-2))` of the ball during the square cut.

Explanation

To solve the problem, we need to find the average acceleration of the cricket ball during the square cut. The ball's speed changes from \(30 \, \text{ms}^{-1}\) to \(40 \, \text{ms}^{-1}\) over a time interval of \(0.01 \, \text{s}\), and it is deflected at an angle of \(90^\circ\).

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities**:
- Initial velocity \(v_1 = 30 \, \text{ms}^{-1}\)
- Final velocity \(v_2 = 40 \, \text{ms}^{-1}\)

2. **Determine the Change in Velocity (\(\Delta V\))**:
Since the velocities are at a \(90^\circ\) angle to each other, we can use the Pythagorean theorem to find the magnitude of the change in velocity:
\[ \Delta V = \sqrt{v_1^2 + v_2^2} \]
Plugging in the values:
\[ \Delta V = \sqrt{(30)^2 + (40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{ms}^{-1} \]

3. **Calculate Average Acceleration**:
The average acceleration \(a\) is given by the formula:
\[ a = \frac{\Delta V}{\Delta t} \]
Where \(\Delta t = 0.01 \, \text{s}\). Substituting the values:
\[ a = \frac{50 \, \text{ms}^{-1}}{0.01 \, \text{s}} = 5000 \, \text{ms}^{-2} \]

4. **Express in Required Format**:
We need to express the answer in the form of \(a \times 10^2 \, \text{ms}^{-2}\):
\[ a = 5000 \, \text{ms}^{-2} = 50 \times 10^2 \, \text{ms}^{-2} \]

### Final Answer:
The magnitude of the average acceleration of the ball during the square cut is:
\[ \boxed{50} \]