A rectangular loop of sides `8cm` and `2 cm` with a small cut is moving out of a region of uniform magnetic field of magnitude `0.3` T directed normal to the loop. The velocity of the loop is `1cm s^(-1)` in the direction normal 'to the (i) longer side, (ii) shorter side of the loop. If the ratio of voltage induced in case (i) to case (ii) is `alpha` and the ratio of time for which voltage induced in case (ti) to case (i) is `beta`, then calculate `(alpha times beta)`.

Answer

To solve the problem, we need to calculate the induced voltage in two different cases and the time for which the voltage is induced. Then we will find the product of the ratios of voltage and time.

### Step-by-step Solution:

1. **Identify the dimensions of the loop and the magnetic field:**
- The rectangular loop has dimensions:
- Length (longer side) = 8 cm = 0.08 m
- Width (shorter side) = 2 cm = 0.02 m
- The magnetic field (B) = 0.3 T, directed normal to the loop.

2. **Calculate the induced voltage for case (i) - moving normal to the longer side:**
- In this case, the width of the loop (2 cm) is moving out of the magnetic field.
- The induced voltage (V1) can be calculated using the formula:
\[ V_1 = B \cdot v \cdot l \]
where \( l \) is the length of the side moving out of the field (which is the width of the loop).
- Substituting the values:
\[ V_1 = 0.3 \, \text{T} \cdot 0.01 \, \text{m/s} \cdot 0.02 \, \text{m} = 0.00006 \, \text{V} \]

3. **Calculate the induced voltage for case (ii) - moving normal to the shorter side:**
- In this case, the length of the loop (8 cm) is moving out of the magnetic field.
- The induced voltage (V2) can be calculated using the same formula:
\[ V_2 = B \cdot v \cdot L \]
where \( L \) is the length of the side moving out of the field (which is the longer side).
- Substituting the values:
\[ V_2 = 0.3 \, \text{T} \cdot 0.01 \, \text{m/s} \cdot 0.08 \, \text{m} = 0.00024 \, \text{V} \]

4. **Calculate the ratio of the induced voltages (alpha):**
\[ \alpha = \frac{V_1}{V_2} = \frac{0.00006}{0.00024} = \frac{1}{4} \]

5. **Calculate the time for which the voltage is induced in both cases:**
- For case (i), the time (T1) taken for the shorter side to exit the magnetic field:
\[ T_1 = \frac{\text{Width}}{\text{Velocity}} = \frac{0.02 \, \text{m}}{0.01 \, \text{m/s}} = 2 \, \text{s} \]
- For case (ii), the time (T2) taken for the longer side to exit the magnetic field:
\[ T_2 = \frac{\text{Length}}{\text{Velocity}} = \frac{0.08 \, \text{m}}{0.01 \, \text{m/s}} = 8 \, \text{s} \]

6. **Calculate the ratio of the times (beta):**
\[ \beta = \frac{T_2}{T_1} = \frac{8}{2} = 4 \]

7. **Calculate the product (alpha times beta):**
\[ \alpha \times \beta = \frac{1}{4} \times 4 = 1 \]

### Final Answer:
\[ \alpha \times \beta = 1 \]