Two taps `A` and `B` supply water at temperature `10^(@)C` and `50^(@)C` respectively . Tap `A` alone fills the tank in `1` hour and tap `B` alone fills the tank in `3` hour. If we open both the taps together in the empty tank , if the final temperature of the water in the completely filled tank is found to be `5alpha ("in" "^(@)C)` . Find the value of `alpha`. Neglect loss of heat to surrounding and heat capacity of the tank.

Explanation

To solve the problem, we will use the principle of heat exchange and the concept of filling rates of the taps.

### Step-by-step Solution:

1. **Determine the filling rates of the taps:**
- Tap A fills the tank in 1 hour, so its rate of filling is \( R_A = 1 \) tank/hour.
- Tap B fills the tank in 3 hours, so its rate of filling is \( R_B = \frac{1}{3} \) tank/hour.

2. **Calculate the combined filling rate when both taps are open:**
- The combined rate when both taps are open is:
\[ R_{total} = R_A + R_B = 1 + \frac{1}{3} = \frac{4}{3} \text{ tanks/hour} \]

3. **Determine the time taken to fill the tank when both taps are open:**
- Since the combined rate is \( \frac{4}{3} \) tanks/hour, the time taken to fill 1 tank is:
\[ t = \frac{1 \text{ tank}}{R_{total}} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \text{ hours} \]

4. **Calculate the volumes of water supplied by each tap:**
- Let the volume of the tank be \( V \).
- Volume supplied by tap A in \( \frac{3}{4} \) hours:
\[ V_A = R_A \times t = 1 \times \frac{3}{4} = \frac{3}{4} V \]
- Volume supplied by tap B in \( \frac{3}{4} \) hours:
\[ V_B = R_B \times t = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} V \]

5. **Calculate the final temperature of the water in the tank:**
- The temperature of water from tap A is \( T_A = 10^\circ C \) and from tap B is \( T_B = 50^\circ C \).
- The final temperature \( T_f \) can be calculated using the formula for the weighted average temperature:
\[ T_f = \frac{(V_A \cdot T_A) + (V_B \cdot T_B)}{V_A + V_B} \]
- Substituting the values:
\[ T_f = \frac{\left(\frac{3}{4} V \cdot 10\right) + \left(\frac{1}{4} V \cdot 50\right)}{V} \]
- Simplifying:
\[ T_f = \frac{\frac{30}{4} + \frac{50}{4}}{1} = \frac{80}{4} = 20^\circ C \]

6. **Relate the final temperature to the given expression:**
- It is given that the final temperature is \( 5\alpha \):
\[ 5\alpha = 20 \]
- Solving for \( \alpha \):
\[ \alpha = \frac{20}{5} = 4 \]

### Final Answer:
The value of \( \alpha \) is \( 4 \).