Liquids A and B are at `30^(@)` C and `20^(@)` C, respectively. When mixed in equal masses,the temperature of the mixture is found to be `26^(@)` C, The specific heats of A and B are in the ratio of `m:n`, where m and n are integers, then find minimum value of `m+n`.

Options

A. 5
B. 2
C. 6
D. 7

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hotter liquid (A) will be equal to the heat gained by the cooler liquid (B) when they are mixed.

### Step-by-Step Solution:

1. **Identify the given data:**
- Temperature of liquid A, \( T_A = 30^\circ C \)
- Temperature of liquid B, \( T_B = 20^\circ C \)
- Final temperature of the mixture, \( T_f = 26^\circ C \)
- Let the mass of each liquid be \( m \).

2. **Calculate the heat lost by liquid A:**
- The heat lost by liquid A when it cools from \( 30^\circ C \) to \( 26^\circ C \) is given by:
\[ Q_A = m \cdot S_A \cdot (T_A - T_f) = m \cdot S_A \cdot (30 - 26) = m \cdot S_A \cdot 4 \]

3. **Calculate the heat gained by liquid B:**
- The heat gained by liquid B when it warms up from \( 20^\circ C \) to \( 26^\circ C \) is given by:
\[ Q_B = m \cdot S_B \cdot (T_f - T_B) = m \cdot S_B \cdot (26 - 20) = m \cdot S_B \cdot 6 \]

4. **Set up the equation using the conservation of energy:**
- According to the principle of conservation of energy:
\[ Q_A = Q_B \]
Substituting the expressions for \( Q_A \) and \( Q_B \):
\[ m \cdot S_A \cdot 4 = m \cdot S_B \cdot 6 \]

5. **Cancel out the mass \( m \) from both sides:**
\[ S_A \cdot 4 = S_B \cdot 6 \]

6. **Rearranging the equation gives us the ratio of specific heats:**
\[ \frac{S_A}{S_B} = \frac{6}{4} = \frac{3}{2} \]

7. **Identify the integers \( m \) and \( n \):**
- From the ratio \( \frac{S_A}{S_B} = \frac{m}{n} = \frac{3}{2} \), we can identify \( m = 3 \) and \( n = 2 \).

8. **Calculate \( m + n \):**
\[ m + n = 3 + 2 = 5 \]

### Final Answer:
The minimum value of \( m + n \) is \( 5 \).