Two wires A and B of same material and mass have their lengths in the ratio `1 : 2`. On connecting them to the same source, the rate of heat dissipation in B is found to be 5 W . The rate of heat dissipation in A is

Options

A. 10W
B. 5W
C. 20W
D. None of these

Answer

To solve the problem step by step, we will analyze the relationship between the lengths of the wires, their resistances, and the power dissipated in each wire.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two wires, A and B, made of the same material and having the same mass.
- The lengths of the wires are in the ratio \(1:2\). Let's denote the length of wire A as \(L_A\) and the length of wire B as \(L_B\). Therefore, we can write:
\[ L_A = L \quad \text{and} \quad L_B = 2L \]

2. **Using the Formula for Power Dissipation:**
- The power dissipated in a resistor (or wire) is given by:
\[ P = \frac{V^2}{R} \]
- Since both wires are connected to the same voltage source, the voltage \(V\) is the same for both wires.

3. **Relating Power to Resistance:**
- The resistance \(R\) of a wire is given by:
\[ R = \rho \frac{L}{A} \]
where \(\rho\) is the resistivity of the material, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.
- Since both wires are made of the same material and have the same mass, we can conclude that the resistivity \(\rho\) is constant.

4. **Finding the Resistance Ratio:**
- The cross-sectional area \(A\) can be expressed in terms of the volume and length. Since the mass \(m\) is the same for both wires, we have:
\[ \text{Volume} = \text{Area} \times \text{Length} = A \times L \]
- Therefore, the area \(A\) can be expressed as:
\[ A = \frac{m}{\rho L} \]
- Substituting this into the resistance formula gives:
\[ R_A = \rho \frac{L_A}{A_A} \quad \text{and} \quad R_B = \rho \frac{L_B}{A_B} \]

5. **Establishing the Relationship Between Resistances:**
- Since the lengths are in the ratio \(1:2\):
\[ R_A \propto L_A^2 \quad \text{and} \quad R_B \propto L_B^2 \]
- Therefore, the ratio of resistances is:
\[ \frac{R_A}{R_B} = \frac{L_A^2}{L_B^2} = \frac{1^2}{2^2} = \frac{1}{4} \]

6. **Finding the Power Dissipation in Wire A:**
- Since power is inversely proportional to resistance:
\[ \frac{P_A}{P_B} = \frac{R_B}{R_A} = 4 \]
- Given that \(P_B = 5 \, \text{W}\), we can find \(P_A\):
\[ P_A = 4 \times P_B = 4 \times 5 = 20 \, \text{W} \]

7. **Conclusion:**
- The rate of heat dissipation in wire A is \(20 \, \text{W}\).

### Final Answer:
The rate of heat dissipation in wire A is **20 W**.

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