A cylindrical rod having temperature `T_(1)` and `T_(2)` at its ends. The rate of flow of heat is `Q_(1) cal//sec`. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat `Q_(2)` will be

Options

A. `4Q_1`
B. `2Q_2`
C. `Q_1/4`
D. `Q_1/2`

Explanation

To solve the problem, we need to analyze how the rate of heat flow changes when the linear dimensions of a cylindrical rod are doubled. We will use the formula for heat conduction through a cylindrical rod.

### Step-by-Step Solution:

1. **Understand the Formula for Heat Flow:**
The rate of heat flow \( Q \) through a cylindrical rod is given by the formula:
\[ Q = \frac{K \cdot (T_1 - T_2) \cdot A}{L} \]
where:
- \( K \) is the thermal conductivity of the material,
- \( T_1 \) and \( T_2 \) are the temperatures at the ends of the rod,
- \( A \) is the cross-sectional area of the rod,
- \( L \) is the length of the rod.

2. **Identify Changes When Dimensions are Doubled:**
When all linear dimensions are doubled:
- The new length \( L_2 = 2L_1 \).
- The radius \( R_2 = 2R_1 \).
- The cross-sectional area \( A \) of a cylinder is given by \( A = \pi R^2 \). Therefore, the new area \( A_2 \) will be:
\[ A_2 = \pi (R_2)^2 = \pi (2R_1)^2 = 4\pi R_1^2 = 4A_1 \]

3. **Substitute New Values into the Heat Flow Formula:**
Now, we can express the new rate of heat flow \( Q_2 \):
\[ Q_2 = \frac{K \cdot (T_1 - T_2) \cdot A_2}{L_2} \]
Substituting \( A_2 \) and \( L_2 \):
\[ Q_2 = \frac{K \cdot (T_1 - T_2) \cdot (4A_1)}{2L_1} \]

4. **Simplify the Expression:**
Simplifying the expression for \( Q_2 \):
\[ Q_2 = \frac{4K \cdot (T_1 - T_2) \cdot A_1}{2L_1} = \frac{2K \cdot (T_1 - T_2) \cdot A_1}{L_1} \]
Notice that \( Q_1 = \frac{K \cdot (T_1 - T_2) \cdot A_1}{L_1} \). Therefore:
\[ Q_2 = 2Q_1 \]

5. **Conclusion:**
The rate of flow of heat \( Q_2 \) when all linear dimensions are doubled is:
\[ Q_2 = 2Q_1 \]

### Final Answer:
The rate of flow of heat \( Q_2 \) will be \( 2Q_1 \).