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. `4 Q_(1)`
B. `2 Q_(1)`
C. `Q_(1)/4`
D. `Q_(1)/2`

Explanation

To solve the problem of how the rate of heat flow changes when all linear dimensions of a cylindrical rod are doubled, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Initial Conditions**:
Let the initial radius of the rod be \( R_1 \) and the initial length be \( L_1 \). The initial cross-sectional area \( A_1 \) can be calculated as:
\[ A_1 = \pi R_1^2 \]
Thus, the initial rate of heat flow \( Q_1 \) is:
\[ Q_1 = K \cdot \pi R_1^2 \cdot \frac{(T_1 - T_2)}{L_1} \]

3. **Doubling the Linear Dimensions**:
When all linear dimensions are doubled:
- The new radius \( R_2 = 2R_1 \)
- The new length \( L_2 = 2L_1 \)
- The new cross-sectional area \( A_2 \) becomes:
\[ A_2 = \pi R_2^2 = \pi (2R_1)^2 = 4\pi R_1^2 \]

4. **Calculate the New Rate of Heat Flow**:
The new rate of heat flow \( Q_2 \) can be expressed as:
\[ Q_2 = K \cdot A_2 \cdot \frac{(T_1 - T_2)}{L_2} \]
Substituting \( A_2 \) and \( L_2 \):
\[ Q_2 = K \cdot (4\pi R_1^2) \cdot \frac{(T_1 - T_2)}{2L_1} \]
Simplifying this gives:
\[ Q_2 = 2K \cdot \pi R_1^2 \cdot \frac{(T_1 - T_2)}{L_1} \]
Thus, we can relate \( Q_2 \) to \( Q_1 \):
\[ Q_2 = 2Q_1 \]

5. **Final Result**:
Therefore, the new rate of flow of heat \( Q_2 \) when all linear dimensions are doubled is:
\[ Q_2 = 2Q_1 \]