Question

Heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio 1 : 2 and the lengths in the ratio 2 : 1. If the temperature difference between the ends is same, then ratio of the rate of flow of heat through them will be

• A. 2:1
• B. 8:1
• C. 1:1
• D. 1:8

Answer 1

The Correct Option is D

To solve this problem, we need to understand how the rate of heat flow through a rod is determined using the formula from thermal conduction.

The rate of heat flow \( Q \) through a cylindrical rod is given by the formula:

Q = \frac{k \cdot A \cdot \Delta T}{L}

Where:

• k is the thermal conductivity of the material (same for both rods).
• A is the cross-sectional area of the rod.
• \Delta T is the temperature difference between the ends (same for both rods).
• L is the length of the rod.

Since the rods are cylindrical, the cross-sectional area A is given by A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}.

Let's denote the diameters of the two rods as d_1 and d_2, and the lengths as L_1 and L_2. According to the problem, we have:

• \frac{d_1}{d_2} = \frac{1}{2}
• \frac{L_1}{L_2} = \frac{2}{1}

Substitute these into the formula for each rod:

For the first rod:

Q_1 = \frac{k \cdot \frac{\pi d_1^2}{4} \cdot \Delta T}{L_1}

For the second rod:

Q_2 = \frac{k \cdot \frac{\pi d_2^2}{4} \cdot \Delta T}{L_2}

Now, to find the ratio of their heat flows \frac{Q_1}{Q_2}:

\frac{Q_1}{Q_2} = \frac{\left(\frac{\pi d_1^2}{4}\right) \cdot \frac{1}{L_1}}{\left(\frac{\pi d_2^2}{4}\right) \cdot \frac{1}{L_2}}

Cancel the common terms and simplify:

\frac{Q_1}{Q_2} = \frac{d_1^2 \cdot L_2}{d_2^2 \cdot L_1}

Substitute the ratios:

• d_1 = \frac{1}{2}d_2 \(\Rightarrow d_1^2 = \frac{1}{4}d_2^2\)
• L_2 = \frac{1}{2}L_1

So:

\frac{Q_1}{Q_2} = \frac{\frac{1}{4}d_2^2 \cdot \frac{1}{2}L_1}{d_2^2 \cdot L_1} = \frac{1}{8}

Thus, the ratio of the rate of flow of heat through the rods is 1:8.