Question:
Two metal rods \(A\) and \(B\) have lengths in the ratio \(1:2\), thermal conductivities in the ratio \(1:2\) and cross-sectional areas in the ratio \(1:4\). The ends of the two rods are maintained between the same temperature difference. Then the ratio of heat currents
\[
\left(\frac{H_A}{H_B}\right)
\]
is
Two metal rods \(A\) and \(B\) have lengths in the ratio \(1:2\), thermal conductivities in the ratio \(1:2\) and cross-sectional areas in the ratio \(1:4\). The ends of the two rods are maintained between the same temperature difference. Then the ratio of heat currents
\[
\left(\frac{H_A}{H_B}\right)
\]
is
Show Hint
For steady-state conduction,
\[
H=\frac{kA\Delta T}{L}.
\]
Heat current is directly proportional to thermal conductivity and area, but inversely proportional to length.
Updated On: Jul 29, 2026
- \(1:4\)
- \(4:1\)
- \(2:1\)
- \(1:2\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The rate of heat flow (heat current) through a rod is given by
\[
H=\frac{kA\Delta T}{L},
\]
where
\[
k=\text{thermal conductivity},
\quad
A=\text{cross-sectional area},
\quad
L=\text{length}.
\]
Step 1: Write the given ratios. \[ L_A:L_B=1:2, \] \[ k_A:k_B=1:2, \] \[ A_A:A_B=1:4. \] Since both rods are maintained under the same temperature difference, \[ \Delta T_A=\Delta T_B. \]
Step 2: Form the ratio of heat currents. \[ \frac{H_A}{H_B} = \frac{\dfrac{k_AA_A\Delta T}{L_A}} {\dfrac{k_BA_B\Delta T}{L_B}}. \] \[ = \frac{k_A}{k_B} \cdot \frac{A_A}{A_B} \cdot \frac{L_B}{L_A}. \]
Step 3: Substitute the given ratios. \[ \frac{H_A}{H_B} = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{1}. \] \[ = \frac14. \] Therefore, \[ H_A:H_B=1:4. \] Hence, \[ \boxed{\frac{H_A}{H_B}=1:4} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Write the given ratios. \[ L_A:L_B=1:2, \] \[ k_A:k_B=1:2, \] \[ A_A:A_B=1:4. \] Since both rods are maintained under the same temperature difference, \[ \Delta T_A=\Delta T_B. \]
Step 2: Form the ratio of heat currents. \[ \frac{H_A}{H_B} = \frac{\dfrac{k_AA_A\Delta T}{L_A}} {\dfrac{k_BA_B\Delta T}{L_B}}. \] \[ = \frac{k_A}{k_B} \cdot \frac{A_A}{A_B} \cdot \frac{L_B}{L_A}. \]
Step 3: Substitute the given ratios. \[ \frac{H_A}{H_B} = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{1}. \] \[ = \frac14. \] Therefore, \[ H_A:H_B=1:4. \] Hence, \[ \boxed{\frac{H_A}{H_B}=1:4} \] \[ \boxed{\text{Answer = (A)}} \]
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