Question:
Two rectangular metal rods, identical in all respects, have been welded end-to-end as shown in figure (I). \(100\) J of heat flows through the combination in \(20\) minutes. If the rods are welded as one on top of the other as shown in figure (II), in how many minutes will the same amount of heat flow through the new combination? In either case, the temperature difference across the entry and exit points of heat is \(100^\circ\text{C}\).
Two rectangular metal rods, identical in all respects, have been welded end-to-end as shown in figure (I). \(100\) J of heat flows through the combination in \(20\) minutes. If the rods are welded as one on top of the other as shown in figure (II), in how many minutes will the same amount of heat flow through the new combination? In either case, the temperature difference across the entry and exit points of heat is \(100^\circ\text{C}\).
Show Hint
For heat conduction:
\[
R=\frac{L}{kA}
\]
Series combination:
\[
R_{\text{eq}}=R_1+R_2
\]
Parallel combination:
\[
\frac1{R_{\text{eq}}}
=
\frac1{R_1}
+
\frac1{R_2}
\]
Heat current is inversely proportional to thermal resistance.
Updated On: Jun 16, 2026
- \(2.5\) min
- \(5\) min
- \(10\) min
- \(10.5\) min
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Rate of heat conduction is given by
\[
H=\frac{kA\Delta T}{L}
\]
For identical rods:
• End-to-end arrangement \(\rightarrow\) series combination.
• One above the other \(\rightarrow\) parallel combination.
Thermal resistance is \[ R=\frac{L}{kA} \] and \[ H=\frac{\Delta T}{R}. \]
Step 1: Find the equivalent thermal resistance in arrangement (I). Let the thermal resistance of each rod be \[ R=\frac{L}{kA}. \] Since the rods are connected end-to-end, \[ R_s=R+R=2R. \] Therefore, \[ H_s=\frac{\Delta T}{2R}. \] Given \(100\) J flows in \(20\) min.
Step 2: Find the equivalent thermal resistance in arrangement (II). In the second arrangement the rods are in parallel. \[ \frac1{R_p} = \frac1R+\frac1R = \frac2R \] Hence, \[ R_p=\frac{R}{2}. \] Therefore, \[ H_p = \frac{\Delta T}{R/2} = \frac{2\Delta T}{R}. \]
Step 3: Compare the rates of heat flow. \[ \frac{H_p}{H_s} = \frac{2\Delta T/R}{\Delta T/(2R)} = 4. \] Thus the second arrangement conducts heat \(4\) times faster.
Step 4: Calculate the new time. For the same heat quantity, \[ t\propto\frac1H. \] Hence, \[ t_p = \frac{20}{4} = 5\ \text{min}. \] \[\begin{aligned} \boxed{5\ \text{min}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
• End-to-end arrangement \(\rightarrow\) series combination.
• One above the other \(\rightarrow\) parallel combination.
Thermal resistance is \[ R=\frac{L}{kA} \] and \[ H=\frac{\Delta T}{R}. \]
Step 1: Find the equivalent thermal resistance in arrangement (I). Let the thermal resistance of each rod be \[ R=\frac{L}{kA}. \] Since the rods are connected end-to-end, \[ R_s=R+R=2R. \] Therefore, \[ H_s=\frac{\Delta T}{2R}. \] Given \(100\) J flows in \(20\) min.
Step 2: Find the equivalent thermal resistance in arrangement (II). In the second arrangement the rods are in parallel. \[ \frac1{R_p} = \frac1R+\frac1R = \frac2R \] Hence, \[ R_p=\frac{R}{2}. \] Therefore, \[ H_p = \frac{\Delta T}{R/2} = \frac{2\Delta T}{R}. \]
Step 3: Compare the rates of heat flow. \[ \frac{H_p}{H_s} = \frac{2\Delta T/R}{\Delta T/(2R)} = 4. \] Thus the second arrangement conducts heat \(4\) times faster.
Step 4: Calculate the new time. For the same heat quantity, \[ t\propto\frac1H. \] Hence, \[ t_p = \frac{20}{4} = 5\ \text{min}. \] \[\begin{aligned} \boxed{5\ \text{min}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
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