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 \(100^\circ\text{C}\).

• A. \(2.5\) min
• B. \(5\) min
• C. \(10\) min
• D. \(10.5\) min

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.

Answer 1

The Correct Option is B

Step 1: Treat heat flow like an electric circuit.
Each rod has a thermal resistance $R_{th} = \frac{L}{kA}$. Heat flow rate is $H = \frac{\Delta T}{R_{th}}$, just like current equals voltage over resistance. Call each rod resistance $R$.
Step 2: Arrangement I is end to end (series).
When welded tip to tip the two resistances add, so $R_{\text{I}} = R + R = 2R$.
Step 3: Find how long arrangement I takes.
The same $100$ J flows in $20$ minutes, so $H_{\text{I}} = \frac{\Delta T}{2R}$, giving time $t_{\text{I}} = 20$ min for $100$ J.
Step 4: Arrangement II is one on top of the other (parallel).
Stacked rods share the heat, so $R_{\text{II}} = \frac{R}{2}$.
Step 5: Compare the two heat flow rates.
The same temperature difference acts in both cases, so the heat rate is inversely proportional to resistance. Going from $2R$ to $\frac{R}{2}$ multiplies the rate by $\frac{2R}{R/2}=4$.
Step 6: Same heat, four times faster.
If the rate is four times larger, the time for the same $100$ J is one fourth: $t_{\text{II}} = \frac{20}{4} = 5$ min.
\[ \boxed{t_{\text{II}} = 5\ \text{min}} \]