Two rods of same length and material are joined end to end. They transfer heat in 8 second. When they are joined in parallel they transfer same amount of heat in same conditions in time
Show Hint
Memorize this useful network shortcut for competitive exams: for $n$ identical elements, the ratio of parallel to series effective resistance is always $\frac{R_p}{R_s} = \frac{1}{n^2}$. Since $n = 2$ here, the parallel combination is always $2^2 = 4$ times faster than the series configuration!
Step 1: The problem.
Two identical rods carry heat from a hot side to a cold side. Joined end to end (series) they pass a fixed amount of heat in $8$ s. We want the time when the same two rods are joined side by side (parallel) under the same hot and cold sides.
Step 2: Heat flow is like electric flow.
Each rod has a thermal resistance $R = L/(KA)$. Heat flows through resistance just like current. For a fixed heat $Q$ and fixed temperature difference, the time taken grows with the total resistance:
\[ t \propto R_{\text{effective}} \]
Step 3: Series resistance.
Two equal resistances in series add up:
\[ R_s = R + R = 2R \]
Step 4: Parallel resistance.
Two equal resistances in parallel give half:
\[ R_p = \frac{R}{2} \]
Step 5: Take the ratio of times.
\[ \frac{t_p}{t_s} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4} \]
Step 6: Find the parallel time.
\[ t_p = \frac{t_s}{4} = \frac{8}{4} = 2\ \text{s} \]
So the parallel rods take $2$ s, which is option (2).
\[ \boxed{t_p = 2\ \text{s}} \]