Question

A cylindrical rod has temperatures '$T_1$' and '$T_2$' at its ends. The rate of flow of heat is '$Q_1$' $\text{cal s}^{-1}$. If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat '$Q_2$' will be

• A. $Q_2 = \frac{Q_1}{2}$
• B. $Q_2 = \frac{Q_1}{4}$
• C. $Q_2 = 4Q_1$
• D. $Q_2 = 2Q_1$

Hint:

Since thermal resistance is defined as $R_{\text{th}} = \frac{L}{KA}$, doubling the length increases resistance by $2\times$, but doubling the radius increases the area by $4\times$, which cuts resistance by $4\times$. The net effect is that the total resistance is halved ($\frac{2}{4} = \frac{1}{2}$), meaning the heat flow current must double ($2\times$).

Answer 1

The Correct Option is D

Step 1: State what changes.
A cylindrical rod conducts heat at rate $Q_1$. Both its length and radius are doubled while the end temperatures stay fixed. We want the new rate $Q_2$.
Step 2: Recall the conduction law.
Steady-state heat flow follows $Q = \dfrac{K A \,\Delta T}{L}$, where $A$ is cross-sectional area and $L$ the length.
Step 3: Bring in the geometry.
For a cylinder $A = \pi r^2$, so $Q = \dfrac{K \pi r^2 \Delta T}{L}$. With $K$ and $\Delta T$ fixed, $Q \propto \dfrac{r^2}{L}$.
Step 4: Apply the scaling.
New radius $r' = 2r$ and new length $L' = 2L$, so \[ Q_2 \propto \frac{(2r)^2}{2L} = \frac{4r^2}{2L} = 2\cdot\frac{r^2}{L} \]
Step 5: Compare to the original.
Since $Q_1 \propto \dfrac{r^2}{L}$, dividing gives $\dfrac{Q_2}{Q_1} = 2$.
Step 6: Conclude.
The doubled radius boosts area fourfold while the doubled length only halves the flow, so the net effect is a factor of $2$: $Q_2 = 2Q_1$.
\[ \boxed{Q_2 = 2Q_1\ \text{(option 4)}} \]