CGPET 2021 Solved Paper

Conduction of heat in rod is given by relation $\frac{Q = K A \Delta \theta t}{l}$ where, K = coefficient of thermal conductivity of heat, A = cross - sectional area, ∆θ = temperature difference, t = time takn by heat flow and l = length of rod. If the radius of rod be r, then $A = \pi r^2 \Rightarrow Q = \frac{K \cdot \pi r^2 \cdot \Delta \theta t}{l}$ Rate of flow of heat through conductor, $\frac{Q}{t} = \frac{K \pi r^2 \Delta \theta}{l}$ Here, $K, \pi \text{ and } \Delta \theta$ are constants. ∴ $Q \propto \frac{r^2}{l}$ For more value of Q, (i) r should be maximum. (ii) l should be minimum. The value of $\frac{r^2}{r}$ for each observation, (a) $r = 1\,\text{cm} = 10^{-2}\,\text{m}, l = 1\,\text{m}$ $Q_1 \propto \frac{r^2}{l} = \frac{(10^{-2})^2}{1} = 10^{-4}$ (b) $r = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m}, l = 2\,\text{m}$ $Q_2 \propto \frac{r^2}{l} = \frac{(2 \times 10^{-2})^2}{2} = 2 \times 10^{-4}$ (c) $r = 1\,\text{cm} = 10^{-2}\,\text{m}, l = \frac{1}{2}\,\text{cm} = \frac{1}{2} \times 10^{-2}\,\text{m}$ $Q_3 \propto \frac{r^2}{l} = \frac{(10^{-2})^2}{\frac{1}{2} \times 10^{-2}} = 2 \times 10^{-2}$ (d) $r = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m}, l = \frac{1}{2}\,\text{m}$ $Q_4 \propto \frac{r^2}{l} = \frac{(2 \times 10^{-2})^2}{\frac{1}{2}} = 8 \times 10^{-4}$ The value of $\frac{r^2}{l}$ is maximum for the dimensions given in option (c), hence it will conduct maximum heat.