Class 11 NEET Physics Properties of Matter Questions

The amount of heat flows in time t through a cylindrical metallic rod of length L and uniform area of cross-section $A = \pi R^2$ with its ends maintained at temperatures $T_1$ and $T_2 (T_1 > T_2)$ is given by$Q = \frac{K A (T_1 - T_2) t}{L}$......(i)where AT is the thermal conductivity of the material of the rodArea of cross-section of new rod $A' = \pi (R/2)^2$$\frac{\pi R^2}{4} = A/4$....(ii)As the volume of the rod remains unchanged$\therefore A L = A' L'$where $L'$ is the length the new rodor $L' = L A / A'$...........(iii)$= 4 L$ (Using (ii))Now, the amount of heat flows in same time t in the new rod with its ends maintained at the same temperatures $T_1$ and $T_2$ is given by $Q' = \frac{K A' (T_1 - T_2) t}{L'}$..........(iv)Substituting the values of $A'$ and $L'$ from equations (ii) and (iii) in the above equation, we get$= \frac{1}{16} Q$ (Using (i))