AIPMT 2010 Physics and Chemistry Paper

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 $Q=\frac{KA(T_1-T_2)}{L}t$ .....(i) where $K$ is the thermal conductivity of the material of the rod. Area of cross-section of new $\text{rod} A'=\pi\left(\frac{R}{2}\right)^2=\frac{\pi R^2}{4}=\frac{A}{4}$.....(ii) As the volume of the rod remains unchanged $\therefore AL = A' L'$ where L' is the length the new rod or $L' = L \frac{A}{A'}$.......(iii) $=4L$ (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 $Q'=\frac{K (A/4) (T_1-T_2) t}{4 L}=\frac{1}{16} \frac{K A (T_1-T_2) t}{L}$ $=\frac{1}{16} Q ($ Using (i))