A) Q/4
B) Q/16
C) 2Q
D) Q/2
Correct Answer: B
Solution :
In steady state the amount of heat flowing from one face to the other face in time t is given by \[Q=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})t}{l}\]where K is coefficient of thermal conductivity of material of rod \[\Rightarrow \] \[\frac{Q}{t}\propto \frac{A}{l}\propto \frac{{{r}^{2}}}{l}\] ?(i) As the metallic rod is melted and the material is formed into a rod of half the radius \[{{V}_{1}}={{V}_{2}}\] \[\pi r_{1}^{2}{{l}_{1}}=\pi r_{2}^{2}{{l}_{2}}\] \[\Rightarrow \] \[{{l}_{1}}=\frac{{{l}_{2}}}{4}\] ?(ii) Now, from Eqs. (i) and (ii) \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{r_{1}^{2}}{{{l}_{1}}}\times \frac{{{l}_{2}}}{r_{2}^{2}}=\frac{r_{1}^{2}}{{{l}_{1}}}\times \frac{4{{l}_{1}}}{{{({{r}_{1}}/2)}^{2}}}\] \[\Rightarrow \] \[{{Q}_{1}}=16\,{{Q}_{2}}\]
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