c
\(\frac{Q}{t}\,\, = \,\,{\text{0}}{\text{.5}}\,\,{\text{cal}}\,\,{{\text{s}}^{{\text{ - 1}}}}\,{\text{;}}\,\,{\text{r}}\,\, = \,\,{\text{1}}\,\,\,{\text{cm }}\)

ક્ષેત્રફળ \( A = \pi r^{2} = 3.142 × 1 cm^{2} = 3.142 cm^{2}\)

\(L = \) સળિયાની લંબાઈ \(= 2{m} = 200 cm, T_1 = 250^oC, T_2 = ?\)

આપણે જાણીએ છીએ કે  \(\frac{Q}{t}\,\, = \,\,\frac{{KA\,\,({T_1} - {T_2})}}{L}\) અથવા  \(({T_1} - {T_2})\,\,\, = \,\,\,\frac{Q}{t}\,\, \times \,\,\frac{L}{{KA}}\,\, = \,\,\frac{{0.5\,\, \times \,\,200}}{{0.26\,\,{C^{ - 1}} \times \,\,\,3.142}}\,\, = \,\,122.\,4{\,^o }C\)

\(\therefore \,\,{T_2} = \,\,{250^o }C\,\, - \,\,{122.4^o }C\,\, = \,\,{127.6^o }C\)