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Three rods $A, B$ and $C$ of thermal conductivities $K, 2\,K$ and $4\,K$, cross-sectional areas $A, 2\,A$ and $2\,A$ and lengths $2l, l$ and $l$ respectively are connected as shown in the figure. If the ends of the rods are maintained at temperatures $100^o\,C, 50^o\,C$, and $0^o\,C$ respectively, then the temperature $\theta$ of the junction is ......... $^oC$
$\frac{300}{7}$
$20$
$\frac{200}{7}$
$\frac{200}{13} $
Solution
Let $R_{A}, R_{B}$ and $R_{C}$ be the thermal resistance of rods
$A,$ $\mathrm{B}$ and $\mathrm{C}$ respectively. Then
$\mathrm{R}_{\mathrm{A}}=\frac{2 l}{\mathrm{KA}} ; \mathrm{R}_{\mathrm{B}}=\frac{l}{2 \mathrm{K} \cdot 2 \mathrm{A}}=\frac{l}{4 \mathrm{KA}}$
and $\mathrm{R}_{\mathrm{C}}=\frac{l}{4 \mathrm{K} \cdot 2 \mathrm{A}}=\frac{l}{8 \mathrm{KA}}$
since thermal current
$\mathrm{H}=\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\theta_{1}-\theta_{2}}{l / \mathrm{KA}}=\frac{\theta_{1}-\theta_{2}}{\mathrm{R}}$
or $\quad \mathrm{RH}=\theta_{1}-\theta_{2}$
$\therefore \quad 100-\theta=\mathrm{R}_{\mathrm{A}} \mathrm{H}_{\mathrm{A}}$ $…(i)$
$\theta-50=\mathrm{R}_{\mathrm{B}} \mathrm{H}_{\mathrm{B}}$ $…(ii)$
$\theta-0=\mathrm{R}_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}=\mathrm{R}_{\mathrm{C}}\left(\mathrm{H}_{\mathrm{A}}-\mathrm{H}_{\mathrm{B}}\right)$ $…(iii)$
On substituting values of $\mathrm{R}_{\mathrm{A}}, \mathrm{R}_{\mathrm{B}}$ and $\mathrm{R}_{\mathrm{C}},$ we get
$100-\theta=\frac{2 l}{\mathrm{KA}} \cdot \mathrm{H}_{\mathrm{A}} \Rightarrow \mathrm{H}_{\mathrm{A}}=\frac{\mathrm{K} \mathrm{A}}{2 l}(100-\theta)$
$\theta-50=\frac{l}{4 \mathrm{KA}} \mathrm{H}_{\mathrm{B}} \Rightarrow \mathrm{H}_{\mathrm{B}}=\frac{4 \mathrm{KA}}{l}(\theta-50)$
$\theta=\frac{l}{8 \mathrm{KA}}\left(\mathrm{H}_{\mathrm{A}}-\mathrm{H}_{\mathrm{B}}\right)$
$=\frac{1}{8 \mathrm{KA}}\left[\frac{\mathrm{KA}}{2 l}(100-\theta)-\frac{4 \mathrm{KA}}{l}(\theta-50)\right]$
$\theta=20^{\circ} \mathrm{C}$
