Question 105 of 126

Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod is $4{cm}^{2}$. End of copper rod is maintained at $100^{\circ}C$. Where as ends of brass and steel are kept at $0^{\circ}C$. Lengths of the copper, brass and steel rods are $46,13$and $12cms$ respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are $0.92,0.26$ and $0.12$ CGS units respectively. Rate of heat flow through copper rod is :

A$1.2Cal/s$B$2.4Cal/s$C$4.8Cal/s$D$6.0Cal/s$

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Answer: (C) $4.8Cal/s$

{L}_{Cu}=46cm;{L}_{B}=13cm&{L}_{S}=12cm.



\begin{matrix}{K}_{Cu}=0.92 \\ {K}_{B}=0.26 \\ {K}_{S}=0.12\end{matrix}}CGSunit

We know that $R=\dfrac{L}{KA}$
${R}_{Cu}=\dfrac{46}{0.92\times 4}=12.5$
${R}_{B}=\dfrac{13}{0.26\times 4}=12.5$

${R}_{S}=\dfrac{12}{0.12\times 4}=25$Now, if $T$ is temperature of junction then

$\dfrac{100-T}{{R}_{Cu}}=\dfrac{T-0}{{R}_{B}}+\dfrac{T-0}{{R}_{S}}$

$\dfrac{100-T}{12.5}=\dfrac{T}{12.5}+\dfrac{T}{25}$

$\text{∴}\dfrac{100-T}{12.5}=\dfrac{3T}{25}$

$200=5T$

$∴T{=40}^{^{\circ}}$

$\Rightarrow {Q}_{Cu}=\dfrac{100-T}{{R}_{Cu}}$

$\text{=}\dfrac{100-40}{12.5}$

$\text{=}\dfrac{60}{12.5}$

$\Rightarrow {Q}_{Cu}=4.8Cal/s$