Three rods of Copper, Brass and Steel are welded together to form a...

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 ° C

. Where as ends of brass and steel are kept at

0 ° 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 : Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each

Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each

Options

A $1 . 2Cal / s$
B $2 . 4Cal / s$
C $4 . 8Cal / s$
D $6 . 0Cal / s$

Correct answer

C. $4 . 8Cal / s$

Step-by-step solution

${L}_{Cu} = 46cm; {L}_{B} = 13cm \text{ and } {L}_{S} = 12cm$ .

$\begin{array}{ll}{K}_{Cu} = 0 .92 \\ {K}_{B} = 0 .26 \\ {K}_{S} = 0 .12\end{array}$ } CGS unit $ 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 $</blockquote><blockquote>Now, if$ 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}^{∘} $ $ ⇒ {Q}_{Cu} = \dfrac{100 - T}{{R}_{Cu}} $ $ \text{=}\dfrac{100 - 40}{12 .5} $ $ \text{=}\dfrac{60}{12 .5} $ $ ⇒ {Q}_{Cu} = 4 .8 Cal / s$

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