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QJEEM14Phf5c4: JEE Main 2014 Physics - Class 11 JEEM14Phf5c4 Solution: Three rods of Copper, Brass and Steel are welded together to…

Three rods of Copper, Brass and Steel are welded together to form a YY shaped structure. Area of cross - section of each rod =4cm2.= 4c{m^2}. End of copper rod is maintained at 100∘C{100^ \circ }C where as ends of brass and steel are kept at 0∘C{0^ \circ }C . Lengths of the copper, brass and steel rods are 46,46, 1313 and 1212 cmscms respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are 0.92,0.260.92, 0.26 and 0.120.12 CGSCGS units respectively. Rate of heat flow through copper rod is:

Multiple Choice Options

  • A 1.21.2 cal/scal/s
  • B 2.42.4 cal/scal/s
  • C 4.84.8 cal/scal/s
  • D 6.06.0 cal/scal/s

Step-by-Step Verified Solution

Rate of heat flow is given by,

Q=KA(θ1−θ2)lQ = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}

Where, K=K= coefficient of thermal conductivity l=l= length of rod and A=A= Area of cross-section of rod
JEE Main 2014 (Offline) Physics - Heat and Thermodynamics Question 388 English Explanation

If the junction temperature is T,T, then

QCopper  =  QBrass  +  QSteel{Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}

0.92×4(100−T)46=0.26×4×(T−0)13+0.12×4×(T−0)12{{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}

⇒200−2T=2T+T\Rightarrow 200 - 2T = 2T + T

⇒T=40∘C\Rightarrow T = {40^ \circ }C

∴\therefore QCopper=0.92×4×6046=4.8 cal/s{Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s

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