Verified Solution • 2025–26 Session moderate

QJEEAdv16Ph4957: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph4957 Solution: The ends Q and R of two thin wires, PQ and RS, are soldered…

The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10oC. Now the end P is maintained at 10oC, while the end S is heated and maintained at 400oC. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 ×\times 10-5 K-1 , the change in length of the wire PQ is

Multiple Choice Options

  • A 0.78 mm
  • B 0.90 mm
  • C 1.56 mm
  • D 2.34 mm

Step-by-Step Verified Solution

JEE Advanced 2016 Paper 2 Offline Physics - Heat and Thermodynamics Question 59 English Explanation 1

Rate of heat flow from P to Q,

dQdt=2KA(T−10)1{{dQ} \over {dt}} = {{2KA(T - 10)} \over 1}

Rate of heat flow from Q to S,

dQdt=KA(4000−T)1{{dQ} \over {dt}} = {{KA(4000 - T)} \over 1}

At steady state, state rate of heat flow is same

∴\therefore 2KA(T−10)1=KA(400−T){{2KA(T - 10)} \over 1} = KA(400 - T)

or 2T−20=400−T2T - 20 = 400 - T or 3T=4203T = 420

∴\therefore T=140∘T = 140^\circ

JEE Advanced 2016 Paper 2 Offline Physics - Heat and Thermodynamics Question 59 English Explanation 2

Temperature of junction is 140∘140^\circ C.

Temperature at a distance x from end P is Tx=(130x+10∘){T_x} = (130x + 10^\circ )

Change in length dx is suppose dy.

Then, dy=αdx(Tx−10)dy = \alpha dx({T_x} - 10)

∫0Δydy=∫01αdx(130x+10−10)\int_0^{\Delta y} {dy} = \int_0^1 {\alpha dx(130x + 10 - 10)}

Δy=[αx22×130]01\Delta y = \left[ {{{\alpha {x^2}} \over 2} \times 130} \right]_0^1

Δy=1.2×10−5×65\Delta y = 1.2 \times {10^{ - 5}} \times 65

Δy=78.0×10−5\Delta y = 78.0 \times {10^{ - 5}} m = 0.78 mm

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