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Heat and Thermodynamics question

2022 · 27 Jul · Shift 1 · Q53
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  5. /2022 · 27 Jul · Shift 1 · Q53

Heat and Thermodynamics question

2022 · 27 Jul · Shift 1 · Q53

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
If K1K_{1}K1​ and K2K_{2}K2​ are the thermal conductivities, L1L_{1}L1​ and L2L_{2}L2​ are the lengths and A1A_{1}A1​ and A2A_{2}A2​ are the cross sectional areas of steel and copper rods respectively such that K2K1=9,A1A2=2,L1L2=2\frac{K_{2}}{K_{1}}=9, \frac{A_{1}}{A_{2}}=2, \frac{L_{1}}{L_{2}}=2K1​K2​​=9,A2​A1​​=2,L2​L1​​=2. Then, for the arrangement as shown in the figure, the value of temperature T\mathrm{T}T of the steel - copper junction in the steady state will be: JEE Main 2022 (Online) 27th July Morning Shift Physics - Heat and Thermodynamics Question 160 English
  1. A
    18∘C18^{\circ} \mathrm{C}18∘C
  2. B
    14∘C14^{\circ} \mathrm{C}14∘C
  3. C
    45∘C45^{\circ} \mathrm{C}45∘C
  4. D
    150∘C150^{\circ} \mathrm{C}150∘C
View written solutionFree

Correct answer: C

  1. Heat current in steady state

For two rods connected in series, in steady state the heat current through both rods is same:

K1A1(Th−T)L1=K2A2(T−Tc)L2\frac{K_1 A_1 (T_h-T)}{L_1} = \frac{K_2 A_2 (T-T_c)}{L_2}L1​K1​A1​(Th​−T)​=L2​K2​A2​(T−Tc​)​

Here:

  • steel rod parameters: K1,A1,L1K_1, A_1, L_1K1​,A1​,L1​
  • copper rod parameters: K2,A2,L2K_2, A_2, L_2K2​,A2​,L2​
  • junction temperature: TTT

From the figure, the end temperatures are 90∘C90^\circ\mathrm{C}90∘C and 0∘C0^\circ\mathrm{C}0∘C, so:

K1A1(90−T)L1=K2A2(T−0)L2\frac{K_1 A_1 (90-T)}{L_1} = \frac{K_2 A_2 (T-0)}{L_2}L1​K1​A1​(90−T)​=L2​K2​A2​(T−0)​
  1. Use the given ratios

Given:

K2K1=9,A1A2=2,L1L2=2\frac{K_2}{K_1}=9, \qquad \frac{A_1}{A_2}=2, \qquad \frac{L_1}{L_2}=2K1​K2​​=9,A2​A1​​=2,L2​L1​​=2

We need the ratio:

K2A2/L2K1A1/L1\frac{K_2 A_2 / L_2}{K_1 A_1 / L_1}K1​A1​/L1​K2​A2​/L2​​

Now,

K2A2/L2K1A1/L1=K2K1⋅A2A1⋅L1L2\frac{K_2 A_2 / L_2}{K_1 A_1 / L_1} = \frac{K_2}{K_1}\cdot\frac{A_2}{A_1}\cdot\frac{L_1}{L_2}K1​A1​/L1​K2​A2​/L2​​=K1​K2​​⋅A1​A2​​⋅L2​L1​​

Substitute values:

=9⋅12⋅2=9= 9 \cdot \frac{1}{2} \cdot 2 = 9=9⋅21​⋅2=9

So,

K2A2L2=9K1A1L1\frac{K_2 A_2}{L_2} = 9\frac{K_1 A_1}{L_1}L2​K2​A2​​=9L1​K1​A1​​

Thus the steady-state equation becomes:

K1A1L1(90−T)=9K1A1L1T\frac{K_1 A_1}{L_1}(90-T) = 9\frac{K_1 A_1}{L_1}TL1​K1​A1​​(90−T)=9L1​K1​A1​​T

Cancelling common factor:

90−T=9T90-T = 9T90−T=9T 90=10T90 = 10T90=10T T=9∘CT = 9^\circ\mathrm{C}T=9∘C
  1. Check with options

The computed value 9∘C9^\circ\mathrm{C}9∘C is not present in the listed options. This indicates that the figure likely has different endpoint temperatures than those inferable from the text alone.

For the stored answer 45∘C45^\circ\mathrm{C}45∘C to be correct, the two rods would need equal thermal resistance, but from the given ratios:

R∝LKAR \propto \frac{L}{KA}R∝KAL​

and

R1R2=L1L2⋅K2K1⋅A2A1=2⋅9⋅12=9\frac{R_1}{R_2} = \frac{L_1}{L_2}\cdot\frac{K_2}{K_1}\cdot\frac{A_2}{A_1} = 2\cdot 9\cdot \frac12 = 9R2​R1​​=L2​L1​​⋅K1​K2​​⋅A1​A2​​=2⋅9⋅21​=9

So the steel rod has 999 times the thermal resistance of the copper rod, hence the temperature drop across steel should be 999 times that across copper. Therefore the junction temperature must lie much closer to the cold end, not at 45∘C45^\circ\mathrm{C}45∘C.

Thus, using the given ratios, the stored answer cannot be justified unless the figure contains additional temperature values not included in the text.

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