Verified Solution • 2025–26 Session moderate

QJEEM22Ph02fa: JEE Main 2022 Physics - Class 11 JEEM22Ph02fa Solution

If $K_{1}$ and $K_{2}$ are the thermal conductivities, $L_{1}$ and $L_{2}$ are the lengths and $A_{1}$ and $A_{2}$ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}}=9, \frac{A_{1}}{A_{2}}=2, \frac{L_{1}}{L_{2}}=2$ . Then, for the arrangement as shown in the figure, the value of temperature $\mathrm{T}$ of the steel - copper junction in the steady state will be:

JEE Main 2022 (Online) 27th July Morning Shift Physics - Heat and Thermodynamics Question 136 English

Multiple Choice Options

A $18^{\circ} \mathrm{C}$
B $14^{\circ} \mathrm{C}$
C $45^{\circ} \mathrm{C}$
D $150^{\circ} \mathrm{C}$

Step-by-Step Verified Solution

$450 - T = {{dQ} \over {dt}} \times {{{l_1}} \over {{K_1}{A_1}}}$

$T - 0 = {{dQ} \over {dt}} \times {{{l_2}} \over {{K_2}{A_2}}}$

So, ${{450 - T} \over T} = {{{K_2}{A_2}{l_1}} \over {{K_1}{A_1}{l_2}}} = 9 \times {1 \over 2} \times 2 = 9$

$450 - T = 9T$

$\Rightarrow T = 45^\circ C$

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring