JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 7)
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:
_27th_July_Morning_Shift_en_7_1.png)
$$18^{\circ} \mathrm{C}$$
$$14^{\circ} \mathrm{C}$$
$$45^{\circ} \mathrm{C}$$
$$150^{\circ} \mathrm{C}$$
Explanation
$$450 - T = {{dQ} \over {dt}} \times {{{l_1}} \over {{K_1}{A_1}}}$$
$$T - 0 = {{dQ} \over {dt}} \times {{{l_2}} \over {{K_2}{A_2}}}$$
So, $${{450 - T} \over T} = {{{K_2}{A_2}{l_1}} \over {{K_1}{A_1}{l_2}}} = 9 \times {1 \over 2} \times 2 = 9$$
$$450 - T = 9T$$
$$ \Rightarrow T = 45^\circ C$$
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