QJEEAdv18Phd2e2: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phd2e2 Solution: Two conducting cylinders of equal length but different radii…

The intermediate temperature is given by the formula

$T = {{{{{k_1}{A_1}{T_1}} \over {{l_1}}} + {{{k_2}{A_2}{T_2}} \over {{l_2}}}} \over {{{{k_1}{A_1}} \over {{l_1}}} + {{{k_2}{A_2}} \over {{l_2}}}}}$

Here, T = 200 k, T₁ = 300 k, T₂ = 100 k, l₁ = l₂ and A₁ = $\pi$ r², A₂ = 4 $\pi$ r²

$\therefore$ $200 = {{{k_1}\pi {r^2} \times 300 + {k_2}\pi (4{r^2})100} \over {{k_1}\pi {r^2} + {k_2}\pi (4{r^2})}}$

$\therefore$ $200 = {{300{k_1} + 400{k_2}} \over {{k_1} + 4{k_2}}}$

$\therefore$ 200 k₁ + 800 k₂ = 300 k₁ + 400 k₂

$\Rightarrow$ 400 k₂ = 100 k₁

$\therefore$ ${{{k_1}} \over {{k_2}}} = 4.00$