Transmission of Heat | JEE Main previous year questions with Solutions

$R_{1} = \frac{2 l}{k_{1} A}, R_{2} = \frac{2 l}{k_{2} A}, R_{3} = \frac{2 l}{k_{3} A}$

$(100 - θ) (\frac{1}{R_{1}} + \frac{1}{R_{2}}) = \frac{(θ - 0)}{R_{3}}$

$(100 - θ) \frac{(k_{1} + k_{2}) A}{2 l} = \frac{θ k_{3} A}{l}$

$(100 - θ) (k_{1} + k_{2} = 2 θ k_{3})$

$θ = \frac{100 (k_{1} + k_{2})}{k_{1} + k_{2} + k_{3}} = \frac{100 \times 180}{450} = 40$

360

$R_{1} = \frac{l_{1}}{K_{1} A_{1}}, R_{2} = \frac{l_{2}}{K_{2} A_{2}}$

$\frac{d Q}{d t} = \frac{△ T}{R}$

${(\frac{d Q}{d t})}_{1} = {(\frac{d Q}{d t})}_{2}$

$\frac{400 - T}{R_{1}} = \frac{T - 200}{R_{2}}$

$\frac{400 - T}{T - 200} = \frac{R_{1}}{R_{2}} = (\frac{l_{2}}{l_{2}}) {(\frac{r_{2}}{r_{2}})}^{2} \times \frac{K_{2}}{K_{1}}$

$\frac{400 - T}{T - 200} = \frac{1}{22} \times {(\frac{1}{2})}^{2} \times 2 = (\frac{1}{4})$

1600 – 4T = T –200

5T = 1800 $\Rightarrow$ T = 360 K

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