Transmission of Heat | JEE Main previous year questions with Solutions

(40)

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

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

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

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

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

(360)

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

$\frac{d Q}{d t} = \frac{△ T}{R}$

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

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

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

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

1600 – 4T = T –200

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

(40)

Let the temperature of the contact surface is T, then

$H_{A} = H_{B}$

$\frac{K_{A} A (T_{A} - T)}{L} = \frac{K_{B} A (T - T_{B})}{L}$

$84 (100 - T) = 126 (T - 0)$

$2 (100 - T) = 3 T$

$\Rightarrow T = 40 ° C$

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