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Question

Two conductors of same crass-section and conductivities $$K, 3K$$ and lengths $$3d$$ and $$d$$ respectively are connected end to end as shown in figure. Temperature of end of first conductor is $$\theta_1$$ and that of second conductor is $$\theta_2$$. The temperature of junction in steady state is $$(\theta_2 > \theta_1)$$.

1611935_b28ec9dbf29748a6ab1d8780bf9adc30.png


A
10θ2+9θ119
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B
θ2+9θ10
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C
9θ2+θ110
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D
9θ2+10θ119
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Solution

The correct option is C $$\dfrac{9\theta_2 + \theta_1}{10}$$
Equating heat current in both slabs
$$\dfrac{K(\theta - \theta_1)}{3d} = \dfrac{3K(\theta_2 - \theta)}{d}$$
$$\theta - \theta_1 = 9\theta_2 - 9\theta$$
$$10\theta = 9\theta_2 + \theta_1$$
$$\theta = \dfrac{9\theta_2 + \theta_1}{10}$$

Physics
NCERT
Standard XII

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