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PhysicsHeat & ThermoConductionMedium2 minQB

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Passage / Comprehension

For heat flow through a cross-section, $\frac{d Q}{d t} = - K S (\frac{d T}{d x})$ , where K is thermal conductivity, S is cross-sectional area and $\frac{d T}{d x}$ is the temperature gradient. In the rod, $T (x) = 100 - 2 x^{2} - 4 x$ kelvin, with x in metres. Length $L = 2 m$ ; $K = 100 W/ (m K)$ ; $S = 0.1 m^{2}$ ; $M = 5 kg$ ; $α = 1 0^{- 4} / K$ . The source diagram marks A at $x = 0$ and B at $x = L$ .

Temperature gradient at A and B would be :

Options:

Answer:

Solution:

$T (x) = 100 - 2 x^{2} - 4 x$
$\frac{d T}{d x} = - 4 x - 4$
$\frac{d T}{d x}_{A} = - 4 K/m (x = 0)$
$\frac{d T}{d x}_{B} = - 12 K/m (x = 2 m)$
Answer B.

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Conduction

⏱ 2mℹ️ Source: QB

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