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A heat flux of $4000$ J/s is to be passed through a copper rod of length $10$ cm and area of cross-section $100$ $c m^{2}$ .The thermal conductivity of copper is $400$ $W / m^{\circ} C .$ The two ends of this rod must be kept at a temperature difference of:

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$100^{\circ} C$

$1^{\circ} C$

$10^{\circ} C$

$1000^{\circ} C$

answer is C.

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Detailed Solution

From $\frac{d Q}{d t} = \frac{K A Δ θ}{l}$

$Δ θ = \frac{l}{K \times A} \times \frac{d Q}{d t}$

$= \frac{0.1}{400 \times (100 \times 10^{- 4})} \times 4000 = 100 ° C$

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