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A nuclear fuel rod generates energy at a rate of $5 \times 10^{8} \mathrm{Watt} / \mathrm{m}^{3}$. It is in the shape of a cylinder of radius $4.0 \mathrm{~mm}$ and length $0.20 \mathrm{~m}$. A coolant of specific heat $4 \times 10^{3} \mathrm{~J} /(\mathrm{kg}-\mathrm{K})$ flows past it at a rate of $0.2 \mathrm{~kg} / \mathrm{s}$. The temperature rise in this coolant is approximately -
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$6^{\circ} \mathrm{C}$
$\frac{\mathrm{dm}}{\mathrm{dt}} \times \mathrm{S} \Delta \mathrm{T}=\frac{\mathrm{d} \theta}{\mathrm{dt}}$
$\frac{\mathrm{d} \theta}{\mathrm{dt}}=5 \times 10^{8} \times$ volume of rod
$=5 \times 10^{8} \times \pi \times(4)^{2} \times 10^{-6} \times \frac{0.2}{10}$
$=5 \times 10 \times \pi \times 16 \times 2$
$=1600 \pi$
$0.2 \times 4 \times 10^{3} \Delta \mathrm{T}=1600 \pi$
$8 \times 10^{2} \Delta \mathrm{T}=16 \times 10^{2} \pi$
$\Delta \mathrm{T}=3.14 \times 2$
$\Rightarrow 6.28^{\circ} \mathrm{C}$
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