KVPY SX SB Exam 2016 Question Paper

Question : 38 of 120

Marks: +1, -0

A nuclear fuel rod generates energy at a rate of

5 × 10^8

Watt/

m^3

. It is in the shape of a cylinder of radius 4.0 mm and length 0.20 m . A coolant of specific heat

4 × 10^3

J/(kg-K) flows past it at a rate of 0.2 kg/s. The temperature rise in this coolant is approximately -

Solution:

{dm}/{dt}×S\ΔT={dθ}/{dt}

{dθ}/{dt}=5×10^8×\volume \of \rod

=5×10^8×π×4^2×10^{-6}×0.2/10

= 1600 π

0.2 × 4 × 10^3\ΔT = 1600π

8 × 10^2\ΔT = 16 × 10^2π

\ΔT = 3.14 × 2= 6.28ºC

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