Properties Of Solids And Liquids Question 438

Question: A cylindrical rod of aluminum is of length 20 cm, and radius 2 cm. The two ends are maintained at temperatures of $ 0{}^\circ C $ and $ 50{}^\circ C $ [the coefficient of thermal conductivity is $ \frac{0.5cal}{cm\times sec{{\times }^{o}}C} $ ]Then the thermal resistance of the rod in $ \frac{cal}{sec{{\times }^{o}}C} $

Options:

A) 318

B) 31.8

C) 3.18

D) 0.318

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Thermal resistance $ R=/lKA $

Where r =20cm. & A(cylindrical rod) $ =\pi r^{2}=40\pi cm^{2} $

So $ R=\frac{20}{0.5\times 40\pi }=0.318\frac{cal}{\sec \times {}^\circ C} $



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