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} $
BETA
