Question

A cylindrical rod of aluminium is of length $20$ cms and radius $2$ cms. The two ends are maintained at temperatures of $0^{\circ }C$ and ${50}^{\circ }C$ the coefficient of thermal conductivity is $0.5cal/\left(cm\times sec{\times }^{\circ }C\right)$Then the thermal resistance of the rod in $cal/\left(sec{\times }^{\circ }C\right)$ is

• A. 318
• B. 31.8
• C. 3.18
• D. 0.318

Answer 1

The correct option is D 0.318

Thermal resistance $R=l/KA$
Where $l=20cm$ &

$A$ (cylindrcal rod) $=\pi r^2=40\pi c{m}^2$
So $R=\frac{20}{0.5\times 40\pi }=0.318\frac{cal}{sec{\times }^{\circ }C}$