JEE Main 12 Jan 2019 Morning Question 14
Question: A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is $ K _1 $ and that of the outer cylinder is $ K _2 $ . Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{2K _1+3K _2}{5} $
B) $ \frac{K _1+K _2}{2} $
C) $ K _1+K _2 $
D) $ \frac{K _1+3K _2}{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
-
We know that thermal resistance $ R=\frac{l}{KA} $
For inner cylinder $ R _1=\frac{l}{K _1\pi R^{2}} $
For outer cylinder $ R _2=\frac{l}{K _2\pi [{{(2R)}^{2}}-{{(R)}^{2}}]} $ $ R _2=\frac{l}{K _2\pi 3R^{2}} $ $ R _{eq} $
is the equivalent thermal resistance of the cylinder
$ \frac{1}{R _{eq}}=\frac{1}{R _1}+\frac{1}{R _2} $ $ \frac{K _{eq}4\pi R^{2}}{l}=\frac{K _{l}\pi R^{2}}{l}+\frac{3K _2\pi R^{2}}{l} $ $ 4K _{eq}=K _1+3K _2 $ $ K _{eq}=\frac{K _1+3K _2}{4} $
