A cylinder A of a material of B thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius equal to the cylinder. The shell is made of a material of thermal conductivity $K_{2}$ . The two ends of the combined system are maintained at two different temperature. If the effective thermal conductivity of the system is $\frac{4 K_{1} + 5 K_{2}}{9}$ the ratio of outer radius of the shell to the radius of the cylinder.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{5}{3}$

$\frac{5}{2}$

2.0

$\frac{3}{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Radius of cylinder = R and outer radius of shell = nR

$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{\frac{1}{K_{1} {πR}^{2}} \times \frac{1}{K_{2} (n^{2} - 1) {πR}^{2}}}{\frac{1}{K_{1} {πR}^{2}} + \frac{1}{K_{2} (n^{2} - 1) {πR}^{2}}}$

But $R_{eq} = \frac{1}{{Kπn}^{2} R^{2}}$

On solving, $K = \frac{[K_{1} + K_{2} (n^{2} - 1)]}{n^{2}}$

= $\frac{4 K_{1} + 5 K_{2}}{9} (given)$

= $\frac{\frac{K_{1} + 5 K_{2}}{4}}{\frac{9}{4}}$

n = $\frac{3}{2}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET