The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are $T_{2}$ and $T_{1} (T_{2} > T_{1})$ The rate of heat transfer through the slab, in a steady state is $(\frac{A (T_{2} - T_{1}) K}{x}) f$ , with $f$ equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2}{3}$

$1$

$\frac{1}{2}$

$\frac{1}{3}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Equation of thermal conductivity of the given combination $K_{e q} = \frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} = \frac{x + 4 x}{\frac{x}{K} + \frac{4 x}{2 K}} = \frac{5}{3} K$
Hence rate of flow of heat through the given combination is $\frac{Q}{t} = \frac{K_{e q} . A (T_{2} - T_{1})}{(x + 4 x)} = \frac{\frac{5}{3} K A (T_{2} - T_{1})}{5 x} = \frac{\frac{1}{3} K A (T_{2} - T_{1})}{x}$
On comparing it with given equation we get $f = \frac{1}{3}$