Question

Find the ratio of thermal conductivity $\frac{K_{1} }{K_{2} }$ if length of all rods are same?

Answer 1

Steady state temperature are shown in diagram.

we have to find the ratio of thermal conductivity k₁/k₂ if length of all rods are same.

solution : let dQ amount of heat flows from left side to right side so each rod gets heat Q in their region.

using formula,

dQ/dt = kA dT/dx

for rod of thermal conductivity k₁

dQ/dt = K₁A (100° - 70°)/l = K₁A × 30/l......(1)

for rod of thermal conductivity k₂

dQ/dt = K₂A(70° - 20°)/l = k₂A50/l ......(2)

from equations (1) and (2) we get,

K₁A × 30/l = k₂A50/l

⇒K₁/K₂ = 5/3

Therefore the ratio of thermal conductivity k₁/k₂ is 3 : 5 ( question included wrong options )

Answer 2

QUESTION:

Steady state temperature are shown in the diagram. Find the ratio of thermal conductivity K₁/K₂, if length of all rods are same.

ANSWER:

• The ratio of K₁/K₂ = 5/3

GIVEN:

• Steady state temperature is given.

TO FIND:

• The ratio of K₁/K₂.

EXPLANATION:

$\boxed{\bold{\large{\gray{Thermal\ Conductivity\ (K) = \dfrac{QL}{A\Delta T}}}}}$

Let Q be the heat through the conductors and l be the length.

Also assume area of conductors as A.

$\sf \leadsto K_1 = \dfrac{QL}{A \ \Delta T_1}$

$\sf \leadsto K_1 = \dfrac{QL}{A \ (100 - 70)}$

$\sf \leadsto K_1 = \dfrac{QL}{30A}$

$\sf \leadsto K_2 = \dfrac{QL}{A \ \Delta T_2}$

$\sf \leadsto K_2 = \dfrac{QL}{A \ (70 - 20)}$

$\sf \leadsto K_2 = \dfrac{QL}{50A}$

$\sf \leadsto \dfrac{K_1}{K_2} = \dfrac{\dfrac{QL}{30A}}{ \dfrac{QL}{50A}}$

$\sf \leadsto \dfrac{K_1}{K_2} = \dfrac{\dfrac{1}{30A}}{ \dfrac{1}{50A}}$

$\sf \leadsto \dfrac{K_1}{K_2} = \dfrac{\dfrac{1}{3}}{ \dfrac{1}{5}}$

$\sf \leadsto \dfrac{K_1}{K_2} = \dfrac{5}{3}$

HENCE THE RATIO OF K₁/K₂ = 5 : 3.