Question

For Physics Legends...

Figure shows the face and interface ...(refer to attachment)

Answer 1

Answer:

option d is correct

Explanation:

dear here is ur correct answer hope u got it

Answer 2

Topic

Heat Transfer

Given

A figure is given which shows the face and interface of composite slab consisting of two materials A and B.

To Find

The relation between Thermal Conductivities of A and B.

Formula to be used

$\dfrac{dQ}{dt}=$  $\dfrac{kA( \sf T_1 - \sf T_2 )}{l}$

where,

• $\dfrac{dQ}{dt} = \sf Rate \:of\:Heat\:Flow$  

• k = Thermal Conductivity

• A = Area of Slab

• l = length of slab

• $\sf T_1 - \sf T_2$ = Change in Temperature

Solution

For interface A of slab,

Length of slab = 2l

Change in temperature = $\sf T_1 - \sf T_2$

100 °C - 70 °C

30 °C

So, we can write,

$\dfrac{dQ}{dt}=$  $\dfrac{\sf k_AA( 30^{\circ} C )}{2l}$

For interface B of slab,

Length of slab = l

Change in temperature = $\sf T_1 - \sf T_2$

70 °C - 20 °C

50°C

So, we can write,

$\dfrac{dQ}{dt}=$  $\dfrac{\sf k_BA( 50^{\circ} C )}{l}$

It must be noted that materials are same for both interfaces as they are part of single slab.

$\sf \textbf{So,}\: \dfrac{dQ}{dt} \: \sf \textbf{is same for both interfaces.}$

It is clearly visible that 'Area' is same for both interfaces.

Now,

Taking ratio,

$\dfrac{\dfrac{dQ}{dt} = \dfrac{\sf k_AA( 30^{\circ} C )}{2l}}{\dfrac{dQ}{dt} = \dfrac{\sf k_BA( 50^{\circ} C )}{l}}$

$1 = \dfrac{\dfrac{\sf k_A 30^{\circ} C}{2}}{\sf k_B 50^{\circ} C}$

Now cross multiply,

 $\sf k_B50=\dfrac{\sf k_A 30 }{2}$

$\sf 100\sf k_B$ = $\sf30 k_A$

$\sf10 k_B$ = $\sf3 k_A$

Answer

So, answer is

$\sf 10k_B$ = $\sf 3k_A$

which is option A.