Q.

A composite slab consists of two slabs  A and  B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of  A and B are  k1 and k2 respectively. A steady temperature difference of 12∘C is maintained across the composite slab. If  k1=k22, the temperature difference across slab  A will be:

Moderate

Unlock the Full Solution and Master the Concept

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept
By Expert Faculty of Sri Chaitanya

a

4∘C

b

8∘C

c

12∘C

d

16∘C

answer is 2.

(Detailed Solution Below)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

Let temperature of junction be  To∘CTemperature difference across slab  A=T1−To∘CTemperature difference across slab  B=To−T2∘CSince both slabs are connected in series, heat current through both will be equal k1AT1−T0l2=k2ATo−T2l2 As k1=k22⇒k22T1−To=To−T2k2⇒T1−To=2To−2T2⇒T1+2T2=3T0….(1) As T1−T2=12∘C⇒T2=T1−12…..(2) From (1) and (2), we get ⇒T1+2T1−12=3To⇒3T1−24=3T0⇒T1−T0=8∘C
ctaimg
Choose your Grade and Exam
edit icon

Connect with our

Expert Counsellors

access to India's best teachers with a record of producing top rankers year on year.

+91

We will send a verification code via OTP.

whats app icon