COMEDK Exam 2018 Solved Paper

The given situation is shown belowThermal conductivity of copper
=9× thermal conductivity of steel
Kc=9Ks
Since, both bar are connected in series, hence heat current along the bar will be same.
‌ i. e ‌‌‌

dQ

dt

=‌

KcA(T1−T)

l1

‌=‌

KsA(T−T2)

l2

⇒‌

9KsA(100−T)

18

=‌

KsA(T−0)

6

⇒3(100−T)=T
⇒300−3T=T
⇒4T=300
‌T=75∘C