Two conducting slabs having same shape and size of heat conductivities k1 and k2 joined as shown in figure. The temperature at ends of the slabs are θ1 and θ2 (θ1>θ2). The final temperature of junction (θm ) is:
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
a
k1θ1+k2θ2k1+k2
b
k1θ2+k2θ1k1+k2
c
k1θ2-k2θ1k1+k2
d
none of these
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Assuming both slabs have identical thicknesses, ∆Q∆t=k1A(θ1-θm)l=k2A(θm-θ2)l⇒θm=k1θ1+k2θ2k1+k2