The two rods show

Let’s take first condition –

Since, rods are connected in series, rate of heat flow remains constant.

= = constant

Where

∆T = is change in temperature between the two sides of the rods.

A= Area of cross section of the rods

K = thermal conductivity of the rods

L = length of the rod

Hence

=

Where ,

K_{A ,} K_B are thermal conductivity of rod A and B

⇒ 30 K_A = 70 K_{B ……………………………}(1)

Now, in second condition, the rods are interchanged as shown in fig below –

Let x°C be the unknown temperature of the junction AB

Now, since rate of heat flow remains constant-

=

That gives

100_B - x_B = _A x

Substituting for _A from(1) and solving the above equation

100 = x +x

⇒

x = 30°C