Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T_1 = 300\,K and T_2 = 100\,K , as shown in the figure. The radius of the bigger cylinder is twice that of the smaller once and the thermal conductivities of the materials of the smaller and the larger cylinders are K_1 and K_2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200K, then K_1/K_2 = \underline\hspace2cm .

Properties of Matter PYQ — JEE Advanced 2018 · Shift 1 · Q48 with Solution

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Correct answer: 4

In steady state, the rate of heat flow through both cylinders connected in series is the same.

Using Fourier’s law,

\frac{Q}{t} = K A \frac{\Delta T}{L}

Since both cylinders have equal length, let each length be $L$ .

Let the radius of the smaller cylinder be $r$ . Then the radius of the bigger cylinder is $2r$ .

So,

A_1 = \pi r^2

for the smaller cylinder, and

A_2 = \pi (2r)^2 = 4\pi r^2 = 4A_1

for the bigger cylinder.

Given:

Hence,

So both cylinders have the same temperature drop.

Let the smaller cylinder have conductivity $K_1$ and area $A_1$ , and the larger cylinder have conductivity $K_2$ and area $A_2$ .

Since the heat current is same through both:

K_1 A_1 \frac{100}{L} = K_2 A_2 \frac{100}{L}

Cancelling common terms $100/L$ ,

K_1 A_1 = K_2 A_2

Using $A_2 = 4A_1$ ,

K_1 A_1 = K_2 (4A_1)

K_1 = 4K_2

Therefore,

\frac{K_1}{K_2} = 4

Derived answer: $4$

Stored correct answer: $4$

They agree.