Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are : \(\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=2\) and \(\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{1}{2}\). The free ends of rods A and B are maintained at \(400 \mathrm{~K}, 200 \mathrm{~K}\), respectively. The temperature of rods interface is ______ K , when equilibrium is established.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Heat & Thermodynamics

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are :

\frac{L _{A}}{L _{B}} = \frac{1}{2}, \frac{r _{A}}{r _{B}} = 2

and

\frac{K _{A}}{K _{B}} = \frac{1}{2}

. The free ends of rods A and B are maintained at

400 K, 200 K

, respectively. The temperature of rods interface is ______ K , when equilibrium is established.

Answer: 360

Solution

Each rod's thermal resistance is

R = \frac{L}{K π r ^{2}}

. Using the given ratios,

\frac{R _{A}}{R _{B}} = \frac{L _{A}}{L _{B}} \cdot \frac{K _{B}}{K _{A}} \cdot \frac{r _{B}^{2}}{r _{A}^{2}} = \frac{1}{2} \times 2 \times \frac{1}{4} = \frac{1}{4}

. In steady state the same heat flows through both rods, so

\frac{400 - T}{R _{A}} = \frac{T - 200}{R _{B}}

, which gives

\frac{400 - T}{T - 200} = \frac{1}{4}

. Then

4 (400 - T) = T - 200

, so

5 T = 1800

. The interface temperature is

360 K

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Heat & Thermodynamics →

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Heat & Thermodynamics

Solution

More JEE Main questions on Heat & Thermodynamics