JEE Main 2024 · Physics · Calorimetry and Heat Transfer · medium

Two rods of equal cross-section are joined end to end. The first has length 10 cm and thermal conductivity K, the second has length 30 cm and…

Numerical answer

Two rods of equal cross-section are joined end to end. The first has length

10\ \mathrm{cm}

and thermal conductivity

K

, the second has length

30\ \mathrm{cm}

and conductivity

2K

. The free end of the first rod is held at

100^\circ\mathrm{C}

and the free end of the second at

0^\circ\mathrm{C}

. In steady state with no side losses, the temperature (in

^\circ\mathrm{C}

) of the junction is

Show answer and solution

Answer: 60

In steady state the heat current through both rods is the same:

\dfrac{KA(100-T)}{10}=\dfrac{2KA(T-0)}{30}

. Simplifying,

3(100-T)=2T

, so

300=5T

and

T=60^\circ\mathrm{C}

. Thermal resistances are in series (

R\propto L/K

10/K

and

15/K

); the temperature drop divides in the ratio

2:3

, giving

40

and

60

degrees.

Original question written in the JEE Main 2024 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake