JEE Main 2017 · Physics · Calorimetry and Heat Transfer · medium

Two metal rods are joined end to end, with identical cross-sections of 4 cm^2. Rod P is 20 cm long with thermal conductivity 200 W m^-1K^-1 and rod Q…

Single correct option

Two metal rods are joined end to end, with identical cross-sections of

4\ \mathrm{cm^2}

. Rod

P

20\ \mathrm{cm}

long with thermal conductivity

200\ \mathrm{W\,m^{-1}K^{-1}}

and rod

Q

10\ \mathrm{cm}

long with conductivity

50\ \mathrm{W\,m^{-1}K^{-1}}

. The free end of

P

is kept at

100^\circ\mathrm{C}

and the free end of

Q

0^\circ\mathrm{C}

; the sides are insulated. In the steady state the temperature of the junction is nearly

Show answer and solution

Answer: (A) $66.7^\circ\mathrm{C}$

In steady state the same heat current flows through both rods, so they act as thermal resistances in series, with

R = \dfrac{L}{KA}

R_P = \dfrac{0.2}{200\times4\times10^{-4}} = 2.5\ \mathrm{K\,W^{-1}}

and

R_Q = \dfrac{0.1}{50\times4\times10^{-4}} = 5\ \mathrm{K\,W^{-1}}

. Heat current

H = \dfrac{100 - 0}{7.5} = 13.3\ \mathrm{W}

. The drop across

P

13.3\times2.5 = 33.3\ \mathrm{K}

, so the junction is at

100 - 33.3 = 66.7^\circ\mathrm{C}

. The junction is nearer the hot end because

P

offers less resistance than

Q

;

50^\circ\mathrm{C}

would need equal resistances.

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake