Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^{\circ} \mathrm{C}$ and $40^{\circ} \mathrm{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y , the temperature at junction points $B$ and $E$ are (close to) :

(A) $89^{\circ} \mathrm{C}$ and $73^{\circ} \mathrm{C}$ respectively

(B) $80^{\circ} \mathrm{C}$ and $60^{\circ} \mathrm{C}$ respectively

(C) $80^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$ respectively

(D) $60^{\circ} \mathrm{C}$ and $45^{\circ} \mathrm{C}$ respectively

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Heat & Thermodynamics

Rods

x

and

y

of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points

A

and

F

are maintained at

10 0^{\circ} C

and

4 0^{\circ} C

respectively. Given the thermal conductivity of rod x is three times of that of rod y , the temperature at junction points

B

and

E

are (close to) :

Answer: (A)

Equal rods have thermal resistance inversely proportional to conductivity, so with

k_{x} = 3 k_{y}

let rod

x

R

and rod

y

3 R

. From

A

B

there is one rod

x

(

R

); between

B

and

E

two branches run in parallel,

x + x = 2 R

and

y + y = 6 R

, combining to

\frac{( 2 R ) ( 6 R )}{8 R} = 1.5 R

; from

E

F

there is one rod

y

(

3 R

). The total is

R + 1.5 R + 3 R = \frac{11 R}{2}

, so the heat current is

H = \frac{100 - 40}{11 R /2} = \frac{120}{11 R}

. Then

T_{B} = 100 - H R = 100 - \frac{120}{11} \approx 8 9^{\circ} C

and

T_{E} = 40 + H (3 R) = 40 + \frac{360}{11} \approx 7 3^{\circ} C .

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option