Rods and of equal dimensions but of different materials are joined as shown in the figure. Temperatures of end points and are maintained at and respectively. [2026]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
Rods $x$ and $y$ of equal dimensions but of different materials are joined as shown in the figure. Temperatures of end points $A$ and $F$ are maintained at $100 ° C$ and $40 ° C$ 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): [2026]

1 89 °C and 73 °C respectively

2 80 °C and 60 °C respectively

3 80 °C and 70 °C respectively

4 60 °C and 45 °C respectively

Ans.
(1)

$Let [R = \frac{ℓ}{3 K A}]$

$H = [\frac{100 - 40}{\frac{11 R}{2}}] . . . (1)$

$H = \frac{100 - T_{B}}{R} . . . (2)$

$H = \frac{T_{E} - 40}{3 R} . . . (3)$

Using (1) and (2)

$120 = 1100 - 11 T_{A}$

$T_{B} = 89 ° C$

Using (1) and (3)

$T_{E} = 73 ° C$

Solution

1 89 °C and 73 °C respectively

2 80 °C and 60 °C respectively

3 80 °C and 70 °C respectively

4 60 °C and 45 °C respectively

Ans.
(1)

$Let [R = \frac{ℓ}{3 K A}]$

$H = [\frac{100 - 40}{\frac{11 R}{2}}] . . . (1)$

$H = \frac{100 - T_{B}}{R} . . . (2)$

$H = \frac{T_{E} - 40}{3 R} . . . (3)$

Using (1) and (2)

$120 = 1100 - 11 T_{A}$

$T_{B} = 89 ° C$

Using (1) and (3)

$T_{E} = 73 ° C$

Want Us to Email you About Special Offers & Updates?

Solution

1 89 °C and 73 °C respectively

2 80 °C and 60 °C respectively

3 80 °C and 70 °C respectively

4 60 °C and 45 °C respectively

Ans. (1) Let [R=ℓ3KA] H=[100-4011R2] ...(1) H=100-TBR ...(2) H=TE-403R ...(3) Using (1) and (2) 120=1100-11TA TB=89°C Using (1) and (3) TE=73°C

Want Us to Email you About Special Offers & Updates?

Ans.
(1)

$Let [R = \frac{ℓ}{3 K A}]$

$H = [\frac{100 - 40}{\frac{11 R}{2}}] . . . (1)$

$H = \frac{100 - T_{B}}{R} . . . (2)$

$H = \frac{T_{E} - 40}{3 R} . . . (3)$

Using (1) and (2)

$120 = 1100 - 11 T_{A}$

$T_{B} = 89 ° C$

Using (1) and (3)

$T_{E} = 73 ° C$