Home
Class 12
PHYSICS
Two rods A and B of different materials ...

Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are `K_(1)` and `K_(2)`. The thermal conductivity of the composite rod will be

A

`(3(K_(1)+K_(2)))/(2)`

B

`K_(1)+K_(2)`

C

`2(K_(1)+K_(2))`

D

`(K_(1)+K_(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D

Thermal resistance `(R)=(kA)/(l)`
`R_("eq")=(R_(1)R_(2))/(R_(1)+R_(2))`
`(because` for parallel)
`l/(2k_("eq")A)=l/(k_(1)A)+l/(k_(2)A)`
Thermal resistance `(R)=l/(kA)`
Rods are in parallel combination
`R_("eq")=(R_(1)R_(2))/(R_(1)+R_(2))`
`l/(K_("eq")(2A))=(l/(K_(1)A)*l/(K_(2)A))/(l/(K_(1)A)+l/(K_(2)A))=(1/(K_(1)K_(2))l/(A))/(1/(K_(1))+1/(K_(2)))`
`=(1/(K_(1)K_(2))l/(A))/((K_(1)+K_(2))/(K_(1)K_(2)))=1/((K_(1)+K_(2)))l/(A)`
`rArrK_("eq")=(K_(1)+K_(2))/(2)`
Promotional Banner

Topper's Solved these Questions

  • THERMAL PROPERTIES OF MATTER

    PHYSICS WALLAH|Exercise Level- 2|30 Videos
  • SYSTEM OF PARTICLES AND ROTATIONAL MOTION

    PHYSICS WALLAH|Exercise NEET PAST 5 YEARS QUESTIONS |22 Videos
  • THERMODYNAMICS

    PHYSICS WALLAH|Exercise NEET Past 5 Years Questions|14 Videos

Similar Questions

Explore conceptually related problems

The thermal conductivity of a rod depends on

A slab consists of two layers of different materials of the same thickness and having thermal conductivities K_(1) and K_(2) . The equivalent thermal conductivity of the slab is

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities K_(1) and K_(2) . The equivalent conductivity of the combination is

Two rods of the same length and diameter, having thermal conductivities K_(1) and K_(2) , are joined in parallel. The equivalent thermal conductivity to the combinationk is

A composite slab consists of two slabs A and B of different materials but of the same thickness placed one on top of the other. The thermal conductivities of A and B are k_(1) and K_(2) respectively. A steady temperature difference of 12^(@)C is maintained across the composite slab. If k_(1) = K_(2)//2 , the temperature difference across slab A will be:

Two slabs A and B of different materials but of the same thicknesss are joined end to end to form a composite slab. The thermal conductivities of A and B are K_(1) and K_(2) respectively. A steady temperature difference of 12^(@) C is maintained across the composite slab. If K_(1) = K_(2)/2 , the temperature difference across slabs A is

A wall has two layers A and B each made of different materials. The thickness of both the layers is the same. The thermal conductivity of A, K_(A) = 3K_(B) . The temperature different across the wall is 20^(@)C in thermal equilibrium

Consider a compound slab consisting of two different material having equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is