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Law Of Thermal Conductivity MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Law of Thermal Conductivity is considered one of the most asked concept.

  • 14 Questions around this concept.

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The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be:

The unit of thermal conductivity is : 

A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100 °C. A block of ice 0 °C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is:

(Given latent heat of fusion of ice = 3.36 x 105 J kg-1)

 Three rods of Copper, Brass and Steel are welded together to form a Y - shaped structure.  Area of cross - section of each    rod = 4 cm2. End of copper rod is maintained at 1000C where as ends of brass and steel are kept at 00C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :

 

Two rods, one semi-circular of thermal conductivity K1 and the other straight of thermal conductivity K2 and of the same cross-sectional area, are joined as shown in the figure. The points A and B are maintained at the same temperature difference. If the rate of flow of heat is the same in two rods, then k1k2 is.

 

 

 

 

Two rods A and B of different materials are welded together. Their thermal conductivities are K1 and K1. The thermal conductivity of the composite rod will be

Temperature difference of 120C is maintained between two ends of a uniform rod AB of length 2 L. The 3L2 is connected across AB (See figure). In another bent rod PQ, of the same cross-section as AB and length 2, is connected across AB (See figure). In steady state, temperature(in 0C ) difference between P and Q will be close to :

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?

Enlargement of cotyledons is an indication of 
 

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Concepts Covered - 1

Law of Thermal Conductivity

Law of Thermal Conductivity - 

 

                                          

Q=KA(θ1θ2)tl

Q= Amount of heat transfer
t= Time of heat flow
K= Thermal conductivity of the material

So, from the above equation we can calculate the- Rate of flow of heat i.e. heat current which can be written as -

Qt=H=KA(θ1θ2)l


In the differential form, this heat current can also be written as -

dQdt=KAdθdx


In the case of a non-steady state or variable cross-section, this is the more general equation that can be used to solve problems.

Relation of thermal conductivity of some material -

KAg>KCu>KAlKSolid >KLiquid >KGas KMetals >KNon-metals 
Thermal resistance (Rth): The thermal resistance of a body is defined as the measure of its opposition to the flow of heat through it. It is defined as the ratio of temperature difference to the heat current (Qt).

Rth=θ1θ2H=θ1θ2KA(θ1θ2)/I=lKA
 

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Law of Thermal Conductivity

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