
The dimensional formula for coefficient of thermal conductivity is:
(A). \[\left[ {MLTK} \right]\]
(B). \[\left[ {MLT{K^{ - 1}}} \right]\]
(C). \[\left[ {ML{T^{ - 1}}{K^{ - 1}}} \right]\]
(D). \[\left[ {ML{T^{ - 3}}{K^{ - 1}}} \right]\]
Answer
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Hint: Coefficient of thermal conductivity of a solid material is the rate of flow of heat per unit area per unit temperature gradient across the solid. So, the formula of the Coefficient of thermal conductivity includes the quantities like heat, length, area, temperature, time.
Complete step-by-step answer:
So, as we know the formula for the Coefficient of thermal conductivity is as follows,
${K_{th}} = \dfrac{{Q\Delta x}}{{A\Delta Tt}}$,
Where,
$Q = $heat, $A$= area, $\Delta x$= change in length, $\Delta T$= change in the temperature, $t = $time.
The dimension of the quantities,
For heat, $Q = \left[ {M{L^2}{T^{ - 2}}} \right]$,
For area, $A = \left[ {{L^2}} \right]$,
For change in length the dimension will be same as that of that length, so $\Delta x = \left[ L \right]$,
For change in the temperature the dimension will be same as that of the temperature so the dimension of the change temperature, $\Delta T = \left[ K \right]$
And the dimension for the time, $t = \left[ T \right]$
So now writing the dimensions of the quantities in the formula of the Coefficient of thermal conductivity,
${K_{th}} = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ {{L^2}} \right]\left[ K \right]\left[ T \right]}}$
$ \Rightarrow {K_{th}} = \dfrac{{\left[ {M{L^{3 - 2}}{T^{ - 2 - 1}}} \right]}}{{\left[ K \right]}}$
$ \Rightarrow {K_{th}} = \dfrac{{\left[ {ML{T^{ - 3}}} \right]}}{{\left[ K \right]}}$
$ \Rightarrow {K_{th}} = \left[ {ML{T^{ - 3}}{K^{ - 1}}} \right]$
So, the dimension of the Coefficient of thermal conductivity is $\left[ {ML{T^{ - 3}}{K^{ - 1}}} \right]$.
So, option (D) is the correct answer.
Note: Thermal conductivity is the rate at which heat is transferred by one of the methods of heat transfer that is conduction, through a cross sectional area of that particular material, when a temperature gradient exists which is perpendicular to the area. Significance of thermal conductivity is that the slower the rate at which temperature differences transmit through the material, and so the more effective it is as an insulator. The value of the coefficient of thermal conductivity will reflect the conductivity of the material. If the value of the coefficient of thermal conductivity is large then the material is a good conductor of heat and if the value of the coefficient of thermal conductivity is small then it is a good thermal insulator.
Complete step-by-step answer:
So, as we know the formula for the Coefficient of thermal conductivity is as follows,
${K_{th}} = \dfrac{{Q\Delta x}}{{A\Delta Tt}}$,
Where,
$Q = $heat, $A$= area, $\Delta x$= change in length, $\Delta T$= change in the temperature, $t = $time.
The dimension of the quantities,
For heat, $Q = \left[ {M{L^2}{T^{ - 2}}} \right]$,
For area, $A = \left[ {{L^2}} \right]$,
For change in length the dimension will be same as that of that length, so $\Delta x = \left[ L \right]$,
For change in the temperature the dimension will be same as that of the temperature so the dimension of the change temperature, $\Delta T = \left[ K \right]$
And the dimension for the time, $t = \left[ T \right]$
So now writing the dimensions of the quantities in the formula of the Coefficient of thermal conductivity,
${K_{th}} = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ {{L^2}} \right]\left[ K \right]\left[ T \right]}}$
$ \Rightarrow {K_{th}} = \dfrac{{\left[ {M{L^{3 - 2}}{T^{ - 2 - 1}}} \right]}}{{\left[ K \right]}}$
$ \Rightarrow {K_{th}} = \dfrac{{\left[ {ML{T^{ - 3}}} \right]}}{{\left[ K \right]}}$
$ \Rightarrow {K_{th}} = \left[ {ML{T^{ - 3}}{K^{ - 1}}} \right]$
So, the dimension of the Coefficient of thermal conductivity is $\left[ {ML{T^{ - 3}}{K^{ - 1}}} \right]$.
So, option (D) is the correct answer.
Note: Thermal conductivity is the rate at which heat is transferred by one of the methods of heat transfer that is conduction, through a cross sectional area of that particular material, when a temperature gradient exists which is perpendicular to the area. Significance of thermal conductivity is that the slower the rate at which temperature differences transmit through the material, and so the more effective it is as an insulator. The value of the coefficient of thermal conductivity will reflect the conductivity of the material. If the value of the coefficient of thermal conductivity is large then the material is a good conductor of heat and if the value of the coefficient of thermal conductivity is small then it is a good thermal insulator.
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