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
595k+ views
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

