Question

The following is the dimensions of:
$\left[ {{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{\theta }^{1}} \right]$

A. Coefficient of thermal conductivity
B. Coefficient of viscosity
C. Modulus of rigidity
D. Thermal resistance

Answer 1

Hint: For each of the options, from their basic formulae, find the dimensional formula for the terms.
Formula used: Formula for Coefficient of thermal conductivity, Coefficient of viscosity, Modulus of rigidity, Thermal resistance:

& \text{coefficient of thermal conductivity}=\dfrac{\text{Rate of heat transfer per unit time}\times \text{distance}}{\text{Change in temperature in Kelvin}\times \text{Area of cross-section}} \\\ & \text{coefficient of viscosity}\left( \eta \right)=\dfrac{Fdz}{Adv} \\\ & \text{Modulus of rigidity }\left( \mu \right)=\dfrac{\text{stress}}{\text{strain}} \\\ & \text{Thermal resistance }(R)=\dfrac{\text{Change in temperature in kelvin}}{\text{Rate of heat transfer per unit time}} \\\ \end{aligned}$$ Complete step-by-step answer: Every quantity can be expressed in the terms of the following seven dimensions Dimension Symbol Length L Mass M Time T Electric charge Q Luminous intensity C Temperature $\theta $ Angle None For option A. Coefficient of thermal conductivity. Coefficient of thermal conductivity of a material is the rate of flow heat per unit area per unit change in temperature across a solid. $$\begin{aligned} & \text{coefficient of thermal conductivity}=\dfrac{\text{Rate of heat transfer per unit time}\times \text{distance}}{\text{Change in temperature in Kelvin}\times \text{Area of cross-section}} \\\ & \Rightarrow k=\dfrac{Qd}{A\left( {{\theta }_{2}}-{{\theta }_{1}} \right)t} \\\ \end{aligned}$$ Therefore, the dimensional formula equals $\left[ {{M}^{1}}{{L}^{1}}{{T}^{-3}}{{\theta }^{-1}} \right]$ Hence, this option is incorrect. For option B. Coefficient of viscosity. Coefficient of viscosity is defined as the tangential force required to maintain a unit velocity gradient in the depth of a unit area of a liquid. $\text{coefficient of viscosity}\left( \eta \right)=\dfrac{Fdz}{Adv}$ Therefore, the dimensional formula equals $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-1}} \right]$. For option C. Modulus of rigidity. Modulus of rigidity is defined as the ratio of shear stress to shear strain. $\text{Modulus of rigidity }\left( \mu \right)=\dfrac{\text{stress}}{\text{strain}}$ Where, the unit of stress is the same as that pressure and strain is a dimensionless quantity. Therefore, the dimensional formula equals $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$. For option D. Thermal resistance. Thermal resistance is defined as the resistance a body provides when heat is transferred. In simple words, it is the ratio of change in temperature to rate of heat transfer. $\text{Thermal resistance }(R)=\dfrac{\text{Change in temperature in kelvin}}{\text{Rate of heat transfer per unit time}}$ Therefore, the dimensional formula equals $\left[ {{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{\theta }^{1}} \right]$. Thus, the answer to this question option D. Thermal resistance. Note: Firstly, taking change for a quantity does not change its dimension because it is essentially the difference of the quantity. Secondly, since there is a dimension of $\theta $ in the question, the term should have a term for temperature. Therefore, option B and C can be neglected.