Answer

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**Hint:**The power to which fundamental units are raised in order to obtain the unit of a physical quantity is called the dimensions of that physical quantity. Dimensions of a physical quantity do not depend on the system of units. We’ll go over the various physical quantities one-by-one and then compare the dimensions.

**Formula Used:**

\[\text{Energy density = }\dfrac{\text{Energy}}{\text{Volume}}\] , \[{\text{Young's Modulus = }}\dfrac{{{\text{Linear Stress}}}}{{{\text{Linear Strain}}}}\]

**Complete step by step solution:**

Energy density can be said to be the energy per unit volume. The energy of a body can be said to be equivalent to the work done, which in turn is equivalent to the product of the force and the displacement. Force is a product of the mass and acceleration of a body. Representing the above analysis in equation form, we get

\[\begin{align}

& \text{Energy density = }\dfrac{\text{Energy}}{\text{Volume}}=\dfrac{\text{Work done}}{\text{Volume}} \\

& \Rightarrow \text{Energy density =}\dfrac{\text{Force}\times \text{Displacement}}{\text{Volume}} \\

& \Rightarrow \text{Energy density =}\dfrac{\text{Mass}\times \text{acceleration}\times \text{Displacement}}{\text{Volume}} \\

\end{align}\]

The dimensions of Energy density can now be given as

\[\begin{align}

& \text{Energy density =}\dfrac{\left[ M \right]\times \left[ L{{T}^{-2}} \right]\times \left[ L \right]}{\left[ {{L}^{3}} \right]} \\

& \Rightarrow \text{Energy density =}\left[ M{{L}^{-1}}{{T}^{-2}} \right] \\

\end{align}\]

Since, the refractive index is a ratio, it is a dimensionless physical quantity. Similarly, the dielectric constant is also a ratio and is thus a dimensionless physical quantity. Both the Refractive index and Dielectric constant have a dimension of \[1\].

Young’s Modulus of a substance is a ratio of the linear stress to the linear strain. Since strain is a ratio and hence dimensionless. Young’s modulus has the same dimensions as linear stress, which is force per unit area. Force is equal to the product of mass and acceleration. Representing Young’s modulus in equation form, we have

\[{\text{Young's Modulus = }}\dfrac{{{\text{Linear Stress}}}}{{{\text{Linear Strain}}}}\]

Dimensions of Young’s Modulus=Dimensions of linear stress ($\because$ Strain is dimensionless)

\[{\text{Linear Stress = }}\dfrac{{{\text{Force}}}}{{{\text{Area}}}}{\text{ = }}\dfrac{{{\text{Mass}} \times {\text{acceleration}}}}{{{\text{Area}}}}\]

The dimensions of Young’s modulus can now be given as

\[ {\text{Linear Stress = }}\dfrac{{{\text{[M]}} \times [L{T^{ - 2}}]}}{{\left[ {{L^2}} \right]}}\]

\[\Rightarrow {\text{Young's Modulus = }}\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\]

From the solution given above, we can say that Energy Density and Young’s Modulus have the same dimensions. We can go forward and find the dimensions of the Magnetic Field but we already have our answer, so you can skip that part and say that:

**Option (C) is the correct answer.**

**Note:**In the given question, we converted the given physical quantities into the basic units. But for that, we must know the dimensions of several basic quantities like mass, acceleration, time, area, volume, velocity, etc. Acceleration has dimensions \[\left[ L{{T}^{-2}} \right]\] , velocity has dimensions \[\left[ L{{T}^{-1}} \right]\] , area has dimensions \[\left[ {{L}^{2}} \right]\] , volume has dimensions \[\left[ {{L}^{3}} \right]\]. Without knowing the dimensions of these basic quantities, you cannot reach the correct answer.

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