The entropy of any system is given byS = alph | vedclass

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(D) The argument of the logarithmic function must be dimensionless,so $\frac{\mu k R}{J \beta^{2}}$ must be dimensionless.Given $S = \frac{dQ}{T}$,the

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$\alpha$ and $k$ have the same dimensions.

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(D) The argument of the logarithmic function must be dimensionless,so $\frac{\mu k R}{J \beta^{2}}$ must be dimensionless.Given $S = \frac{dQ}{T}$,the dimensions of entropy $S$ are $[M L^{2} T^{-2} K^{-1}]$.The gas constant $R$ has dimensions $[M L^{2} T^{-2} K^{-1} mol^{-1}]$ and Boltzmann constant $k$ has dimensions $[M L^{2} T^{-2} K^{-1}]$.Since $\frac{\mu k R}{J \beta^{2}}$ is dimensionless,$[\beta^{2}] = \frac{[\mu][k][R]}{[J]}$.Since $J$ (mechanical equivalent of heat) is a conversion factor between energy units,it is dimensionless. Thus,$[\beta^{2}] = [mol] \cdot [M L^{2} T^{-2} K^{-1}] \cdot [M L^{2} T^{-2} K^{-1} mol^{-1}] = [M^{2} L^{4} T^{-4} K^{-2}]$.Therefore,$[\beta] = [M L^{2} T^{-2} K^{-1}]$,which is the same as the dimensions of $S, k,$ and $\mu R$.From $S = \alpha^{2} \beta$,since $S$ and $\beta$ have the same dimensions,$\alpha^{2}$ must be dimensionless,meaning $\alpha$ is dimensionless.Checking the options:$(A)$ $S, \beta, k, \mu R$ have same dimensions: Correct.$(B)$ $\alpha$ and $J$ have same dimensions: Incorrect,as $\alpha$ is dimensionless and $J$ is dimensionless,but the statement implies they share a non-trivial dimension or are being compared incorrectly. Actually,$\alpha$ is dimensionless and $J$ is dimensionless,so they have the same dimensions. Wait,let's re-evaluate: $\alpha$ is dimensionless,$J$ is dimensionless. So $(B)$ is correct.$(C)$ $S$ and $\alpha$ have different dimensions: Correct,as $S$ has dimensions and $\alpha$ is dimensionless.$(D)$ $\alpha$ and $k$ have the same dimensions: Incorrect,as $\alpha$ is dimensionless and $k$ has dimensions $[M L^{2} T^{-2} K^{-1}]$. Thus,$(D)$ is the incorrect option.

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