The entropy of any system is given by
$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$ where $\alpha$ and $\beta$ are the constants. $\mu$, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.
[Take $S = {{dQ} \over T}$]
Choose the incorrect option from the following :

Since, entropy of the system is given by<br/><br/>$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$ .... (i)<br/><br/>As, $S = {Q \over {\Delta T}}$ [given]<br/><br/>$\Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ .... (ii)<br/><br/>$\because$ Dimensions of Q = [ML²T^$-$2]<br/><br/>Dimension of T = [K]<br/><br/>Boltzmann constant, $k = {{energy} \over T}$ [$\because$ Dimensions of energy = [ML²T^$-$2]]<br/><br/>$\Rightarrow [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ ..... (iii)<br/><br/>From Eqs. (ii) and (iii), we can write,<br/><br/>$[S] = [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$ ..... (iv)<br/><br/>$\because$ Gas constant, $[R] = {{[Energy]} \over {[nT]}} = {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}}$ .... (v)<br/><br/>and mechanical equivalent of heat<br/><br/>[J] = [M⁰L⁰T⁰] .... (vi)<br/><br/>As, [$\mu$kR] = [J$\beta$]² <br/><br/>Using Eqs. (iii), (v) and (vi), we get<br/><br/>$$ \Rightarrow [mol] \times {{[M{L^2}{T^{ - 2}}]} \over {[K]}} \times {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}} = [{\beta ^2}]$$<br/><br/>$\Rightarrow [\beta ] = [M{L^2}{T^{ - 2}}{K^{ - 1}}]$ ..... (vii)<br/><br/>Using Eq. (i), we can write,<br/><br/>$$[{\alpha ^2}] = {{[S]} \over {[\beta ]}} = {{[M{L^2}{T^{ - 2}}{K^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}{K^{ - 1}}]}} \Rightarrow \alpha = [{M^0}{L^0}{T^0}]$$ .... (viii)<br/><br/>So, from Eqs. (iii) and (viii), we can say that $\alpha$ and k have different dimensions.