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(D) Efficiency $\eta$ is a dimensionless quantity. The argument of the logarithmic function must also be dimensionless,so $\frac{\beta x}{kT} = 1$ (di

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Dimensions of $\alpha$ are same as that of $\beta$

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(D) Efficiency $\eta$ is a dimensionless quantity. The argument of the logarithmic function must also be dimensionless,so $\frac{\beta x}{kT} = 1$ (dimensionless).Since $\eta$ is dimensionless,$[\eta] = [M^0 L^0 T^0]$.From $\frac{\beta x}{kT} = 1$,we get $\beta = \frac{kT}{x}$.Given $k = [M L^2 T^{-2} K^{-1}]$ and $T = [K]$,we have $\beta = \frac{[M L^2 T^{-2} K^{-1}][K]}{[L]} = [M L T^{-2}]$.This is the dimension of force,so option $A$ is correct.Since $\eta = \frac{\alpha \beta}{\sin 	heta} \cdot 	ext{constant}$,and $\eta$ and $\sin 	heta$ are dimensionless,$[\alpha \beta] = [1]$,which implies $[\alpha] = [\beta^{-1}] = [M^{-1} L^{-1} T^2]$.Option $D$ states $[\alpha] = [\beta]$,which is incorrect.For option $B$,$[\alpha^{-1} x] = [\beta x] = [M L T^{-2}][L] = [M L^2 T^{-2}]$,which is the dimension of energy. Thus,$B$ is correct.For option $C$,$[\eta^{-1} \sin 	heta] = [1] = [\alpha \beta]$,which is correct.

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