The efficiency of an engine is given by eta = | vedclass

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

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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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