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(D) Efficiency $\eta$ is a dimensionless quantity. Therefore,$[\eta] = [M^0 L^0 T^0]$.In the expression $\eta = \frac{\alpha \beta}{\sin 	heta} \log_

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Dimensions of $\alpha$ is same as that of $\beta$.

Vedclass · Answer

(D) Efficiency $\eta$ is a dimensionless quantity. Therefore,$[\eta] = [M^0 L^0 T^0]$.In the expression $\eta = \frac{\alpha \beta}{\sin 	heta} \log_{e} \frac{\beta x}{k T}$,the argument of the logarithm must be dimensionless. Thus,$[\frac{\beta x}{k T}] = [M^0 L^0 T^0]$.Since $[k T] = 	ext{Energy} = [M L^2 T^{-2}]$ and $[x] = [L]$,we have $[\beta] = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$,which is the dimension of force.Now,since $\eta$ is dimensionless and $\sin 	heta$ is dimensionless,the term $\frac{\alpha \beta}{\sin 	heta}$ must be dimensionless. Thus,$[\alpha \beta] = [M^0 L^0 T^0]$.Since $[\beta] = [M L T^{-2}]$,then $[\alpha] = [M^{-1} L^{-1} T^2]$.Checking the options:$A$: $[\beta] = [M L T^{-2}]$ (Force),which is correct.$B$: $[\alpha^{-1} x] = [M L T^{-2} \cdot L] = [M L^2 T^{-2}]$ (Energy),which is correct.$C$: $[\eta^{-1} \sin 	heta] = [1] = [\alpha \beta]$,which is correct.$D$: $[\alpha] = [M^{-1} L^{-1} T^2]$ and $[\beta] = [M L T^{-2}]$. These are not the same. Thus,this is the incorrect option.

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