Consider the efficiency of engine is given by $\eta=\frac{\alpha \beta}{\sin heta} \log _{e} \frac

Question

Consider the efficiency of engine is given by $\eta=\frac{\alpha \beta}{\sin 	heta} \log _{e} \frac{\beta x}{k T}$, where $\alpha$ and $\beta$ are constants. If $T$ is temperature, $k$ is Boltzmann constant, $	heta$ is angular displacement and $x$ has the dimensions of length. Then, choose the incorrect option.\

Professor PC Thomas Classes · Answer

(d) : Efficiency, $\eta=\frac{\alpha \beta}{\sin 	heta} \log _e \frac{\beta x}{k T}$ Dimension of $\left[\frac{\beta x}{k T}ight]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0ight]$ \ $ \beta=\frac{C k T}{x}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{[\mathrm{L}]}=\left[\mathrm{MLT}^{-2}ight] \quad 	ext { (where } C 	ext { is constant) } \ $ So, $\beta$ has dimension of force. $$ \begin{aligned} & {[\eta]=\left[\frac{\alpha \beta}{\sin 	heta}ight] ;[\alpha]=\left[\frac{1}{\beta}ight]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^2ight]} \ & {\left[\frac{x}{\alpha}ight]=[x \beta]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]} \end{aligned} $$ So, $\alpha^{-1} x$ has the dimensions of energy. Since $\eta$ and $\sin 	heta$ are dimensionless and $\alpha \beta$ is dimensionless, so, dimensions of $\eta^{-1} \sin 	heta$ is same as that of $\alpha \beta$.