Consider the efficiency of Carnot's engine is given by η =...

Given that, $η = \dfrac{\alpha\beta}{sin\theta}\log \left(\dfrac{\betax}{kT}\right)$. Since expression inside the logarithm is dimensionless

\left[\betax\right] = \left[kT\right]$</p><p>Since the dimensions of$kT$are equivalent to that of energy,</p><p>\left[\beta\right] = \dfrac{\left[kT\right]}{\left[x\right]} = \dfrac{\left[E\right]}{\left[x\right]} = \dfrac{\left[{ML}^{2}{T}^{- 2}\right]}{\left[L\right]} = \left[{MLT}^{- 2}\right]$

Therefore, $\beta$ has dimensions of force.

Since efficiency is a dimensionless ratio,

$\left[\alpha\right]\left[\beta\right] = \left[{M}^{0}{L}^{0}{T}^{0}\right]$

$⇒ \left[\beta\right] = \dfrac{1}{\left[\alpha\right]}$

Therefore, dimensions of $\alpha$ is not same as that of $\beta$. Hence option D is incorrect statement and the answer.

$\dfrac{\left[x\right]}{\left[\alpha\right]} = \left[\beta\right]\left[x\right] = \left[{ML}^{2}{T}^{- 2}\right]$. Therefore, ${\alpha}^{- 1}x$ has dimensions of energy.