Physics - Unit & Dimension Question with Solution | TestHub

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

$\left[\beta x\right]=\left[kT\right]$

Since the dimensions of $kT$ are equivalent to that of energy,

$\left[\beta \right]=\frac{\left[kT\right]}{\left[x\right]}=\frac{\left[E\right]}{\left[x\right]}=\frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}{\left[\mathrm{L}\right]}=\left[{\mathrm{MLT}}^{-2}\right]$

Therefore, $\beta$ has dimensions of force.

Since efficiency is a dimensionless ratio,

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

$\Rightarrow \left[\beta \right]=\frac{1}{\left[\alpha \right]}$

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

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