Q122 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
The efficiency of an engine is given by $\eta=\frac{\alpha \beta}{\sin \theta} \cdot \log _{\mathrm{e}} \frac{\beta \mathrm{x}}{\mathrm{kT}}$, where $\alpha$ and $\beta$ are constants. If $T$ is the absolute temperature, $k$ - Boltzmann constant, $\theta$-angular displacement and $\mathbf{x}$ is distance, then the incorrect statement is
- ADimensions of $\beta$ are same as that of force
- BDimensions of $\alpha^{-1} \mathrm{x}$ are same as that of energy
- CDimensions of $\eta^{-1} \sin \theta$ are same as that of $\alpha \beta$
- Correct answerDDimensions of $\alpha$ are same as that of $\beta$
Solution
The efficiency of an engine is given by
$\begin{aligned}<br />& \eta=\frac{\alpha \beta}{\sin \theta} \log _{\mathrm{e}} \frac{\beta \mathrm{x}}{\mathrm{kT}} \\<br />& {[\alpha \beta]=[\eta]=[\sin \theta]=\text { Dimension less }} \\<br />& {\left[\eta^{-1} \sin \theta\right]=[\alpha \cdot \beta]=\text { Dimension less }} \\<br />& \frac{\beta \mathrm{x}}{\mathrm{kt}}=1 \\<br />& \beta=\frac{\mathrm{kT}}{\mathrm{x}}<br />\end{aligned}$
Here, Boltzmann constant, $\mathrm{K}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
$\begin{aligned}<br />& \beta=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{K}]}{[\mathrm{L}]} \\<br />& =\left[\mathrm{ML} \mathrm{T}^{-2}\right]<br />\end{aligned}$
Force, $\mathrm{F}=\mathrm{ma}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$
Force, $\mathrm{F}=\left[\mathrm{ML} \mathrm{T}^{-2}\right]$
Hence, Dimension of $\beta$ are same as that of force.
$\begin{aligned}<br />& \beta=\alpha^{-1}=\left[\mathrm{ML} \mathrm{T}^{-2}\right] \\<br />& \alpha^{-1} \mathrm{x}=\left[\mathrm{ML} \mathrm{T}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]<br />\end{aligned}$
Energy, $\mathrm{E}=\mathrm{mc}^2=\mathrm{ML}^2 \mathrm{~T}^{-2}$
Hence, dimension of $\alpha^{-1} \mathrm{x}$ are same as that of energy.
$\begin{aligned}<br />& \eta=\frac{\alpha \beta}{\sin \theta} \log _{\mathrm{e}} \frac{\beta \mathrm{x}}{\mathrm{kT}} \\<br />& {[\alpha \beta]=[\eta]=[\sin \theta]=\text { Dimension less }} \\<br />& {\left[\eta^{-1} \sin \theta\right]=[\alpha \cdot \beta]=\text { Dimension less }} \\<br />& \frac{\beta \mathrm{x}}{\mathrm{kt}}=1 \\<br />& \beta=\frac{\mathrm{kT}}{\mathrm{x}}<br />\end{aligned}$
Here, Boltzmann constant, $\mathrm{K}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
$\begin{aligned}<br />& \beta=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{K}]}{[\mathrm{L}]} \\<br />& =\left[\mathrm{ML} \mathrm{T}^{-2}\right]<br />\end{aligned}$
Force, $\mathrm{F}=\mathrm{ma}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$
Force, $\mathrm{F}=\left[\mathrm{ML} \mathrm{T}^{-2}\right]$
Hence, Dimension of $\beta$ are same as that of force.
$\begin{aligned}<br />& \beta=\alpha^{-1}=\left[\mathrm{ML} \mathrm{T}^{-2}\right] \\<br />& \alpha^{-1} \mathrm{x}=\left[\mathrm{ML} \mathrm{T}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]<br />\end{aligned}$
Energy, $\mathrm{E}=\mathrm{mc}^2=\mathrm{ML}^2 \mathrm{~T}^{-2}$
Hence, dimension of $\alpha^{-1} \mathrm{x}$ are same as that of energy.
Hence, the correct answer is option: 'D'.