JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 1)
Consider the efficiency of carnot's engine is given by $$\eta=\frac{\alpha \beta}{\sin \theta} \log_e \frac{\beta x}{k T}$$, where $$\alpha$$ and $$\beta$$ are constants. If T is temperature, k is Boltzmann constant, $$\theta$$ is angular displacement and x has the dimensions of length. Then, choose the incorrect option :
Dimensions of $$\beta$$ is same as that of force.
Dimensions of $$\alpha^{-1} x$$ is same as that of energy.
Dimensions of $$\eta^{-1} \sin \theta$$ is same as that of $$\alpha \beta$$.
Dimensions of $$\alpha$$ is same as that of $$\beta$$.
Explanation
Since, dimensions trigonometric function and logarithmic function are dimensionless quantities.
$$ \therefore[\eta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]
$$
Also, dimensions of temperature, $[T]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}{ }^0 \mathrm{~K}\right]$
Dimensions of Boltzmann constant, $[k]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
Dimension of $x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]$
$$ \therefore[\eta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]
$$
Also, dimensions of temperature, $[T]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}{ }^0 \mathrm{~K}\right]$
Dimensions of Boltzmann constant, $[k]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
Dimension of $x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]$
(A) $$[\beta ] = \left[ {{{kT} \over x}} \right] = \left[ {{E \over x}} \right] = [ML{T^{ - 2}}] = [F]$$
(B) $$[\alpha \beta ] = [{M^0}{L^0}{T^0}]$$
$${[\alpha ]^{ - 1}} = [\beta ] = \left[ {{{kT} \over x}} \right]$$
So $${[\alpha ]^{ - 1}}[x] = [kT] = [M{L^2}{T^{ - 2}}]$$
(C) $$\eta \sin \theta = \alpha \beta $$
So $$[\eta \sin \theta ] = [\alpha \beta ]$$
$$[\eta ] = [{M^0}{L^0}{T^0}]$$ it is dimensionless quantity
(D) $$[\alpha ] \ne [\beta ]$$
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