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QJEEM22Ph9ba8: JEE Main 2022 Physics - Class 11 JEEM22Ph9ba8 Solution: Consider the efficiency of carnot's engine is given by , where…

Consider the efficiency of carnot's engine is given by η=αβsin⁡θlog⁡eβxkT\eta=\frac{\alpha \beta}{\sin \theta} \log_e \frac{\beta x}{k T} , where α\alpha and β\beta are constants. If T is temperature, k is Boltzmann constant, θ\theta is angular displacement and x has the dimensions of length. Then, choose the incorrect option :

Multiple Choice Options

  • A Dimensions of β\beta is same as that of force.
  • B Dimensions of α−1x\alpha^{-1} x is same as that of energy.
  • C Dimensions of η−1sin⁡θ\eta^{-1} \sin \theta is same as that of αβ\alpha \beta .
  • D Dimensions of α\alpha is same as that of β\beta .

Step-by-Step Verified Solution

Since, dimensions trigonometric function and logarithmic function are dimensionless quantities.

∴[η]=[M0 L0 T0]

Also, dimensions of temperature, [T]=[M0 L0 T0 K][T]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}{ }^0 \mathrm{~K}\right]

Dimensions of Boltzmann constant, [k]=[ML2 T−2 K−1][k]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]

Dimension of x=[M0LT0]x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]

(A) [β]=[kTx]=[Ex]=[MLT−2]=[F][\beta ] = \left[ {{{kT} \over x}} \right] = \left[ {{E \over x}} \right] = [ML{T^{ - 2}}] = [F]

(B) [αβ]=[M0L0T0][\alpha \beta ] = [{M^0}{L^0}{T^0}]

[α]−1=[β]=[kTx]{[\alpha ]^{ - 1}} = [\beta ] = \left[ {{{kT} \over x}} \right]

So [α]−1[x]=[kT]=[ML2T−2]{[\alpha ]^{ - 1}}[x] = [kT] = [M{L^2}{T^{ - 2}}]

(C) ηsin⁡θ=αβ\eta \sin \theta = \alpha \beta

So [ηsin⁡θ]=[αβ][\eta \sin \theta ] = [\alpha \beta ]

[η]=[M0L0T0][\eta ] = [{M^0}{L^0}{T^0}] it is dimensionless quantity

(D) [α]≠[β][\alpha ] \ne [\beta ]

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