JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 15)
The entropy of any system is given by
$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ where $$\alpha$$ and $$\beta$$ are the constants. $$\mu$$, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.
[Take $$S = {{dQ} \over T}$$]
Choose the incorrect option from the following :
$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ where $$\alpha$$ and $$\beta$$ are the constants. $$\mu$$, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively.
[Take $$S = {{dQ} \over T}$$]
Choose the incorrect option from the following :
$$\alpha$$ and J have the same dimensions.
S and $$\alpha$$ have different dimensions
S, $$\beta$$, k and $$\mu$$R have the same dimensions
$$\alpha$$ and k have the same dimensions
Explanation
Since, entropy of the system is given by
$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ .... (i)
As, $$S = {Q \over {\Delta T}}$$ [given]
$$ \Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ .... (ii)
$$\because$$ Dimensions of Q = [ML2T$$-$$2]
Dimension of T = [K]
Boltzmann constant, $$k = {{energy} \over T}$$ [$$\because$$ Dimensions of energy = [ML2T$$-$$2]]
$$ \Rightarrow [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ ..... (iii)
From Eqs. (ii) and (iii), we can write,
$$[S] = [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ ..... (iv)
$$\because$$ Gas constant, $$[R] = {{[Energy]} \over {[nT]}} = {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}}$$ .... (v)
and mechanical equivalent of heat
[J] = [M0L0T0] .... (vi)
As, [$$\mu$$kR] = [J$$\beta$$]2
Using Eqs. (iii), (v) and (vi), we get
$$ \Rightarrow [mol] \times {{[M{L^2}{T^{ - 2}}]} \over {[K]}} \times {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}} = [{\beta ^2}]$$
$$ \Rightarrow [\beta ] = [M{L^2}{T^{ - 2}}{K^{ - 1}}]$$ ..... (vii)
Using Eq. (i), we can write,
$$[{\alpha ^2}] = {{[S]} \over {[\beta ]}} = {{[M{L^2}{T^{ - 2}}{K^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}{K^{ - 1}}]}} \Rightarrow \alpha = [{M^0}{L^0}{T^0}]$$ .... (viii)
So, from Eqs. (iii) and (viii), we can say that $$\alpha$$ and k have different dimensions.
$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ .... (i)
As, $$S = {Q \over {\Delta T}}$$ [given]
$$ \Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ .... (ii)
$$\because$$ Dimensions of Q = [ML2T$$-$$2]
Dimension of T = [K]
Boltzmann constant, $$k = {{energy} \over T}$$ [$$\because$$ Dimensions of energy = [ML2T$$-$$2]]
$$ \Rightarrow [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ ..... (iii)
From Eqs. (ii) and (iii), we can write,
$$[S] = [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ ..... (iv)
$$\because$$ Gas constant, $$[R] = {{[Energy]} \over {[nT]}} = {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}}$$ .... (v)
and mechanical equivalent of heat
[J] = [M0L0T0] .... (vi)
As, [$$\mu$$kR] = [J$$\beta$$]2
Using Eqs. (iii), (v) and (vi), we get
$$ \Rightarrow [mol] \times {{[M{L^2}{T^{ - 2}}]} \over {[K]}} \times {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}} = [{\beta ^2}]$$
$$ \Rightarrow [\beta ] = [M{L^2}{T^{ - 2}}{K^{ - 1}}]$$ ..... (vii)
Using Eq. (i), we can write,
$$[{\alpha ^2}] = {{[S]} \over {[\beta ]}} = {{[M{L^2}{T^{ - 2}}{K^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}{K^{ - 1}}]}} \Rightarrow \alpha = [{M^0}{L^0}{T^0}]$$ .... (viii)
So, from Eqs. (iii) and (viii), we can say that $$\alpha$$ and k have different dimensions.
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