JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 1)
An expression for a dimensionless quantity P is given by $$P = {\alpha \over \beta }{\log _e}\left( {{{kt} \over {\beta x}}} \right)$$; where $$\alpha$$ and $$\beta$$ are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of $$\alpha$$ will be :
[M0 L$$-$$1 T0]
[M L0 T$$-$$2]
[M L T$$-$$2]
[M L2 T$$-$$2]
Explanation
Given, $P=\frac{\alpha}{\beta} \log _{e}\left[\frac{k t}{\beta x}\right]$
The logarithmic term is dimensionless.
Thus, $[k t / \beta x]$ is also dimensionless.
i.e. $\frac{[k][t]}{[\beta][x]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ .......(i)
We have, $E=k t$
Thus, Eq. (i) becomes,
$$ \begin{array}{r} \frac{\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{\beta}]\left[\mathrm{L}^{1}\right]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ {[\beta]=\left[\mathrm{MLT}^{-2}\right]} \end{array} $$
Since, $P$ is also a dimensionless quantity.
$$ \begin{aligned} &\frac{[\alpha]}{[\beta]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ &{[\alpha]=[\beta]\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]} \\\\ &{[\alpha]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]} \end{aligned} $$
The logarithmic term is dimensionless.
Thus, $[k t / \beta x]$ is also dimensionless.
i.e. $\frac{[k][t]}{[\beta][x]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ .......(i)
We have, $E=k t$
Thus, Eq. (i) becomes,
$$ \begin{array}{r} \frac{\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{\beta}]\left[\mathrm{L}^{1}\right]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ {[\beta]=\left[\mathrm{MLT}^{-2}\right]} \end{array} $$
Since, $P$ is also a dimensionless quantity.
$$ \begin{aligned} &\frac{[\alpha]}{[\beta]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ &{[\alpha]=[\beta]\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]} \\\\ &{[\alpha]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]} \end{aligned} $$
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