(D) The argument of the $\sin$ function must be dimensionless,so $[\frac{\alpha x}{kt}] = [M^{0}L^{0}T^{0}]$.Given $x = [L]$,$k = [ML^{2}T^{-2}\theta^

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(D) The argument of the $\sin$ function must be dimensionless,so $[\frac{\alpha x}{kt}] = [M^{0}L^{0}T^{0}]$.Given $x = [L]$,$k = [ML^{2}T^{-2}\theta^{-1}]$,and $t = [\theta]$.Thus,$[\alpha] = [\frac{kt}{x}] = \frac{[ML^{2}T^{-2}\theta^{-1}][\theta]}{[L]} = [MLT^{-2}]$.Energy density $u$ is energy per unit volume,so $[u] = [ML^{-1}T^{-2}]$.From the expression $u = \frac{\alpha}{\beta}$,we have $[\beta] = \frac{[\alpha]}{[u]}$.$[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^{2}]$.Therefore,the dimensions of $\beta$ are $[M^{0}L^{2}T^{0}]$.

Class 11 Physics Units, Dimensions and Measurement Dimensional Analysis, Uses and Limitations

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An expression of energy density is given by $u = \frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$,where $\alpha, \beta$ are constants,$x$ is displacement,$k$ is Boltzmann constant,and $t$ is the temperature. The dimensions of $\beta$ will be.

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$[ML^{2}T^{-2}\theta^{-1}]$
B
$[M^{0}L^{2}T^{-2}]$
C
$[M^{0}L^{0}T^{0}]$
D
$[M^{0}L^{2}T^{0}]$

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Dimensional Analysis, Uses and Limitations

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An expression of energy density is given by $u = \frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$,where $\alpha, \beta$ are constants,$x$ is displacement,$k$ is Boltzmann constant,and $t$ is the temperature. The dimensions of $\beta$ will be.

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