An expression of energy density is given by u | vedclass

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(D) The argument of the sine function must be dimensionless,so the dimensions of $\frac{\alpha x}{kt}$ must be $[M^0L^0T^0]$.Since $[kt] = [	ext{Ener

Vedclass · Accepted Answer

$[M^0L^2T^0]$

Vedclass · Answer

(D) The argument of the sine function must be dimensionless,so the dimensions of $\frac{\alpha x}{kt}$ must be $[M^0L^0T^0]$.Since $[kt] = [	ext{Energy}] = [ML^2T^{-2}]$,we have $[\alpha] = \frac{[kt]}{[x]} = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]$.The energy density $u$ has dimensions of energy per unit volume,so $[u] = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.From the given equation $u = \frac{\alpha}{\beta} \sin(\dots)$,the dimensions of $\beta$ are given by $[\beta] = \frac{[\alpha]}{[u]}$.Substituting the dimensions: $[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [M^0L^2T^0]$.Therefore,the correct option is $D$.

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

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