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

Option: 1
$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right]$

Option: 2
$\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$

Option: 3
$\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$

Option: 4
$\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]$

Answers (1)

best_answer

$\mathrm{u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right) }$

$\mathrm{{[x]=\left[L^{\prime}\right]} }$

$\mathrm{{[k t]=[\text { Energy }]=\left[M^{1} L^2 T^{-2}\right]} }$

$\mathrm{{\left[\frac{\alpha x}{k t}\right]=\left[M^{\circ} L^{\circ} T^0\right]} }$

$\mathrm{\therefore[\alpha]=\left[M^{1} L^{1} T^{-2}\right] \rightarrow(1) }$

$\mathrm{{[u]=\left[\frac{\alpha}{\beta}\right]=\left[M^{1} L^{-1} T^{-2}\right]} }$

$\mathrm{{[\beta]=\left[L^2\right] \rightarrow(2) }}$

Hence 4 is correct option.

Posted by

Ritika Kankaria

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