Question 139: Units , Dimensions and Error Analysis

Question

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 :

Accepted Answer

$u = {\alpha  \over \beta }\sin \left( {{{\alpha x} \over {kt}}} ight)$
$[\alpha] = \left[ {{{kt} \over x}} ight] = {{[Energy]} \over {[Dis	an ce]}}$
$[\beta ] = {{[\alpha ]} \over {[u]}}$
$ = {{[Energy]/[Dis	an ce]} \over {[Energy]/[Volume]}}$
$ = [{L^2}]$

Answer

$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} 	heta^{-1}ight]$

Answer

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

Answer

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

Answer

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

Units , Dimensions and Error Analysis

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 :

Choose the correct answer: