JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 1)
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 :
$$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right]$$
$$\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$$
$$\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$$
$$\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]$$
Explanation
$$u = {\alpha \over \beta }\sin \left( {{{\alpha x} \over {kt}}} \right)$$
$$[\alpha] = \left[ {{{kt} \over x}} \right] = {{[Energy]} \over {[Dis\tan ce]}}$$
$$[\beta ] = {{[\alpha ]} \over {[u]}}$$
$$ = {{[Energy]/[Dis\tan ce]} \over {[Energy]/[Volume]}}$$
$$ = [{L^2}]$$
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