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.
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.
- [JEE MAIN 2022]
- A
$\left[ ML ^{2} T ^{-2} \theta^{-1}\right]$
- B
$\left[ M ^{0} L ^{2} T ^{-2}\right]$
- C
$\left[ M ^{0} L ^{0} T ^{0}\right]$
- D
$\left[ M ^{0} L ^{2} T ^{0}\right]$
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Given below are two statements: One is labelled as Assertion $(A)$ and other is labelled as Reason $(R)$.
Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
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Assertion $(A)$ : Time period of oscillation of a liquid drop depends on surface tension $(S)$, if density of the liquid is $p$ and radius of the drop is $r$, then $T = k \sqrt{ pr ^{3} / s ^{3 / 2}}$ is dimensionally correct, where $K$ is dimensionless.
Reason $(R)$: Using dimensional analysis we get $R.H.S.$ having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
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