JEE Main 2022 · 27 Jul · Shift 2 · Physics Q31

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PhysicsSingle correct+4 / -1Question 31 of 90

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

[ML^2T^{-2}\theta^{-1}]

[M^0L^2T^{-2}]

[M^0L^0T^0]

[M^0L^2T^0]

Answer and worked solution

Official answer: D

The argument of sin is dimensionless, so $\alpha=\frac{kt}x$ has the dimensions of energy over length.

$\beta=\frac\alpha u$ $=\frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]}=[M^0L^2T^0]$ .

(D)

Topics:Dimensional analysis

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JEE Main 2022 · 27 Jul · Shift 2, Physics question 31