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

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

(A) $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right]$

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

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

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Units & Measurement

An expression of energy density is given by

u = \frac{α}{β} sin (\frac{α x}{kt})

, where

α, β

are constants, x is displacement, k is Boltzmann constant and t is the temperature. The dimensions of

β

will be :

Answer: (D)

Energy density has dimensions

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

. The argument of sine must be dimensionless, so

\frac{α x}{k t}

is dimensionless. Since

k

(Boltzmann constant) has dimensions

[ML^{2} T^{- 2} θ^{- 1}]

and

t

has dimensions

[θ]

, we get

k t

has dimensions

[ML^{2} T^{- 2}]

. For the argument to be dimensionless,

α

must have dimensions

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

. Now,

\frac{α}{β}

must have dimensions of energy density

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

. Therefore,

β = \frac{[ ML ^{1} T ^{- 2} ]}{[ ML ^{- 1} T ^{- 2} ]} = [L^{2}] = [M^{0} L^{2} T^{0}]

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option