Verified Solution • 2025–26 Session moderate

QJEEM22Phc3d4: JEE Main 2022 Physics - Class 11 JEEM22Phc3d4 Solution: An expression of energy density is given by , where are…

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

Multiple Choice Options

  • A [ML2 T−2θ−1]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right]
  • B [M0 L2 T−2]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]
  • C [M0 L0 T0]\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]
  • D [M0 L2 T0]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]

Step-by-Step Verified Solution

u=αβsin⁡(αxkt)u = {\alpha \over \beta }\sin \left( {{{\alpha x} \over {kt}}} \right)

[α]=[ktx]=[Energy][Distan⁡ce][\alpha] = \left[ {{{kt} \over x}} \right] = {{[Energy]} \over {[Dis\tan ce]}}

[β]=[α][u][\beta ] = {{[\alpha ]} \over {[u]}}

=[Energy]/[Distan⁡ce][Energy]/[Volume]= {{[Energy]/[Dis\tan ce]} \over {[Energy]/[Volume]}}

=[L2]= [{L^2}]

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