PhysicsEasy156×since 2002Q8260An expression of energy density is given by u=αβsin(αxkt)u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)u=βαsin(ktαx), where α,β\alpha, \betaα,β are constants, xxx is displacement, kkk is Boltzmann constant and t is the temperature. The dimensions of β\betaβ will be :A[ML2 T−2θ−1]\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right][ML2 T−2θ−1]B[M0 L2 T−2]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right][M0 L2 T−2]C[M0 L0 T0]\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right][M0 L0 T0]D[M0 L2 T0]\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right][M0 L2 T0]Check answerSkip