Units & Dimensions

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86 of 149

Question 86 of 149

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

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Answer: (D) [M0L2T0]\left[{M}^{0}{L}^{2}{T}^{0}\right]

Given that energy density U=αβsin⁡αxktU=\dfrac{\alpha }{\beta }\sin \dfrac{\alpha x}{kt}
As αxkt\dfrac{\alpha x}{kt} should be dimensionless, the dimensions of αx\alpha x and ktkt should be same.
⇒[α][x]=[k][t]⇒[α]=[k][t][x]\Rightarrow \left[\alpha \right]\left[x\right]=\left[k\right]\left[t\right]\Rightarrow \left[\alpha \right]=\dfrac{\left[k\right]\left[t\right]}{\left[x\right]}
Now,
[U]=[α][β]\left[U\right]=\dfrac{\left[\alpha \right]}{\left[\beta \right]}
⇒[β]=[α][U]=[k][t][x][EV]=[Et][t][x][EV]\Rightarrow \left[\beta \right]=\dfrac{\left[\alpha \right]}{\left[U\right]}=\dfrac{\left[k\right]\left[t\right]}{\left[x\right]\left[\dfrac{E}{V}\right]}=\dfrac{\left[\dfrac{E}{t}\right]\left[t\right]}{\left[x\right]\left[\dfrac{E}{V}\right]}
⇒[β]=[L3][L]=[M0L2T0]\Rightarrow \left[\beta \right]=\dfrac{\left[{L}^{3}\right]}{\left[L\right]}=\left[{M}^{0}{L}^{2}{T}^{0}\right]

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