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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2022
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\frac{\alpha[ L ]}{\left[ ML ^{2} T ^{-2}\right]}=\left[ M ^{0} L ^{0} T ^{0}\right]\)

\(\alpha=\left[ ML ^{1} T ^{-2}\right]\)

\(\frac{\alpha}{\beta}=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ L ^{3}\right]} \Rightarrow \beta=\frac{\left[ ML ^{1} T ^{-2}\right]\left[ L ^{3}\right]}{ ML ^{2} T ^{-2}}\)

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