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.
Pick an option to check your answer.
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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