TS EAMCET · Physics · Units and Dimensions
The dimension of \(\frac{E^2}{\mu_0}\) in mass \((M)\), length \((L)\) and time \((T)\) is ( \(E=\) electric field, \(\mu_0=\) permeability of free space)
- A \(\left[\mathrm{M}^2 \mathrm{~L}^3 \mathrm{~T}^{-2} \mathrm{~A}^2\right]\)
- B \(\left[\mathrm{MLT}^{-4}\right]\)
- C \(\left[\mathrm{ML}^3 \mathrm{~T}^{-2}\right]\)
- D \(\left[\mathrm{ML}^4 \mathrm{~T}^{-4}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\mathrm{MLT}^{-4}\right]\)
Step-by-step Solution
Detailed explanation
Dimension of \(E = [MLT^{-3}A^{-1}]\) Dimension of \(\mu_0 = [MLT^{-2}A^{-2}]\) Dimension of \(\frac{E^2}{\mu_0} = \frac{([MLT^{-3}A^{-1}])^2}{[MLT^{-2}A^{-2}]}\) \(= [M^{2-1}L^{2-1}T^{-6-(-2)}A^{-2-(-2)}]\) \(= [MLT^{-4}]\)
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