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GUJCET · Physics · Electromagnetic Waves
The dimensional formula of \(\mu_{0} \varepsilon_{0}\) is __________ .
- A \(\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{2}\)
- B \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)
- C \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\)
- D \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{1}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{2}\)
Step-by-step Solution
Detailed explanation
(A) \(\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{2}\)
\(c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\frac{1}{c^{2}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\frac{1}{\left(M^{0} L^{1} T^{-1}\right)^{2}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{2}\)
\(c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\frac{1}{c^{2}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\frac{1}{\left(M^{0} L^{1} T^{-1}\right)^{2}}\)
\(\therefore \mu_{0} \varepsilon_{0}=\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{2}\)
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