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