A) 48\[\left[ M{{L}^{0}}T{{A}^{2}} \right]\]
B) 24\[\left[ M{{L}^{0}}T{{A}^{-2}} \right]\]
C) 56\[y=5\sin 2\pi \left( \frac{t}{0.04}-\frac{x}{40} \right)\]
D) 36\[c=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]
Correct Answer: A
Solution :
Using the formula \[n=\frac{c}{\upsilon }\] or \[n=\frac{c}{\upsilon }\] (Here: \[\sqrt{\frac{{{\mu }_{0}}{{\varepsilon }_{0}}}{\mu \varepsilon }}\] and \[\sqrt{\frac{{{\mu }_{0}}{{\varepsilon }_{0}}}{\mu \varepsilon }}\]) Now from eq (i), we get \[\sqrt{\frac{\mu \varepsilon }{{{\mu }_{0}}{{\varepsilon }_{0}}}}\] or \[\sqrt{\frac{{{\mu }_{0}}{{\varepsilon }_{0}}}{\mu \varepsilon }}\] or \[6\times {{10}^{16}}\] So, \[1.67\times {{10}^{14}}Hz\]
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