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1.Units, Dimensions and Measurement
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$\frac{1}{{{\mu _0}{\varepsilon _0}}}$ की विमा होगी, जहाँ प्रतीकों का सामान्य अर्थ है
A$[L{T^{ - 1}}]$
B$[{L^{ - 1}}T]$
C$[{L^{ - 2}}{T^2}]$
D$[{L^2}{T^{ - 2}}]$
Solution
(d) $c = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} \Rightarrow \frac{1}{{{\mu _0}{\varepsilon _0}}} = {c^2} = [{L^2}{T^{ – 2}}]$
Standard 11
Physics
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