In SI units, the dimensions of $\sqrt{\frac{\epsilon_{0}}{\mu_{0}}}$ is

Question:

In SI units, the dimensions of $\sqrt{\frac{\epsilon_{0}}{\mu_{0}}}$ is

  1. (a) $\mathrm{A}^{-1} \mathrm{TML}^{3}$

  2. $\mathrm{AT}^{2} \mathrm{M}^{-1} \mathrm{~L}^{-1}$

  3. $\mathrm{AT}^{-3} \mathrm{ML}^{3 / 2}$

  4. $\mathrm{A}^{2} \mathrm{~T}^{3} \mathrm{M}^{-1} \mathrm{~L}^{-2}$

JEE Main Previous Year Single Correct Question of JEE Main from Chemistry Laws of Motion chapter.

JEE Main Previous Year 2019


Correct Option: 4

Solution:

$\left[\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}\right]=\sqrt{\frac{\varepsilon_{0}^{2}}{\mu_{0} \varepsilon_{0}}}=\left[\frac{\varepsilon_{0}}{\sqrt{\mu_{0} \varepsilon_{0}}}\right]=\varepsilon_{0} C\left[L T^{-1}\right] \times\left[\varepsilon_{0}\right]$

$\because F=\frac{q^{2}}{4 \pi \varepsilon_{0} r^{2}}$

$\left.\Rightarrow\left[\varepsilon_{0}\right]=\frac{1}{[A T]^{2}}=C\right]$

$\therefore\left[\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}\right]=\left[L T^{-1}\right] \times\left[A^{2} M^{-1} L^{-3} T^{4}\right]$

$=\left[M^{-1} L^{-2} T^{3} A^{2}\right]$

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