JEE Main 2015 · 11 Apr · Shift 1 · Physics Q1

← JEE Main 2015 · 11 Apr · Shift 1

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If electronic charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental quantities, the permeability of vacuum $\mu_0$ can be expressed in units of :

\left(\dfrac{hc}{me^2}\right)

\left(\dfrac h{me^2}\right)

\left(\dfrac h{ce^2}\right)

\left(\dfrac{mc^2}{he^2}\right)

Answer and worked solution

Official answer: C

The fine-structure constant $\alpha=\dfrac{e^2}{2\epsilon_0hc}$ is dimensionless, so $\epsilon_0$ has the units of $\dfrac{e^2}{hc}$ . Then
$\mu_0=\frac1{\epsilon_0c^2}\ \text{has the units of}\ \frac{hc}{e^2c^2}=\frac h{ce^2}.$

Topics:Dimensional analysis

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 1