If the capacitance of a nanocapacitor is measured in terms of a unit $u$ made by combining the electric charge $e,$ Bohr radius $a_0,$ Planck's constant $h$ and speed of light $c$ then
If the capacitance of a nanocapacitor is measured in terms of a unit $u$ made by combining the electric charge $e,$ Bohr radius $a_0,$ Planck's constant $h$ and speed of light $c$ then
- [JEE MAIN 2015]
- A
$u\, = \,\frac{{{e^2}h}}{{{a_0}}}$
- B
$u\, = \,\frac{{hc}}{{{e^2}{a_0}}}$
- C
$u\, = \,\frac{{{e^2}c}}{{h{a_0}}}$
- D
$u\, = \,\frac{{{e^2}{a_0}}}{{hc}}$
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