Physics · 2015
JEE Main · 10 April 2015, Shift 1 · Q16
If the capacitance of a nanocapacitor is measured in terms of a unit ' u ' made by combining the electronic charge 'e', Bohr radius ' a_0 ', Planck's…
If the capacitance of a nanocapacitor is measured in terms of a unit ' $\displaystyle u$ ' made by combining the electronic charge 'e', Bohr radius ' $\displaystyle \mathrm{a}_0$ ', Planck's constant ' h ' and speed of light 'c' then :
Official answer
(3)
$\displaystyle u=\frac{\mathrm{e}^2 \mathrm{a}_0}{\mathrm{hc}}$
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JEE Main 2015 Physics question, with the answer from NTA’s final answer key. Where our answers come from.