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If the capacitance of a nanocapacitor is...

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

A

`u=(e^(2)h)/(a_(0))`

B

`u=(hc)/(e^(2)a_(0))`

C

`u=(e^(2)c)/(ha_(0)`

D

`u=(e^(2)a_(0))/(hc)`

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