If the capacitance of a nanocapacitor is measured in terms of a unit u made by combining the electric charge e, Bohr radius a

Q: 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

d The unit of capacitance is $\frac {Coulomb}{V} = \frac {Coulomb}{\text{work done per unit charge}}$=$\frac{\text { Coulomb }}{\frac{ W }{\text { Coulomb }}}=\frac{\text { Coulomb }^{2}}{ kgm ^{2} s ^{-2}}= C ^{2} kg ^{-1} m ^{-2} s ^{2}$ Checking units of each option: $\mu=\frac{ e ^{2} c }{\text { ha }} \Rightarrow=\frac{\text { Coulomb }^{2} ms ^{-1}}{ kgm ^{2} s ^{-1} m }= C ^{2} kg ^{-1} m ^{-2} s ^{0}$ $\mu=\frac{ e ^{2} h }{ ca _{\circ}} \Rightarrow=\frac{\text { Coulomb }^{2} kgm ^{2} s ^{-1}}{ ms ^{-1} m }= C ^{2} kg ^{1} m ^{0} s ^{0}$ $\mu=\frac{\text { hc }}{ e ^{2} a _{\circ}} \Rightarrow=\frac{ kgm ^{2} s ^{-1} ms ^{-1}}{\text { Coulomb }^{2} m }= C ^{-2} kg ^{1} m ^{1} s ^{-2}$ $\mu=\frac{ e ^{2} a _{\circ}}{ hc } \Rightarrow=\frac{\text { Coulomb }^{2} m }{ kgm ^{2} s ^{-1} ms ^{-1}}= C ^{2} kg ^{-1} m ^{-2} s ^{-2}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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$u\, = \,\frac{{{e^2}h}}{{{a_0}}}$
B$u\, = \,\frac{{hc}}{{{e^2}{a_0}}}$
C$u\, = \,\frac{{{e^2}c}}{{h{a_0}}}$
D$u\, = \,\frac{{{e^2}{a_0}}}{{hc}}$

JEE MAIN 2015, Diffcult

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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