JEE Main 2015 · 10 Apr · Shift 1 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

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 constant 'h' and speed of light 'c' then :

u=\dfrac{e^2c}{ha_0}

u=\dfrac{e^2h}{ca_0}

u=\dfrac{e^2a_0}{hc}

u=\dfrac{hc}{e^2a_0}

Answer and worked solution

Official answer: C

Capacitance has the units of $\dfrac{\text{charge}^2}{\text{energy}}$ . Now $hc$ has units of energy × length, so
$\frac{e^2a_0}{hc}\ \text{has units}\ \frac{\text{C}^2\cdot\text{m}}{\text{J}\cdot\text{m}}=\frac{\text{C}^2}{\text{J}}=\text{F}.$
None of the other combinations reduces to $\frac{\text{C}^2}{\text{J}}$ .

Topics:Dimensional analysis Capacitors and capacitance

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 16