If the capacitance of a nanocapacitor is measured in terms of a unit ' $u$ ' made by combining the electric…

If the capacitance of a nanocapacitor is measured in terms of a unit ' $u$ ' made by combining the electric charge ' $e$ ', Bohr radius ' $\mathrm{a}_{0}$ ', Planck's constant 'h'and speed of light ' $c$ ' then
  1. $\mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{~h}}{\mathrm{a}_{0}}$
  2. $\mathrm{u}=\frac{\mathrm{hc}}{\mathrm{e}^{2} \mathrm{a}_{0}}$
  3. $u=\frac{e^{2} c}{h a_{0}}$
  4. $\mathrm{u}=\frac{\mathrm{e}^{2} \mathrm{a}_{0}}{\mathrm{hc}}$

Solution

Let unit ' $u$ ' related with $e, a_{0}, h$ and $c$ as follows. $[u]=[e]^{a}\left[a_{0}\right]^{b}[h]^{c}[C]^{d}$
Using dimensional method, $\left[M^{-1} L^{-2} T^{+4} A^{+2}\right]=\left[A^{1} T^{1}\right]^{a}[L]^{b}\left[M L 2 T^{-1}\right]^{c}\left[L T^{-1}\right]^{d}$
$\left[M^{-1} L^{-2} T^{+4} A^{+2}\right]=\left[M^{c} L^{b+2 c+d} T^{a-c-d} A^{a}\right]$
$a=2, b=1, c=-1, d=-1$
$\therefore \quad u=\frac{e^{2} a_{0}}{h c}$ ,

Asked in: JEE Mains - Units and Dimensions - Test 1

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