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JEE Main 2015 (10 Apr, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2015 (10 Apr, Shift 1)

Q70JEE Main 2015 · 10 APR · SHIFT 1Medium

Applications of Dimensional Analysis

If the capacitance of a nanocapacitor is measured in terms of a unit 'u' made by combining the electronic charge 'e', Bohr radius 'a0 ', Planck's constant 'h' and speed of light 'c' then:

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Answer: (c)

Capacitance of a sphere $C = 4\pi\varepsilon_0 r$ , so $[C] = [\varepsilon_0 a_0]$ .
$\dfrac{e^2}{4\pi\varepsilon_0\hbar c}$ is dimensionless, so $[\varepsilon_0] = \left[\dfrac{e^2}{hc}\right]$ .
$[u] = \left[\dfrac{e^2a_0}{hc}\right]$ .
So the answer is (c).

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