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NEET Physics QUESTION #7234
Question 1

A physical quantity having the dimension of length can be formed from the speed of light $c$, the gravitational constant $G$, and the quantity $\dfrac{e^2}{4\pi\varepsilon_0}$ (where $e$ is electric charge). Which combination gives the dimension of length?

  • $c^2\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$
  • $\dfrac{1}{c^2}\left[\dfrac{e^2}{G4\pi\varepsilon_0}\right]^{1/2}$
  • $\dfrac{1}{c}G\dfrac{e^2}{4\pi\varepsilon_0}$
  • $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$✔️
Correct Answer Explanation

Let $k = \dfrac{e^2}{4\pi\varepsilon_0}$. Find $[k]$: from Coulomb's law $F = k\dfrac{q^2}{r^2}$, so $[k] = \dfrac{[F][r^2]}{[q^2]} = \dfrac{MLT^{-2}\cdot L^2}{A^2T^2} = ML^3T^{-4}A^{-2}$

$[G] = M^{-1}L^3T^{-2}$, $[c] = LT^{-1}$

$\left[Gk\right] = M^{-1}L^3T^{-2} \cdot ML^3T^{-4}A^{-2} = L^6T^{-6}A^{-2}$

$\left[\sqrt{Gk}\right] = L^3T^{-3}A^{-1}$

$\left[\dfrac{1}{c^2}\sqrt{Gk}\right] = \dfrac{L^3T^{-3}A^{-1}}{L^2T^{-2}} = LT^{-1}A^{-1}$...

The correct combination (known as the classical electron radius analogue) is $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$, which gives dimension $[L]$.