Question

A physical quantity of the dimensions of length that can be formed out of c, G and $\frac{e^2}{4 \pi \varepsilon_0} $ is [c is velocity of light, G is universal constant of gravitation and e is charge] :-

• A. $c^{2}\left[G \frac{e^{2}}{4\pi \varepsilon_{0}}\right]^{1/ 2 } $
• B. $\frac{1}{c^{2}}\left[\frac{e^{2}}{G 4\pi \varepsilon_{0}}\right]^{1/ 2 } $
• C. $\frac{1}{c^{2}} G \frac{e^{2}}{4\pi \varepsilon_{0}} $
• D. $\frac{1}{c^{2}}\left[G\frac{e^{2}}{ 4\pi \varepsilon_{0}}\right]^{1/ 2 } $

Answer 1

The Correct Option is D

To solve this problem, we need to derive a physical quantity with the dimensions of length using the given constants: the speed of light \(c\), the gravitational constant \(G\), and the electrostatic constant \(\frac{e^2}{4 \pi \varepsilon_0}\).

First, let us consider the dimensional formulae of the given quantities:

• The speed of light, \(c\), has dimensions \([c] = [LT^{-1}]\).
• The gravitational constant, \(G\), has dimensions \([G] = [M^{-1}L^3T^{-2}]\).
• The electrostatic constant, \(\frac{e^2}{4 \pi \varepsilon_0}\), has dimensions \([\frac{e^2}{4 \pi \varepsilon_0}] = [ML^3T^{-2}]\).

Now, let's assume a quantity \(\text{Length} = c^a G^b \left(\frac{e^2}{4 \pi \varepsilon_0}\right)^c\) and equate its dimensions to length \([L]\):

Based on the dimensions:

\([L] = [L^aT^{-a}] \cdot [M^{-b}L^{3b}T^{-2b}] \cdot [ML^{3c}T^{-2c}]\)

Combining the terms, we have:

• Mass: \(-b + c = 0\)
• Length: \(a + 3b + 3c = 1\)
• Time: \(-a - 2b - 2c = 0\)

Solving these equations:

1. From equation \(-b + c = 0\), we get \(b = c\).
2. Substitute \(b = c\) into \(a + 3b + 3c = 1\), we get \(a + 6b = 1\).
3. From \(-a - 2b - 2c = 0\), substitute \(b = c\) to obtain \(-a - 4b = 0\), which gives \(a = 4b\).

Substituting \(a = 4b\) into \(a + 6b = 1\):

\(4b + 6b = 1 \Rightarrow 10b = 1 \Rightarrow b = \frac{1}{10}\)

Thus, \(c = b = \frac{1}{10}\) and \(a = 4b = \frac{4}{10} = \frac{2}{5}\).

Hence the dimensional quantity is:

\(c^{\frac{2}{5}} G^{\frac{1}{10}} \left(\frac{e^2}{4 \pi \varepsilon_0}\right)^{\frac{1}{10}}\)

Combining the terms gives us the correct answer:

The length is given by the option \( \frac{1}{c^2} \left[G \frac{e^2}{4 \pi \varepsilon_0}\right]^{1/2} \), because it balances the exponents correctly to yield a dimensionally consistent length.

Therefore, the correct answer is:

\( \frac{1}{c^2}\left[G\frac{e^2}{4\pi \varepsilon_0}\right]^{1/2} \)