A physical quantity of the dimensions of length that can be formed out of c, G and e^2/4πε0
A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{e^2}{4\pi \epsilon_0}\) is [c is velocity of light, G is universal constant of gravitation and e is charge]:
(A) \(\frac{1}{c^2}[G\frac{e^2}{4\pi \epsilon_0}]\)
(B) \(c^2[G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)
(c) \(\frac{1}{c^2}[\frac{e^2}{G4\pi \epsilon_0}]^{1/2}\)
(D) \(\frac{1}{c}G\frac{e^2}{4\pi \epsilon_0}\)
1 Answers
(A) \(\frac{1}{c^2}[G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)
Let the physical quantity formed of the dimensions of length be given as.
[L] = [c]x [G]y \([\frac{e^2}{4\pi \epsilon_0}]^z\) …………….. (i)
Now,
Dimensions of velocity of light [c]x = [LT-1]x Dimensions of universal gravitational constant [G]y = [L3T2M-1]y
Dimensions of \([\frac{e^2}{4\pi \epsilon_0}]^z\) = [ML3T-2]z
Substitrning these in equation (i)
[L] [LT-1]x [M-1L3T-2]y [ML3T-2]z
= Lx+3y+3z M-y+z T-x-2y-2z
Solving for x, y, z
x + 3y + 3z = 1
-y + z = 0
x + 2y + 2z = O
Solving the above equation,
x = -2, y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
∴ L = \(\frac{1}{c^2}\)\([G\frac{e^2}{4\pi \epsilon_0}]^{1/2}\)