Question

A physical quantity of the dimensions of length that
can be formed out of c, G and $\sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$ is [c is velocity
of light, G is the universal constant of gravitation
and e is charge]


$\sf a) \: c^2\Bigg[G \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\Bigg]^{1/2}$


$\sf b) \: \dfrac{1}{c^2}\Bigg[ \dfrac{ {e}^{2} }{G4\pi\varepsilon_{0}}\Bigg]^{1/2}$
$\sf c) \: \dfrac{1}{c}G \dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}$
$\sf d) \: \dfrac{1}{c^2}\Bigg[ G\dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}\Bigg]^{1/2}$
​

Answer 1

Explanation:

I hope it will be help full for you

Answer 2

Question:

A physical quantity of the dimensions of length that can be formed out of c, G and $\sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$ is.

[c is velocity of light, G is the universal constant of gravitation and e is charge]

• $\sf a) \: c^2\Bigg[G \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\Bigg]^{1/2}$
• $\sf b) \: \dfrac{1}{c^2}\Bigg[ \dfrac{ {e}^{2} }{G4\pi\varepsilon_{0}}\Bigg]^{1/2}$
• $\sf c) \: \dfrac{1}{c}G \dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}$
• $\sf d) \: \dfrac{1}{c^2}\Bigg[ G\dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}\Bigg]^{1/2}$

Solution:

Let's suppose the physical quantity formed by c, G and $\sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$ is $[c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c$

According to the question, it is a physical quantity for length.

So,

$L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c$  

Use dimensional analysis,

Dimensions of the following are:

• L= [L]
• $c=[LT^-1]$
• $G=[M^{-1}L^3T{-2}]$
• $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]$

Why $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]?$

Use dimensional analysis for each quantity in $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$

Here we have

• $e^2=[q] ^2=[AT]^2$
• $\epsilon_0=[M^{-1}L^{-3}T^4A^2]$
• Rest are Constants
• Combining these two according to the formula we get,
• $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}=\frac{[AT]^2}{[M^{-1}L^{-3}T^4A^2]}=\frac{1}{[M^{-1}L^{-3}T^2]} = [ML^3T^{-2}]$

Now, equating the dimensional formulas for

$L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c \\\\ L=[LT^-1]^a[M^{-1}L^3T{-2}]^b[ML^3T^{-2}] ^c \\\\L=[L^aT^{-a}] [M^{-b} L^{3b}T^{-2b}][M^cL^{3c}T^{-2c}] \\\\ L=[L^{a+3b+3c}][M^{-b+c}][T^{-a-2b-2c}]$

Now,equating the exponent power we have,

$a+3b+3c=1 - - - - [i]\\\\ -b+c=0 - - - - [ii] \\\\ - a-2b-2c=0 - - - - [iii] \\\\ From [ii], we~have~b=c,~using~this~value~in~[iii] ~we, ~get \\\\ - a-2c-2c=0 \\\\ - a-4c=0 a=-4c \\\\ Using~a=-4c~and~b=c~in~[i]~we~have~\\\\ - 4c+3c+3c=1 \\\\ 2c=1 \\\\ c=\frac{1}{2} =b \\\\ We~ know~that~a=-4c ~so, ~a=-4 \times \frac{1}{2} ~, a=-2$  

The physical quantity so obtained is

$L= [c]^-2[G] ^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ => \frac{1}{c^2} G^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ =\frac{1}{c^2} \bigg[G \frac{e^{2} }{4\pi \varepsilon_{0}} \bigg]^{\frac{1}{2}}$