Question

A physical quantity of the dimensions of length that can be formed out of$c$, $G$ and $\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}$ is ($c$is the velocity of light, $G$is the universal constant of gravitation and $e$ is a charge, ${\varepsilon _0}$ is electrical permittivity):
$\left( a \right)$ ${c^2}{\left[ {G\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\dfrac{1}{2}}}$
$\left( b \right)$ $\dfrac{1}{{{c^2}}}{\left[ {\dfrac{{{e^2}}}{{G4\pi {\varepsilon _0}}}} \right]^{\dfrac{1}{2}}}$
$\left( c \right)$ $\dfrac{1}{c}G\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}$
$\left( d \right)$ $\dfrac{1}{{{c^2}}}{\left[ {G\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\dfrac{1}{2}}}$

Answer 1

Hint To tackle these inquiries initially compose dimensions of all the given amounts in terms of a basic amount and afterward discover the connection between them. And then by using the dimensions we will make the relationship and check the result.
Formula used:
Time period,
$T = 2\pi \sqrt {l/g} $
Here,
$T$ , will be the time period
$l$ , will be the length and
$g$ , will be the acceleration due to gravity.

Complete Step by Step Solution As we know, the time period
$T = 2\pi \sqrt {l/g} $
And,
$\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}} = \left[ {F \times {d^2}} \right]$
And the dimension of the above quantity will be written as
$ \Rightarrow M{c^3}{T^{ - 2}}$
And also similarly the dimension of rest given quantity will be
$G = M{L^3}{T^{ - 2}}$
And that $c$will be
$c = {L^{ - 1}}$
As the length is directly proportional to given three physical quantities.
So mathematically it can be written as
$ \Rightarrow L = {\left[ c \right]^x}{\left[ G \right]^y}{\left[ {\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^z}$
So substituting the dimensions for all the quantity, we get
\[ \Rightarrow L = {\left[ {L{T^{ - 1}}} \right]^x}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^y}{\left[ {M{L^3}{T^{ - 2}}} \right]^z}\]
Now we will make similar variables at one place, so by solving we will get
$ \Rightarrow \left[ L \right] = \left[ {{L^{z + 3y + 3z}}{M^{ - y + z}}{T^{ - x - 2y - 2z}}} \right]$
Now on doing the comparison for both the sides of the above equation, we get
$ \Rightarrow - y + z = 0$
And it can be written as
$ \Rightarrow y = z$
Another equation will be,
$ \Rightarrow x + 3y + 3z = 1$
And there will be one more equation, so it will be
$ \Rightarrow - x - 4z = 0$
Now from all these above three equations, on solving for the value of $x,y,z$
We get,
$ \Rightarrow z = y = \dfrac{1}{2},x = - 2$
On putting the value $x,y,z$in the first equation
Therefore the $L$will be
$L = \dfrac{1}{{{c^2}}}{\left[ {G\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{\dfrac{1}{2}}}$

Hence, the option $D$ will be the right option.

Note Dimensions have a lot of uses which makes it very significant. When we know about the dimensions of a quantity then we can form formulas related to it and also check a particularly given formula. If you want to know how?? Then you can question me again.
Dimensions are also used for converting a quantity into its another unit (either CGS, SI, or even a given unit in the question.)
Dimensions provide us with detailed information about derived quantities in a precise manner.