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If $y=\dfrac{t}{{{\varepsilon }_{o}}LV}$ where t is time, L is length, V is potential and ${{\varepsilon }_{o}}$ ​ is permittivity, then dimension of y is same as that of:

Answer Verified Verified
Hint: We have to find the dimensional formula of the given value. For that we first find the dimensional formulas for each value in the equation separately. Then we combine those dimensional formulas to get the dimension of the given equation.

Formula used:
Potential difference,
$V=\dfrac{W}{Q}$
Work done,
$W=F\times Displacement$
Force,
$F=ma$

Complete step by step answer:
We are given the question $y=\dfrac{t}{{{\varepsilon }_{o}}LV}$ .
It is given that, t is time.
We know that dimension of time; t is given as [T].
And ‘L’ is said to be length.
Dimension of length; L is given as [L]
It is given, ‘V’ is potential.
We know that potential difference is the ratio of work done to the charge moved.
$V=\dfrac{W}{Q}$ , Where ‘W’ is the work done and ‘Q’ is the charge.
Work done is the product of force and displacement.
$W=F\times Displacement$
From Newton’s second law, we know that,
$F=m\times a$
Dimensional formula of,
$\begin{align}
  & \text{Mass=}\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right] \\
 & \text{Displacement=}\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right] \\
 & \text{Acceleration=}\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right] \\
\end{align}$
From this information, we can calculate the dimensional formula of work done.
Work done$=\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]$
Work done $=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]$
And, since charge is the product of current and time its dimensional formula can be written as
$\text{Charge=}\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{I}^{1}} \right]$
Therefore, the dimensional formula of potential difference is,
$\begin{align}
  & V=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-2}} \right]\times {{\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{I}^{1}} \right]}^{-1}} \\
 & V=\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right] \\
\end{align}$
Given,${{\varepsilon }_{o}}$ is the permittivity of free space or vacuum.
Permittivity (${{\varepsilon }_{0}}$) $=\left[ Ch\arg {{e}^{2}} \right]\times {{\left[ Force \right]}^{-1}}\times \left[ Dis\tan c{{e}^{2}} \right]$
Dimensional formula of charge$=\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{I}^{1}} \right]$
Dimensional formula of distance=$\left[ L \right]$
Dimensional formula of force$=\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]$
Therefore, dimensional formula of permittivity of free space,
$\begin{align}
  & {{\varepsilon }_{o}}={{\left[ {{M}^{0}}{{L}^{0}}{{T}^{1}}{{I}^{1}} \right]}^{2}}\times {{\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]}^{-1}}\times {{\left[ {{M}^{0}}{{L}^{1}}{{T}^{0}} \right]}^{-2}} \\
 & {{\varepsilon }_{o}}=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}} \right] \\
\end{align}$
Now let us substitute all these dimensional formulas in y.
We get,
$y=\dfrac{\left[ T \right]}{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{I}^{2}} \right]\left[ L \right]\left[ {{M}^{1}}{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right]}$
$y=\dfrac{\left[ T \right]}{\left[ T \right]\left[ A \right]}$
$y=\dfrac{1}{\left[ A \right]}$
$y={{\left[ A \right]}^{-1}}$
Therefore, the dimensional formula of y is = ${{\left[ A \right]}^{-1}}$

Note:
The power to which the fundamental quantities are raised to express the quantity is called the dimension of that quantity. The expression in which the dimensions of the quantity are represented in terms of the fundamental quantity is the dimensional formula of that quantity. There are a total of seven fundamental dimensions.