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# If C be the capacitance and V be the electric potential, then the dimensional formula of $C{V^2}$ is:A) $\left[ {M{L^{ - 3}}TA} \right]$B) $\left[ {{M^0 }L{T^{ - 2}}{A^0 }} \right]$C) $\left[ {M{L^1}{A^{ - 1}}} \right]$D) $\left[ {M{L^2}{T^{ - 2}}{A^0 }} \right]$

Last updated date: 12th Aug 2024
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Hint: The dimensional formula of any physical quantity is defined as an expression which tells us about the fundamental quantities which are involved in the physical quantity. The dimensional formula of any physical quantity can be useful in checking the correctness of the formula of the physical quantity.

Formula used:
The formula of the capacitance is given by,
$\Rightarrow C = \dfrac{Q}{V}$
Where capacitance is C, the charge is Q and the potential difference is V.
The formula of the potential difference is given by,
$\Rightarrow V = \dfrac{W}{Q}$
Where potential difference is V, the work done is W and the charge is Q.

Complete step by step solution:
It is given in the problem that if C be the capacitance and V be the electric potential, then we need to find the dimensional formula of $C{V^2}$.
The formula of the capacitance is given by,
$\Rightarrow C = \dfrac{Q}{V}$
Where capacitance is C, the charge is Q and the potential difference is V.
Here the capacitance is equal to$\Rightarrow V = \dfrac{W}{Q}$.
The formula of the potential difference is given by,
$\Rightarrow V = \dfrac{W}{Q}$
Where potential difference is V, the work done is W and the charge is Q.
Replacing the potential difference is above relation.
$\Rightarrow C = \dfrac{Q}{V}$
$\Rightarrow C = \dfrac{Q}{{\dfrac{W}{Q}}}$
$\Rightarrow C = \dfrac{{{Q^2}}}{W}$
Also charge is equal to,
$\Rightarrow Q = I \times T$
So we get,
$\Rightarrow C = \dfrac{{{{\left( {I \times T} \right)}^2}}}{W}$
$\Rightarrow C = \dfrac{{{I^2} \times {T^2}}}{W}$
The dimension of I is A, and the dimension of work is $\left[ {M{L^2}{T^{ - 2}}} \right]$.
$\Rightarrow C = \dfrac{{\left[ {{A^2} \times {T^2}} \right]}}{{\left[ {M{L^2}{T^{ - 2}}} \right]}}$
$\Rightarrow C = \left[ {{M^{ - 1}}{A^2}{L^{ - 2}}{T^4}} \right]$………eq (1)
The formula of the potential difference is given by,
$\Rightarrow V = \dfrac{W}{Q}$
Where potential difference is V, the work done is W and the charge is Q.
The charge is equal to,
$Q = I \times T$
Also the dimension of work is$\left[ {M{L^2}{T^{ - 2}}} \right]$.
So we get,
$\Rightarrow V = \dfrac{W}{Q}$
$\Rightarrow V = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {A \times T} \right]}}$
Since the unit of current is A.
$\Rightarrow V = \left[ {M{A^{ - 1}}{L^2}{T^{ - 3}}} \right]$………eq. (2)
Applying,
$\Rightarrow C{V^2}$
Replacing the values from the equation (1) and equation (2) we get,
$\Rightarrow C{V^2}$
$\Rightarrow C{V^2} = \left[ {{M^{ - 1}}{A^2}{L^{ - 2}}{T^4}} \right] \times {\left[ {M{A^{ - 1}}{L^2}{T^{ - 3}}} \right]^2}$
$\Rightarrow C{V^2} = \left[ {{M^{ - 1}}{A^2}{L^{ - 2}}{T^4}} \right] \times \left[ {{M^2}{A^{ - 2}}{L^4}{T^{ - 6}}} \right]$
$\Rightarrow C{V^2} = \left[ {M{A^0 }{L^{ - 2}}{T^{ - 2}}} \right]$.
The dimensional formula of $C{V^2}$ is equal to$\left[ {M{A^0 }{L^{ - 2}}{T^{ - 2}}} \right]$.

The correct answer for this problem is option D.

Note: It is advisable for students to understand and remember the concept of dimensional formula as it is very useful in solving the problems like these. Also it is advised to students to remember the dimensional formula of the standard physical quantities as sometimes we have to use them for calculating the dimensional formula.