Question

If L,C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for C$^{2}$ LR is :

(A). $[{{M}^{{}}}{{L}^{-2}}{{T}^{-1}}{{I}^{0}}]$
(B). $[{{M}^{0}}{{L}^{0}}{{T}^{3}}{{I}^{0}}]$
(C). $[{{M}^{-1}}{{L}^{-2}}{{T}^{6}}{{I}^{2}}]$
(D). $[{{M}^{-1}}{{L}^{-2}}{{T}^{6}}{{I}^{2}}]$

Answer 1

Hint: At first break all the values in the LCR circuit into its most simplest form, then put it into dimensional analysis that will bring the terms in the form of MLT, then take the formula for which dimensional analysis is required, then place the derived dimensional values of LCR, and equate to find the answer.

Complete step-by-step answer:
For capacitance,
We know that Potential energy of a capacitor is = $\dfrac{{{Q}^{2}}}{2C}$
Now,
C=$\dfrac{{{Q}^{2}}}{2\times P.E}$
In dimensional analysis,
C= $\dfrac{{{(I.T)}^{2}}}{M{{L}^{2}}{{T}^{-2}}}$
On simplifying,
C=${{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{I}^{2}}$

For Inductance,
We know that Inductance L=(EMF$\times $ t)/I
Now,
L=((W/Q)$\times $ t))/I
L=(W$\times $ t)/(I$\times $Q)
L= (W$\times $ t)/((I$\times $t) $\times $I)
L=$\dfrac{W}{{{I}^{2}}}$
Therefore in dimensional form,
=$\dfrac{M{{L}^{2}}{{T}^{-2}}}{{{I}^{2}}}$
L=$M{{L}^{2}}{{T}^{-2}}{{I}^{-2}}$
For resistance,
R=V/I,
R=(W/Q)/I,
R=W/Q$\times $I,
R=W/(I$\times $t) $\times $I,
R=W/(I$^{2}$ $\times $t),
In dimensional form,
$\dfrac{M{{L}^{2}}{{T}^{-2}}}{{{I}^{2}}.T}$ ,
R=$M{{L}^{2}}{{T}^{-3}}{{I}^{-2}}$ .
Now according to question,
C$^{2}$ LR=$({{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{I}^{2}})\times M{{L}^{2}}{{T}^{-2}}{{I}^{-2}}\times M{{L}^{2}}{{T}^{-3}}{{I}^{-2}}$
Therefore final result is,
C$^{2}$ LR=${{M}^{0}}{{L}^{0}}{{T}^{3}}{{I}^{0}}$
Therefore option B is the correct answer.

Additional Information:
The equation that we get, when we equate a physical quantity with its dimensional formula is known as Dimensional analysis.
While doing dimensional analysis one of the most important things to remember is that the unit you wish to cancel out must be on the opposite side of the fraction.

Note:Charge Q can be broken into current and time, and EMF is work done by charge, Voltage is also Work done by charge, hence we can see that ultimately most of the dimensional values come at the same place. Try to remember all the formulas , and their simplest outcome , because it will be required at most of the cases while doing Dimensional analysis.