Question 38: Units , Dimensions and Error Analysis

Question

$L, C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}$ corresponds to $\_\_\_\_$ .

Accepted Answer

Write the dimensions of each quantity first.
We know:
Resistance, $R$:
$ [R] = M L^2 T^{-3} A^{-2} $
Inductance, $L$:
$ [L] = M L^2 T^{-2} A^{-2} $
Capacitance, $C$:
$ [C] = M^{-1} L^{-2} T^4 A^2 $
We need:
$ M L^2 T^{-4} A^{-2} $
Now check each option.
Option A: $\dfrac{R}{\sqrt{LC}}$
First find $[LC]$:
$ [LC] = \left(M L^2 T^{-2} A^{-2}ight)\left(M^{-1} L^{-2} T^4 A^2ight)=T^2 $
So,
$ [\sqrt{LC}] = T $
Hence,
$ \left[\frac{R}{\sqrt{LC}}ight] = \frac{M L^2 T^{-3} A^{-2}}{T} = M L^2 T^{-4} A^{-2} $
This matches the required dimension.
Option B: $\dfrac{R}{LC}$
Since
$ [LC]=T^2 $
we get
$ \left[\frac{R}{LC}ight] = \frac{M L^2 T^{-3} A^{-2}}{T^2} = M L^2 T^{-5} A^{-2} $
Not matching.
Option C: $\dfrac{C}{\sqrt{LR}}$
First,
$ [LR] = \left(M L^2 T^{-2} A^{-2}ight)\left(M L^2 T^{-3} A^{-2}ight) = M^2 L^4 T^{-5} A^{-4} $
So,
$ [\sqrt{LR}] = M L^2 T^{-5/2} A^{-2} $
Then,
$ \left[\frac{C}{\sqrt{LR}}ight] = \frac{M^{-1} L^{-2} T^4 A^2}{M L^2 T^{-5/2} A^{-2}} = M^{-2} L^{-4} T^{13/2} A^4 $
Not matching.
Option D: $\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$
First,
$ \left[\frac{L}{C}ight] = \frac{M L^2 T^{-2} A^{-2}}{M^{-1} L^{-2} T^4 A^2} = M^2 L^4 T^{-6} A^{-4} $
So,
$ \left[\sqrt{\frac{L}{C}}ight] = M L^2 T^{-3} A^{-2} $
Also,
$ \left[\frac{1}{R}ight] = M^{-1} L^{-2} T^3 A^2 $
Therefore,
$ \left[\frac{1}{R}\sqrt{\frac{L}{C}}ight] = 1 $
This is dimensionless, so not matching.
Hence, the correct answer is
$ \boxed{	ext{Option A } \frac{R}{\sqrt{LC}}} $

Answer

$\frac{R}{\sqrt{L C}}$

Answer

$\frac{R}{L C}$

Answer

$\frac{C}{\sqrt{L R}}$

Answer

$\frac{1}{R} \sqrt{\frac{L}{C}}$

Units , Dimensions and Error Analysis

L, C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}$ corresponds to $\_\_\_\_$ .

Choose the correct answer: