Question 10: 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: