Question

$L, C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $M L^2 T^{-4} A^{-2}$ corresponds to _________.

• A. $\frac{R}{\sqrt{LC}}$
• B. $\frac{R}{LC}$
• C. $\frac{C}{\sqrt{LR}}$
• D. $\frac{1}{R} \sqrt{\frac{L}{C}}$

Hint:

Instead of recalling a resonance formula, try writing R, L and C from their own defining equations (V=IR, EMF=L dI/dt, Q=CV) and combine.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find which combination of circuit components matches the given dimensional formula.
Step 2: Key Formula or Approach:
Know the base dimensions:
$R = [M L^2 T^{-3} A^{-2}]$
$\sqrt{LC} = [T]$ (Time constant or period property)
$LC = [T^2]$
Step 3: Detailed Explanation:
Let's analyze option (A): $\frac{R}{\sqrt{LC}}$.
Dimensions of Resistance $R$:
From $P = I^2 R \implies R = \frac{Power}{I^2} = \frac{[M L^2 T^{-3}]}{[A^2]} = [M L^2 T^{-3} A^{-2}]$.
Dimensions of $\sqrt{LC}$:
We know resonance frequency $\omega = \frac{1}{\sqrt{LC}}$.
So, $[\frac{1}{\sqrt{LC}}] = [\omega] = [T^{-1}] \implies [\sqrt{LC}] = [T]$.
Now, dimensions of $\frac{R}{\sqrt{LC}}$:
$[ \frac{M L^2 T^{-3} A^{-2}}{T} ] = [M L^2 T^{-4} A^{-2}]$.
This matches the target formula exactly.
Step 4: Final Answer:
The formula corresponds to $\frac{R}{\sqrt{LC}}$.

Answer 2

Concept:

• Every electrical quantity has a defining equation. Instead of using the resonance-frequency relation, we can get the dimensions of $L$, $C$ and $R$ straight from the equations that define them.
• Once $[L]$, $[C]$ and $[R]$ are known individually, we simply substitute them into each option and check which one gives $[M L^2 T^{-4} A^{-2}]$.

Step 1: Find the dimension of resistance $R$.
From $V = IR$, $R = V/I$. Potential difference $V = \frac{\text{Work}}{\text{Charge}} = \frac{[M L^2 T^{-2}]}{[AT]} = [M L^2 T^{-3} A^{-1}]$.
So $[R] = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}]$.

Step 2: Find the dimension of inductance $L$.
From the EMF equation $\varepsilon = L\frac{dI}{dt}$, $[L] = \frac{[V][T]}{[A]} = \frac{[M L^2 T^{-3} A^{-1}][T]}{[A]} = [M L^2 T^{-2} A^{-2}]$.

Step 3: Find the dimension of capacitance $C$.
From $Q = CV$, $[C] = \frac{[Q]}{[V]} = \frac{[AT]}{[M L^2 T^{-3} A^{-1}]} = [M^{-1} L^{-2} T^4 A^2]$.

Step 4: Substitute into the given expression.
$[LC] = [M L^2 T^{-2} A^{-2}] \times [M^{-1} L^{-2} T^4 A^2] = [T^2]$, so $[\sqrt{LC}] = [T]$.
$\left[\dfrac{R}{\sqrt{LC}}\right] = \dfrac{[M L^2 T^{-3} A^{-2}]}{[T]} = [M L^2 T^{-4} A^{-2}]$.
This matches the given formula exactly.

Final Answer: $\dfrac{R}{\sqrt{LC}}$