Physics Q27 – JEE-MAIN 2026

Home Physics Units & MeasurementsQuestion #11

L, C and R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula

M L^{2} T^{- 4} A^{- 2}

corresponds to _______.

Clue 1 of 3: Fundamental Dimensions

Recall the dimensional formulas for basic physical quantities like force, work, charge, and current.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine Dimensional Formulas of L, C, R✦ Active

First, we find the dimensional formulas for resistance ( $R$ ), inductance ( $L$ ), and capacitance ( $C$ ).

R = \frac{V}{I} = \frac{Work/Charge}{Current} = \frac{M L^{2} T^{- 2} / (A T)}{A} = [M L^{2} T^{- 3} A^{- 2}]

L = \frac{V}{d I / d t} = \frac{M L^{2} T^{- 3} A^{- 1}}{A / T} = [M L^{2} T^{- 2} A^{- 2}]

C = \frac{Q}{V} = \frac{A T}{M L^{2} T^{- 3} A^{- 1}} = [M^{- 1} L^{- 2} T^{4} A^{2}]

Step 2: Calculate Dimensions of LC○ Expand

Next, we calculate the dimensions of the product $L C$ and its square root, as these appear in the options.

[L C] = [M L^{2} T^{- 2} A^{- 2}] \times [M^{- 1} L^{- 2} T^{4} A^{2}] = [M^{1 - 1} L^{2 - 2} T^{- 2 + 4} A^{- 2 + 2}] = [T^{2}]

∴ [\sqrt{L C}] = [T]

Step 3: Evaluate Option 1○ Expand

Now, we evaluate the dimensional formula for Option 1: $\frac{R}{\sqrt{L C}}$ .

[\frac{R}{\sqrt{L C}}] = \frac{[M L^{2} T^{- 3} A^{- 2}]}{[T]} = [M L^{2} T^{- 4} A^{- 2}]

This matches the given dimensional formula $M L^{2} T^{- 4} A^{- 2}$ . Therefore, Option 1 is the correct answer. (Other options can be similarly checked and will not match).

💡 Teacher's Secret Hint

Always double-check your fundamental dimensional formulas for L, C, and R to avoid errors in subsequent calculations.

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Physics Question 27 – JEE-MAIN 2026

Clue 1 of 3: Fundamental Dimensions

🧠 Full Solution Path