Q52 / 75 · JEE Main 2026 (05 April Shift 1) · Physics
$L$, $C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $ML^2 T^{-4} A^{-2}$ corresponds to __________.
- Correct answerA$\dfrac{R}{\sqrt{LC}}$
- B$\dfrac{R}{LC}$
- C$\dfrac{C}{\sqrt{LR}}$
- D$\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$
Solution
Dimensional formula of inductance $L$ is obtained from $E = \dfrac{1}{2}LI^2$:
$[L] = \dfrac{[E]}{[I]^2} = \dfrac{ML^2T^{-2}}{A^2} = ML^2T^{-2}A^{-2}$
Dimensional formula of capacitance $C$ is obtained from $E = \dfrac{1}{2}\dfrac{Q^2}{C}$:
$[C] = \dfrac{[Q]^2}{[E]} = \dfrac{A^2T^2}{ML^2T^{-2}} = M^{-1}L^{-2}T^4A^2$
Dimensional formula of resistance $R$ is obtained from $P = I^2R$:
$[R] = \dfrac{[P]}{[I]^2} = \dfrac{ML^2T^{-3}}{A^2} = ML^2T^{-3}A^{-2}$
The dimensional formula of $\sqrt{LC}$ is:
$[\sqrt{LC}] = \sqrt{(ML^2T^{-2}A^{-2})(M^{-1}L^{-2}T^4A^2)} = \sqrt{T^2} = T$
Evaluating the dimensional formula of $\dfrac{R}{\sqrt{LC}}$:
$\left[\dfrac{R}{\sqrt{LC}}\right] = \dfrac{ML^2T^{-3}A^{-2}}{T} = ML^2T^{-4}A^{-2}$
This matches the given dimensional formula.
Answer: $\dfrac{R}{\sqrt{LC}}$
$[L] = \dfrac{[E]}{[I]^2} = \dfrac{ML^2T^{-2}}{A^2} = ML^2T^{-2}A^{-2}$
Dimensional formula of capacitance $C$ is obtained from $E = \dfrac{1}{2}\dfrac{Q^2}{C}$:
$[C] = \dfrac{[Q]^2}{[E]} = \dfrac{A^2T^2}{ML^2T^{-2}} = M^{-1}L^{-2}T^4A^2$
Dimensional formula of resistance $R$ is obtained from $P = I^2R$:
$[R] = \dfrac{[P]}{[I]^2} = \dfrac{ML^2T^{-3}}{A^2} = ML^2T^{-3}A^{-2}$
The dimensional formula of $\sqrt{LC}$ is:
$[\sqrt{LC}] = \sqrt{(ML^2T^{-2}A^{-2})(M^{-1}L^{-2}T^4A^2)} = \sqrt{T^2} = T$
Evaluating the dimensional formula of $\dfrac{R}{\sqrt{LC}}$:
$\left[\dfrac{R}{\sqrt{LC}}\right] = \dfrac{ML^2T^{-3}A^{-2}}{T} = ML^2T^{-4}A^{-2}$
This matches the given dimensional formula.
Answer: $\dfrac{R}{\sqrt{LC}}$
Hence, the correct answer is option: 'A'.