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JEE MAIN_2026
05-04-26_S1
Question
$\mathrm{L}, \mathrm{C}$ and R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}$ corresponds to $\_\_\_\_$ .
Select the correct option:
A
$\frac{R}{\sqrt{L C}}$
B
$\frac{R}{L C}$
C
$\frac{C}{\sqrt{L R}}$
D
$\frac{1}{R} \sqrt{\frac{L}{C}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{\mathrm{w}}{\mathrm{qI}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{I}^2 \mathrm{~T}}=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-2} \\ & \mathrm{U}=\frac{1}{2} \mathrm{LI}^2 \rightarrow \mathrm{ML}^2 \mathrm{~T}^{-2}=\mathrm{LA}^2 \\ & \mathrm{~L}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2} \\ & \mathrm{Q}=\mathrm{CV} \\ & \mathrm{C}=\frac{\mathrm{IT} \times \mathrm{q}}{\mathrm{ML}^2 \mathrm{~T}^{-2}}=\frac{\mathrm{A}^2 \mathrm{~T}^2}{\mathrm{ML}^2 \mathrm{~T}^{-2}} \\ & \mathrm{C}=\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~A}^{-2} \mathrm{~T}^4 \\ & \mathrm{LC}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2} \times \mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~A}^2 \mathrm{~T}^4 \\ & \mathrm{LC}=\mathrm{T}^2 \\ & \frac{\mathrm{R}}{\sqrt{\mathrm{LC}}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}}{\mathrm{~T}}=\left(\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}\right)\end{aligned}$
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