Question:medium

The unit of \( \sqrt{LC} \) (where \( L \) is inductance and \( C \) is capacitance) is:

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Remember the LC-oscillation relation \( \omega = 1/\sqrt{LC} \) (or \( T = 2\pi\sqrt{LC} \)); \( \sqrt{LC} \) has the dimension of time.
Updated On: Jul 10, 2026
  • second
  • henry
  • farad
  • ampere
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The Correct Option is A

Solution and Explanation

Step 1: Use the time-period formula of an LC circuit.
The natural time period of oscillation of an inductor-capacitor circuit is \(T = 2\pi\sqrt{LC}\). Since \(T\) is a time, it is measured in seconds.

Step 2: Isolate \(\sqrt{LC}\).
Dividing by the pure number \(2\pi\) does not change the physical dimension, so \(\sqrt{LC} = \dfrac{T}{2\pi}\) still carries the unit of time.

Step 3: Confirm dimensionally.
The dimension of \(L\) is \([ML^2T^{-2}A^{-2}]\) and of \(C\) is \([M^{-1}L^{-2}T^{4}A^{2}]\). Multiplying, the mass, length and current dimensions cancel, leaving \([T^{4}\cdot T^{-2}] = [T^{2}]\). Taking the square root gives \([T]\).

Step 4: Conclude.
A quantity with dimension of time is measured in seconds, so the unit of \(\sqrt{LC}\) is the second.

\[\boxed{\text{second}}\]
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