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NEET PHYSICSUNITS AND MEASUREMENTSEasy

Question

If L denotes the inductance of an inductor through which a current i is flowing, the dimensions of Li² are:

A

ML²T⁻²

B

Not expressible in MLT

C

MLT⁻²

D

LT

Step-by-Step Solution

To determine the dimensions of the quantity Li2Li^2, we can relate it to the energy stored in an inductor.

  1. Formula: The work done (WW) required to build up a current II in an inductor of inductance LL is stored as magnetic potential energy. The expression is given by W=12LI2W = \frac{1}{2}LI^2 .
  2. Dimensional Analysis: Since the factor 12\frac{1}{2} is a dimensionless constant, the dimensions of Li2Li^2 are identical to the dimensions of Work or Energy.
  3. Dimensions of Energy: The dimensional formula for Work/Energy is [ML2T−2][ML^2T^{-2}] .
  4. Verification: Inductance (LL): Dimensions are [ML2T−2A−2][ML^2T^{-2}A^{-2}] . Current (ii): Dimensions are [A][A].
  • Li2=[ML2T−2A−2]×[A]2=[ML2T−2]Li^2 = [ML^2T^{-2}A^{-2}] \times [A]^2 = [ML^2T^{-2}].

Therefore, the dimensions of Li2Li^2 are [ML2T−2][ML^2T^{-2}].

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from UNITS AND MEASUREMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSUNITS AND MEASUREMENTSdenotesinductanceinductorthroughcurrent

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