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Question

Dimensions of resistance in an electrical circuit, in terms of dimension of mass MM, length LL, time TT, and current II, would be:

A

[ML2T−3I−1][M L^2 T^{-3} I^{-1}]

B

[ML2T−2][M L^2 T^{-2}]

C

[ML2T−1I−1][M L^2 T^{-1} I^{-1}]

D

[ML2T−3I−2][M L^2 T^{-3} I^{-2}]

Step-by-Step Solution

  1. Identify the formula for Resistance: According to Ohm's law, Resistance (RR) is given by the ratio of potential difference (VV) to current (II): R=VIR = \frac{V}{I}.
  2. Find the dimensions of Potential Difference (VV): Potential difference is defined as work done per unit charge, V=WQV = \frac{W}{Q}.
  • The dimensional formula for Work (WW) is [ML2T−2][M L^2 T^{-2}].
  • The dimensional formula for Charge (Q=I×tQ = I \times t) is [IT][I T].
  • Therefore, the dimensions of V=[ML2T−2][IT]=[ML2T−3I−1]V = \frac{[M L^2 T^{-2}]}{[I T]} = [M L^2 T^{-3} I^{-1}].
  1. Calculate the dimensions of Resistance (RR): Substitute the dimensional formula of VV and II back into the resistance equation: R=[ML2T−3I−1][I]=[ML2T−3I−2]R = \frac{[M L^2 T^{-3} I^{-1}]}{[I]} = [M L^2 T^{-3} I^{-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 MEASUREMENTSdimensionsresistanceelectricalcircuitdimension

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