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NEET
Exam
NEET
Subject
Physics
Chapter
Units and Dimensions

NEET · Physics · Units and Dimensions

Which of the following combinations of physical quantities has the...

Which of the following combinations of physical quantities has the dimensional formula [M−1L3T−2][\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}]?

Options

  1. AEnergy×DistanceMass2\dfrac{\text{Energy} \times \text{Distance}}{\text{Mass}^2}
  2. BForce×VolumeMass\dfrac{\text{Force} \times \text{Volume}}{\text{Mass}}
  3. CPressure×VolumeDensity\dfrac{\text{Pressure} \times \text{Volume}}{\text{Density}}
  4. DPower×AreaMass2\dfrac{\text{Power} \times \text{Area}}{\text{Mass}^2}

Correct answer

A. Energy×DistanceMass2\dfrac{\text{Energy} \times \text{Distance}}{\text{Mass}^2}

Step-by-step solution

Let us evaluate the dimensional formula for each given combination of physical quantities.

Option 1:
Energy×DistanceMass2=[ML2T−2]×[L][M2]=[ML3T−2][M2]=[M−1L3T−2]\dfrac{\text{Energy} \times \text{Distance}}{\text{Mass}^2} = \dfrac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}] \times [\mathrm{L}]}{[\mathrm{M}^2]} = \dfrac{[\mathrm{M} \mathrm{L}^3 \mathrm{T}^{-2}]}{[\mathrm{M}^2]} = [\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}]
This matches the given dimensional formula.

Option 2:
Force×VolumeMass=[MLT−2]×[L3][M]=[L4T−2]\dfrac{\text{Force} \times \text{Volume}}{\text{Mass}} = \dfrac{[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}] \times [\mathrm{L}^3]}{[\mathrm{M}]} = [\mathrm{L}^4 \mathrm{T}^{-2}]

Option 3:
Pressure×VolumeDensity=[ML−1T−2]×[L3][ML−3]=[ML2T−2][ML−3]=[L5T−2]\dfrac{\text{Pressure} \times \text{Volume}}{\text{Density}} = \dfrac{[\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-2}] \times [\mathrm{L}^3]}{[\mathrm{M} \mathrm{L}^{-3}]} = \dfrac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-2}]}{[\mathrm{M} \mathrm{L}^{-3}]} = [\mathrm{L}^5 \mathrm{T}^{-2}]

Option 4:
Power×AreaMass2=[ML2T−3]×[L2][M2]=[M−1L4T−3]\dfrac{\text{Power} \times \text{Area}}{\text{Mass}^2} = \dfrac{[\mathrm{M} \mathrm{L}^2 \mathrm{T}^{-3}] \times [\mathrm{L}^2]}{[\mathrm{M}^2]} = [\mathrm{M}^{-1} \mathrm{L}^4 \mathrm{T}^{-3}]

Answer: Energy×DistanceMass2\dfrac{\text{Energy} \times \text{Distance}}{\text{Mass}^2}
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