Back to Directory
NEET PHYSICSEasy

If the dimensions of a physical quantity are given by [MaLbTc][M^a L^b T^c], then the physical quantity will be:

A

pressure if a=1a=1, b=−1b=-1, c=−2c=-2

B

velocity if a=1a=1, b=0b=0, c=−1c=-1

C

acceleration if a=1a=1, b=1b=1, c=−2c=-2

D

force if a=0a=0, b=−1b=-1, c=−2c=-2

Step-by-Step Solution

Let us check the dimensions of each given physical quantity:

  1. Pressure: Pressure=ForceArea=[MLT−2][L2]=[M1L−1T−2]\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{[M L T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]. Comparing with [MaLbTc][M^a L^b T^c], we get a=1,b=−1,c=−2a=1, b=-1, c=-2. This matches the first option.
  2. Velocity: Velocity=[M0L1T−1]\text{Velocity} = [M^0 L^1 T^{-1}]. Here a=0,b=1,c=−1a=0, b=1, c=-1.
  3. Acceleration: Acceleration=[M0L1T−2]\text{Acceleration} = [M^0 L^1 T^{-2}]. Here a=0,b=1,c=−2a=0, b=1, c=-2.
  4. Force: Force=[M1L1T−2]\text{Force} = [M^1 L^1 T^{-2}]. Here a=1,b=1,c=−2a=1, b=1, c=-2.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started