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If the dimension of a physical quantity are given by MaLbTcM^a L^b T^c, then the physical quantity will be

A

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

B

velocity if a=1,b=0,c=−1a=1, b=0, c=-1

C

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

D

force if a=0,b=−1,c=−2a=0, b=-1, c=-2

Step-by-Step Solution

Let us evaluate the dimensional formula for each physical quantity given in the options and compare it with [MaLbTc][M^a L^b T^c]:

  1. Pressure: Pressure = ForceArea=[MLT−2][L2]=[M1L−1T−2]\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 perfectly.

  2. Velocity: Velocity = DisplacementTime=[LT−1]=[M0L1T−1]\frac{\text{Displacement}}{\text{Time}} = [L T^{-1}] = [M^0 L^1 T^{-1}]. Here a=0,b=1,c=−1a=0, b=1, c=-1 (Mismatch with option 2).

  3. Acceleration: Acceleration = VelocityTime=[LT−2]=[M0L1T−2]\frac{\text{Velocity}}{\text{Time}} = [L T^{-2}] = [M^0 L^1 T^{-2}]. Here a=0,b=1,c=−2a=0, b=1, c=-2 (Mismatch with option 3).

  4. Force: Force = Mass ×\times Acceleration =[M1L1T−2]= [M^1 L^1 T^{-2}]. Here a=1,b=1,c=−2a=1, b=1, c=-2 (Mismatch with option 4).

Therefore, the correct physical quantity for the given dimensions is pressure.

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