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If the dimensions of a physical quantity are given by $[M^aL^bT^c]$, then the physical quantity will be
A
Force if a = 0, b = –1, c = –2
B
Pressure if a = 1, b = –1, c = –2
C
Velocity if a = 1, b = 0, c = –1
D
Acceleration if a = 1, b = 1, c = –2
Detailed Solution
Force = mass × acceleration: $[F] = [MLT^{-2}]$, i.e., a = 1, b = 1, c = –2. So the first statement is wrong.
Pressure = force/area: $[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$, i.e., a = 1, b = –1, c = –2. This matches.
Velocity: $[v] = [LT^{-1}]$, i.e., a = 0, b = 1, c = –1. So the third statement is wrong.
Acceleration: $[LT^{-2}]$, i.e., a = 0, b = 1, c = –2. So the fourth statement is wrong.
Hence the quantity is pressure if a = 1, b = –1, c = –2.
Pressure = force/area: $[P] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$, i.e., a = 1, b = –1, c = –2. This matches.
Velocity: $[v] = [LT^{-1}]$, i.e., a = 0, b = 1, c = –1. So the third statement is wrong.
Acceleration: $[LT^{-2}]$, i.e., a = 0, b = 1, c = –2. So the fourth statement is wrong.
Hence the quantity is pressure if a = 1, b = –1, c = –2.