If the dimensions of a physical quantity are given by MᵃLᵇTᶜ, then the physical quantity will be

The correct answer is Pressure if a = 1, b = -1, c = -2.

To determine which physical quantity corresponds to the given dimensions, let us analyze the dimensional formulas of the options provided. The dimensions of a physical quantity are represented in the form $M^a{L}^b{T}^c$, where $M$ represents Mass, $L$ represents Length, and $T$ represents Time.

1. Velocity:
Velocity is defined as displacement per unit time.
Dimension of Velocity = $\frac{\left[L\right]}{\left[T\right]}=\left[M^0{L}^1{T}^{-1}\right]$
Comparing with $M^a{L}^b{T}^c$, we get $a=0,b=1,c=-1$. This does not match the first option.

2. Acceleration:
Acceleration is defined as the rate of change of velocity.
Dimension of Acceleration = $\frac{\left[L{T}^{-1}\right]}{\left[T\right]}=\left[M^0{L}^1{T}^{-2}\right]$
Comparing with $M^a{L}^b{T}^c$, we get $a=0,b=1,c=-2$. This does not match the second option.

3. Force:
Force is defined as the product of mass and acceleration.
Dimension of Force = $\left[M\right]\times \left[L{T}^{-2}\right]=\left[M^1{L}^1{T}^{-2}\right]$
Comparing with $M^a{L}^b{T}^c$, we get $a=1,b=1,c=-2$. This does not match the third option.

4. Pressure:
Pressure is defined as force applied per unit area.
Dimension of Area = $\left[L^2\right]$
Dimension of Pressure = $\frac{\left[\text{Force}\right]}{\left[\text{Area}\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M^1{L}^{-1}{T}^{-2}\right]$
Comparing this with the general formula $M^a{L}^b{T}^c$, we obtain:
$a=1$
$b=-1$
$c=-2$
This perfectly matches the values given in the option: Pressure if a = 1, b = -1, c = -2.