According to Newton, the viscous force acting between liquid layers of area A and velocity gradient Δv/Δx is given by F= -ηA(Δv/Δx), where η is constant called coefficient of viscosity. The dimensional formula of η is

The correct option is [ML⁻¹T⁻¹].

To find the dimensional formula of the coefficient of viscosity ($\eta$), we start with the given formula for the viscous force:
$F=-\eta A\left(\frac{\Delta v}{\Delta x}\right)$

Rearranging the equation to solve for the coefficient of viscosity $\eta$ (ignoring the negative sign, which only indicates the direction of the force opposing the relative motion):
$\eta =\frac{F}{A\left(\frac{\Delta v}{\Delta x}\right)}$

Now, let us determine the dimensions of each physical quantity involved in this expression:

1. Force ($F$): Force is mass times acceleration, so its dimensional formula is:
$\left[F\right]=\left[ML{T}^{-2}\right]$

2. Area ($A$): Area is length squared, so its dimensional formula is:
$\left[A\right]=\left[L^2\right]$

3. Velocity gradient ($\frac{\Delta v}{\Delta x}$): Velocity ($v$) has dimensions of distance over time ($\left[L{T}^{-1}\right]$), and distance ($x$) has dimensions of length ($\left[L\right]$). Therefore:
$\left[\frac{\Delta v}{\Delta x}\right]=\frac{\left[L{T}^{-1}\right]}{\left[L\right]}=\left[T^{-1}\right]$

Substituting these dimensions back into the expression for $\eta$:
$\left[\eta \right]=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]·\left[T^{-1}\right]}$

Simplifying the denominator:
$\left[\eta \right]=\frac{\left[ML{T}^{-2}\right]}{\left[L^2{T}^{-1}\right]}$

Now, simplifying the division by subtracting the powers of like bases (L and T):
$\left[\eta \right]=\left[M\right]·\left[L^{1-2}\right]·\left[T^{-2-\left(-1\right)}\right]$

This gives:
$\left[\eta \right]=\left[M{L}^{-1}{T}^{-1}\right]$

Thus, the dimensional formula of the coefficient of viscosity $\eta$ is [ML⁻¹T⁻¹].