Question

The unit of impulse per unit area is the same as that of:
A. Coefficient of viscosity
B. Surface tension
C. Bulk modulus
D. None of the above

Answer 1

The change in momentum, that is, the product of force and change in time is called the impulse. The ratio of the shearing stress to the velocity gradient of the fluid is called the coefficient of viscosity. Using these definitions and the dimensional formulae, we will answer the question.

Complete answer:
From the given information, we have the data as follows.
The change in momentum, that is, the product of force and change in time is called the impulse. The ratio of the shearing stress to the velocity gradient of the fluid is called the coefficient of viscosity. OR, the measure of the resistance offered by the fluid to deform. The tendency of the liquid surface to shrink into the minimum surface area possible is called the surface tension. The ratio of direct stress to the volumetric strain is called the bulk modulus.
The dimensional formula of impulse per unit area is given as follows.
The formula of impulse and the area is, $\dfrac{I}{A}=\dfrac{F\times t}{A}$
The dimensional formula is, $\begin{aligned} & \dfrac{I}{A}=\dfrac{[ML{{T}^{-2}}][T]}{[{{L}^{2}}]} \\\ & \therefore \dfrac{I}{A}=[M{{L}^{-1}}{{T}^{-1}}] \\\ \end{aligned}$
The dimensional formula of surface tension is given as follows.
The formula of the surface tension is, $\sigma =\dfrac{F}{l}$
The dimensional formula is, $\begin{aligned} & \sigma =\dfrac{[ML{{T}^{-2}}]}{[L]} \\\ & \therefore \sigma =[M{{L}^{0}}{{T}^{-2}}] \\\ \end{aligned}$
The dimensional formula of the coefficient of viscosity is given as follows.
The formula of the coefficient of viscosity is, $\eta =\dfrac{F}{A\times \left( {}^{dv}/{}_{dx} \right)}$
The dimensional formula is, $\begin{aligned} & \eta =\dfrac{[ML{{T}^{-2}}]}{[{{L}^{2}}]\times [{{T}^{-1}}]} \\\ & \therefore \eta =[M{{L}^{-1}}{{T}^{-1}}] \\\ \end{aligned}$
The dimensional formula of the bulk modulus is given as follows.
The formula of the bulk modulus is, $k=\dfrac{dp}{{}^{dv}/{}_{v}}$
The dimensional formula is, $\begin{aligned} & k=\dfrac{[M{{L}^{-1}}{{T}^{-2}}]}{[{{M}^{0}}{{L}^{0}}{{T}^{0}}]} \\\ & \therefore k=[M{{L}^{-1}}{{T}^{-2}}] \\\ \end{aligned}$
$\therefore$ The unit of impulse per unit area is the same as that of the coefficient of viscosity.

Thus, option (A) is correct.

Note:
The coefficient of viscosity is a property of the fluid. The bulk modulus is a property of a material that deals with elasticity in terms of volume. The higher the cohesive nature of the molecules of the fluid, the higher will be the surface tension of the fluid.