Question

A layer of glycerine of thickness $1mm$ is enclosed between a big plate and another plane of area ${10}^{-2}{m}^2$. If the coefficient of viscosity of glycerine is $1kg·m^{-1}·s^{-1}$, then the force in $N$ required to move the plate with a velocity of $0.07m·s^{-1}$.

• A. $7$
• B. $0.7$
• C. $70$
• D. $0.07$

Answer 1

The correct option is B

$0.7$

Step 1: Given data

The given velocity of the plate, $du=0.07m·s^{-1}$.

The thickness of the glycerine layer, $dy=1mm=0.001m$.

The viscosity of glycerine, $\mu =1kg·m^{-1}·s^{-1}$.

Step 2: Formula used

According to Newton's law of viscosity the shear stress of the fluid, $\tau =\mu \frac{du}{dy}$.

Step 3: Calculate the shear stress of the fluid

On substituting the given values in the formula for shear stress of the fluid, we get

$$\begin{gathered} \Rightarrow \tau =1\times \frac{0.07}{0.001} \\ \Rightarrow \tau =70N·m^{-2} \end{gathered}$$

Thus, the shear stress of the fluid, $\tau =70N·m^{-2}$.

Step 4: Calculate force

The given area, $A={10}^{-2}{m}^2$.

Let the required force be $F$.

It is known that, $\tau =\frac{F}{A}$.

$$\begin{gathered} \Rightarrow F=\tau A \\ \Rightarrow F=\left(70\times {10}^{-2}\right)N \\ \Rightarrow F=0.7N \end{gathered}$$

Therefore, the force required to move the plate with a velocity of $0.07m·s^{-1}$ is $0.7N$.

Hence, the correct option is (B).