Question

A plate of area $100c{{m}^{2}}$ is placed on the upper surface of castor oil, 2mm thick. Taking the coefficient of viscosity to be 15.5 poise, calculate the horizontal force necessary to move the plate with a velocity, $3cm{{s}^{-1}}$ .
$\begin{align}
  & (A)0.2800N \\
 & (B)0.2455N \\
 & (C)0.2325N \\
 & (D)0.1000N \\
\end{align}$

Answer 1

Hint: Since, the plate is kept on a viscous fluid, the fluid will exert a viscous force on the sheet opposite to the direction in which the sheet tends to move. This force is directly proportional to the area of contact of the liquid with the object and velocity of the object. Also, it is inversely proportional to the thickness of the viscous liquid on which it is made to move.

Complete step-by-step answer:
Let us first assign some terms that we are going to use in our equations later.
Let the area of the plate be given by A. Then, it has been given to us in the problem that:
$\Rightarrow A=100c{{m}^{2}}$
Converting it into standard units, we get:
$\begin{align}
  & \Rightarrow A=100\times {{10}^{-4}}{{m}^{2}} \\
 & \therefore A={{10}^{-2}}{{m}^{2}} \\
\end{align}$
Now, let the coefficient of viscosity to be given by $\eta $ . Then, the value of $\eta $is given to us as:
$\Rightarrow \eta =15.5poise$
Converting it into standard units, we get:
$\begin{align}
  & \Rightarrow \eta =15.5\times {{10}^{-1}}Pa \\
 & \therefore \eta =1.55Pa \\
\end{align}$
The velocity (say $\vartriangle v$ )of the metal sheet is given to us:
$\Rightarrow \vartriangle v=3cm{{s}^{-1}}$
Converting it into standard units, we get:
$\Rightarrow \vartriangle v=3\times {{10}^{-2}}m{{s}^{-1}}$
And the thickness of the oil layer (say $\vartriangle x$) is given to us as:
$\Rightarrow \vartriangle x=2mm$
Converting it into standard units, we get:
$\Rightarrow \vartriangle x=2\times {{10}^{-3}}m$
We know the formula for viscous force due to a liquid on an object at the surface of the liquid is given by:
$\Rightarrow F=\eta A\dfrac{\vartriangle v}{\vartriangle x}$
Putting the values of all the terms in the right-hand side of the equation, we get:
$\begin{align}
  & \Rightarrow F=\dfrac{1.55\times {{10}^{-2}}\times 3\times {{10}^{-2}}}{2\times {{10}^{-3}}}N \\
 & \Rightarrow F=\dfrac{23.25\times {{10}^{-7}}}{{{10}^{-5}}} \\
 & \therefore F=0.2325N \\
\end{align}$
Hence, the horizontal force necessary to move the plate with a velocity, $3cm{{s}^{-1}}$ is equal to 0.2325N.

So, the correct answer is “Option C”.

Note: One can comprehend from the above solution that the viscous force due to a liquid is proportional to its area, but the frictional force existing between two solid blocks is independent of their respective areas. This is the main difference between a viscous force and a frictional force. That’s also the reason why they were given different names.