# A force of 10 N is required to draw a rectangular glass plate on the surface of a liquid with some velocity. Force needed to draw another glass plate of 3 times length and 2 times width is:

A. \[\dfrac{5}{3}\,{\text{N}}\]

B. $10 N$

C. $60 N$

D. $30 N$

Answer

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**Hint:**The resistance force acting on the glass plate is the viscous force due to the liquid. Recall the formula for the viscous force and determine its dependence on the area of the glass plate. Find out the area of the second plate and thus determine the viscous force acting on the second glass plate.

**Formula used:**

\[{F_v} = \eta A\dfrac{{dv}}{{dx}}\]

Here, \[\eta \] is the coefficient of viscosity, A is the area of the object and \[\dfrac{{dv}}{{dx}}\] is the velocity gradient.

**Complete step by step answer:**

When we draw out an object from the viscous liquid, the resistance provided by the liquid is known as viscous force. In this case, the viscous force acting on the glass plate is 10 N.We have the expression for the viscous force,

\[{F_v} = \eta A\dfrac{{dv}}{{dx}}\] ….. (1)

Here, \[\eta \] is the coefficient of viscosity, A is the area of the object and \[\dfrac{{dv}}{{dx}}\] is the velocity gradient.

Let the length of the glass plate is l and its width is w. Therefore, the area of the first glass plate is,

\[{A_1} = lw\]

Now, we have given that another plate has 3 times the length and 2 times the width of the first glass plate. Therefore, the area of the second plate is,

\[{A_2} = 3l \times 2w\]

\[ \Rightarrow {A_2} = 6lw\]

\[ \Rightarrow {A_2} = 6{A_1}\]

From equation (1), we can write,

\[{F_v} \propto A\]

Therefore, we can write,

\[\dfrac{{{F_2}}}{{{F_1}}} = \dfrac{{{A_2}}}{{{A_1}}}\]

\[ \Rightarrow \dfrac{{{F_2}}}{{{F_1}}} = \dfrac{{6{A_1}}}{{{A_1}}}\]

\[ \Rightarrow {F_2} = 6{F_1}\]

Substituting 10 N for \[{F_1}\] in the above equation, we get,

\[{F_2} = 6\left( {10} \right)\]

\[ \therefore {F_2} = 60\,{\text{N}}\]

**So, the correct answer is option C.**

**Note:**The velocity gradient is the relative velocity between the layers of the same liquid. Since the second glass place is drawn from the same liquid, the velocity gradient remains the same and therefore, we have taken the proportionality only between the viscous force and area of the glass plate. The given glass plate is rectangular; therefore, the area of the glass plate is the product of its length and width.

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