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

B. $10 N$

C. $60 N$

D. $30 N$

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

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

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,

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

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

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