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(B) Let the velocity of the thin plate be $v$ and its area be $A$. The force required to pull the plate is the sum of the viscous forces from both liq

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$2$

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(B) Let the velocity of the thin plate be $v$ and its area be $A$. The force required to pull the plate is the sum of the viscous forces from both liquids:$F = F_1 + F_2 = \eta A \frac{v}{d_1} + 4\eta A \frac{v}{d_2}$Given that $d_1 + d_2 = d$ (constant total distance),we can write $d_2 = d - d_1$. Substituting this:$F = \eta A v \left( \frac{1}{d_1} + \frac{4}{d - d_1} \right)$To find the minimum force,we differentiate $F$ with respect to $d_1$ and set it to zero:$\frac{dF}{dd_1} = \eta A v \left( -\frac{1}{d_1^2} + \frac{4}{(d - d_1)^2} \right) = 0$$\frac{1}{d_1^2} = \frac{4}{(d - d_1)^2}$Taking the square root on both sides:$\frac{1}{d_1} = \frac{2}{d - d_1}$$d - d_1 = 2d_1$$d = 3d_1 \Rightarrow d_1 = \frac{d}{3}$Then $d_2 = d - \frac{d}{3} = \frac{2d}{3}$.Therefore,the ratio $\frac{d_2}{d_1} = \frac{2d/3}{d/3} = 2$.

$A$ thin plate is placed between two fixed plates,separating two liquids with coefficients of viscosity $\eta$ and $4\eta$,as shown in the figure. If the plate is to be pulled by applying the minimum force,then the ratio $\frac{d_2}{d_1}$ is:

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