A thin square plate of side 2 m is moving at | vedclass

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(B) The viscous force $F$ required to move the plate is the sum of the viscous drag forces from both liquids. The formula for viscous force is $F = \e

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The viscous force $F$ required to move the plate is the sum of the viscous drag forces from both liquids. The formula for viscous force is $F = \eta A \frac{dv}{dx}$.Given: side of square plate $L = 2\ m$, so area $A = L^2 = 4\ m^2$. Velocity $v = 10\ m/s$. Viscosities in $SI$ units: $\eta_1 = 1 	ext{ poise} = 0.1 	ext{ Pa} \cdot 	ext{s}$ and $\eta_2 = 4 	ext{ poise} = 0.4 	ext{ Pa} \cdot 	ext{s}$.Total force $F = F_1 + F_2 = \eta_1 A \frac{v}{h_1} + \eta_2 A \frac{v}{h_2}$.Substituting the values: $F = 0.1 	imes 4 	imes \frac{10}{h_1} + 0.4 	imes 4 	imes \frac{10}{h_2} = \frac{4}{h_1} + \frac{16}{h_2}$.Since $h_1 + h_2 = 3$, we have $h_1 = 3 - h_2$. Thus, $F(h_2) = \frac{4}{3 - h_2} + \frac{16}{h_2}$.To find the minimum force, set $\frac{dF}{dh_2} = 0$:$\frac{d}{dh_2} [4(3 - h_2)^{-1} + 16(h_2)^{-1}] = 4(3 - h_2)^{-2} - 16(h_2)^{-2} = 0$.$\frac{4}{(3 - h_2)^2} = \frac{16}{h_2^2} \implies \frac{1}{(3 - h_2)^2} = \frac{4}{h_2^2}$.Taking the square root: $\frac{1}{3 - h_2} = \frac{2}{h_2} \implies h_2 = 6 - 2h_2 \implies 3h_2 = 6 \implies h_2 = 2\ m$.Then $h_1 = 3 - 2 = 1\ m$.Substituting back: $F_{\min} = \frac{4}{1} + \frac{16}{2} = 4 + 8 = 12\ N$.

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