Through a very narrow gap of height h a thin plate of large extension is pulled. At a velocity v, on one side of the plate there exists oil of viscosity $μ_{1}$ and on the other side, oil of viscosity $μ_{2}$ as shown. The position (y) of the plate so that (i) the shear force on the two sides of the plate is equal . (ii) the pull required to drag the plate is minimum respectively are

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$\frac{μ_{2} h}{μ_{1} + μ_{2}}, \frac{h}{1 + \sqrt{\frac{μ_{1}}{μ_{2}}}}$

$\frac{μ_{1} h}{μ_{1} + μ_{2}}, \frac{h}{1 + \sqrt{\frac{μ_{1}}{μ_{2}}}}$

$\frac{μ_{2} h}{μ_{1} + μ_{2}}, \frac{h}{1 + \sqrt{\frac{μ_{2}}{μ_{1}}}}$

$\frac{μ_{1} h}{μ_{1} + μ_{2}}, \frac{h}{1 + \sqrt{\frac{μ_{2}}{μ_{1}}}}$

answer is B.

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Detailed Solution

Given $F_{1} = F_{2} \Rightarrow μ_{1} A \frac{V}{(h - y)} = μ_{2} A \frac{V}{y} \Rightarrow y = (\frac{μ_{2} h}{μ_{1} + μ_{2}})$
$\begin{array}{l} F_{net} = F_{1} + F_{2} \Rightarrow F_{net} = \frac{μ_{1} AV}{h - y} + \frac{μ_{2} AV}{y} ∵ \frac{{dF}_{net}}{dy} = 0 \Rightarrow \frac{d}{dy} [\frac{μ_{1}}{h - y} + \frac{μ_{2}}{y}] = 0 \\ \Rightarrow y = \frac{h}{1 + \sqrt{\frac{μ_{1}}{μ_{2}}}} \end{array}$