A metal block of base area 0.20 m ² is placed on a table, as shown in… | JEE Main 2023 (29 Jan, Shift 2)

Q16JEE Main 2023 · 29 JAN · SHIFT 2Medium

Viscosity & Stokes' Law

A metal block of base area 0.20 m $^{2}$ is placed on a table, as shown in figure. A liquid film of thickness 0.25 mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. If the viscosity of the liquid is $5.0 \times 10^{-3}\text{ Pl}$ , the speed of block is ____________ $\times 10^{-3}$ m/s.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

Constant speed, so applied force = viscous force: $F = \eta A\frac{v}{d}$ .
$v = \frac{Fd}{\eta A} = \frac{0.1\times0.25\times10^{-3}}{5\times10^{-3}\times0.20}$
$v = \frac{2.5\times10^{-5}}{10^{-3}} = 25\times10^{-3}$ m/s
So the answer is $25$ .

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