Question

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 \(5.0×10^{−3}Pl\), the speed of block is  ________ \(×10^{−3}m/s\)

Answer 1

Correct Answer: 25

Using the equation for viscous force: \[ F = \eta A \frac{\Delta v}{\Delta h}. \] Substitute: \[ 0.1 = (5 \times 10^{-3}) (0.2) \frac{v}{0.25 \times 10^{-3}}. \] Solve for \(v\): \[ v = \frac{0.1 \cdot 0.25 \times 10^{-3}}{5 \times 10^{-3} \cdot 0.2}. \] Simplify: \[ v = 25 \times 10^{-3} \, \text{m/s}. \] Thus, the speed is \(25 \times 10^{-3} \, \text{m/s}\).