Question 23

A metal block of base area $$0.20$$ m$$^2$$ is placed on a table, as shown in the 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 $$5.0 \times 10^{-3}$$ Pl, the speed of the block is ______ $$\times 10^{-3}$$ m s$$^{-1}$$.

Correct Answer: 25

Viscous Force $$F = \eta A \frac{v}{l}$$

$$v = \frac{F \times l}{\eta \times A}$$

$$v = \frac{0.1 \times (0.25 \times 10^{-3})}{(5.0 \times 10^{-3}) \times 0.20}$$

$$v = \frac{0.025}{1.0} = 0.025\text{ m s}^{-1}$$

$$0.025\text{ m s}^{-1} = 25 \times 10^{-3}\text{ m s}^{-1}$$

The value is 25.

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