A metal block of base area $$0.2$$ $$m2$$ is connected to a $$0.02$$ kg mass via a string that passes over an ideal pulley as shown in figure. A liquid film of thickness $$0.6$$ mm is placed between the block and the table. When released the block moves to the right with a constant speed of $$0.17$$ $$m/s$$. The $$co-efficient$$ of viscosity of the liquid is :

Question

A metal block of base area $$0.2$$ $$m2$$ is connected to a $$0.02$$ kg mass via a string that passes over an ideal pulley as shown in figure. A liquid film of thickness $$0.6$$ mm is placed between the block and the table. When released the block moves to the right with a constant speed of $$0.17$$ $$m/s$$. The $$co-efficient$$ of viscosity of the liquid is :

Infinity Learn · Accepted Answer

$$T=FVmg=$$η A  $$dVdx0.02$$  ×  $$9.8$$  $$=$$  η  $$0.2$$  ×   $$0.17$$  −  $$00.6$$  ×  $$10$$−$$3$$η  $$=$$  $$3.45$$  ×  $$10$$−$$3$$   Pa−S

Viscosity

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$\begin{array}{l} T = F_{V} \\ m g = η A \frac{d V}{d x} \\ 0 .02 \times 9 .8 = η 0 .2 \times \frac{0 .17 - 0}{0 .6 \times 10^{- 3}} \\ η = 3 .45 \times 10^{- 3} P a - S \end{array}$

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Viscosity

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T=FVmg=η A dVdx0.02 × 9.8 = η 0.2 × 0.17 − 00.6 × 10−3η = 3.45 × 10−3 Pa−S

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$\begin{array}{l} T = F_{V} \\ m g = η A \frac{d V}{d x} \\ 0 .02 \times 9 .8 = η 0 .2 \times \frac{0 .17 - 0}{0 .6 \times 10^{- 3}} \\ η = 3 .45 \times 10^{- 3} P a - S \end{array}$