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A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley as show in figure. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 ms^-1. What is the coefficient of viscosity of the liquid.

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$3.46 \times 10^{- 5} Pa - s$

$3.46 \times 10^{- 6} Pa - s$

$3.46 \times 10^{- 4} Pa - s$

$3.46 \times 10^{- 3} Pa - s$

answer is A.

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Detailed Solution

Shear stress on the fluid = $\frac{F}{A} = \frac{0.010 X 9.8}{0.10}$
Strain rate $= \frac{V}{l} = \frac{0.085}{0.30 X 10^{- 3}}$
$η = \frac{s t r e s s}{s t r a i n r a t e} = 3.46 X 10^{- 3} P a s$

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