A metal block of film area $0.10 m^{2}$ is connected to a $0.010 k g$ mass via a string that passes over an ideal pulley (considered massless & frictionless). A liquid with a film thickness of $0.30 m m$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.085 m / s$ . Find the coefficient of viscosity of the liquid. $(g = 9.8 m / s^{2})$

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$3.45 \times 10^{- 6} P a . s$

$3.45 \times 10^{- 5} P a . s$

$3.45 \times 10^{- 3} P a . s$

$6 \times 10^{- 2} P a . s$

answer is C.

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Detailed Solution

As the acceleration is zero, shearing force is provided by the tension in the string which is equal to weight of the hanging mass. $coefficient of viscosity = η = \frac{s h e a r s t r e s s}{s t r a i n R a t e} = \frac{F / A}{V / l} = \frac{m g l}{A V}$