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Question: A metal block of area \(0.10m^{2}\) is connected to a 0.01 kg mass via a string that passes over a m...

A metal block of area 0.10m20.10m^{2} is connected to a 0.01 kg mass via a string that passes over a mass less and frictionless pulley as shown in figure. A liquid with a film thickness of 0.3 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.08ms−10.08ms^{- 1}The coefficient of viscosity of the liquid is (Take g=10ms−2g = 10ms^{- 2})

A

2.5×10−3 Pa s2.5 \times 10^{- 3}\text{ Pa s}

B

3.5×10−3 Pa s3.5 \times 10^{- 3}\text{ Pa s}

C

4.5×10−3 Pa s4.5 \times 10^{- 3}\text{ Pa s}

D

6.5×10−3 Pa s6.5 \times 10^{- 3}\text{ Pa s}

Answer

3.5×10−3 Pa s3.5 \times 10^{- 3}\text{ Pa s}

Explanation

Solution

Here, m=0.01kg,l=03mm=0.3×10−3m,m = 0.01kg,l = 03mm = 0.3 \times 10^{- 3}m,

g=10ms−2,v=0.085ms−1,A=0.1m2g = 10ms^{- 2},v = 0.085ms^{- 1},A = 0.1m^{2}

The metal block moves to the right due to tension T of the string which is equal to the weight of the mass suspended at the end of the string.

Thus,

Shear force, F=T=mg=0.0kg×10ms−2F = T = mg = 0.0kg \times 10ms^{- 2}

=0.1N= 0.1N

Shear stress on the fluid =FA=0.1N0.1m2= \frac{F}{A} = \frac{0.1N}{0.1m^{2}}

Strain rate =vl=0.085ms−10.3×10−3m= \frac{v}{l} = \frac{0.085ms^{- 1}}{0.3 \times 10^{- 3}m}

Coefficient of viscosity, η=ShearstressStrainrate\eta = \frac{Shearstress}{Strainrate}

=(0.1N0.1m2)×(0.3×10−3m)0.085ms−1=3.5×10−3Pas= \left( \frac{0.1N}{0.1m^{2}} \right) \times \frac{(0.3 \times 10^{- 3}m)}{0.085ms^{- 1}} = 3.5 \times 10^{- 3}Pas