. A metal block of area 0.1m2 is connected to a 0.01kg mass via a string that passes over an ideal pulley. A liquid with a film thickness of 0.3mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085ms-1. Find the coefficient of viscosity of the liquid.
Answers
Since the body is moving with constant velocity it indicates that the net force acting are balanced
Answer:3.5 ( 10^-3)
Explanation:
Here, m = 0.01 kg, l = 0.3 mm = 0.3 \times 10^{-3}m,m=0.01kg,l=0.3mm=0.3×10
−3
m,
g = 10 m s^{-2}, v = 0.085 m s^{-1}, A = 0.1 m^2g=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 = 0.01 kg \times 10 m s^{-2} = 0.1 NF=T=mg=0.01kg×10ms
−2
=0.1N
Shear stress on the fluid = \displaystyle \frac{F}{A} = \frac{0.1 N}{0.1 m^2}
A
F
=
0.1m
2
0.1N
Strain rate = \displaystyle \frac{v}{l} = \frac{0.085 m s^{-1}}{0.3 \times 10^{-3}m}
l
v
=
0.3×10
−3
m
0.085ms
−1
Coefficient of viscosity, \displaystyle \eta = \frac{Shear \,\, stress}{Strain \,\, rate}η=
Strainrate
Shear stress
= \displaystyle \left( \frac{0.1N}{ 0.1 m^2} \right) \times \frac{0.3 \times 10^{-3} m}{0.085 ms^{-1}} = 3.5 \times 10^{-3} Pa \, s(
0.1 m
2
0.1N
)×
0.085ms
−1
0.3×10
−3
m
=3.5×10
−3
Pas