Question

A spherical ball of radius 0.3 mm falls vertically through water. Find the coefficient of viscosity of water given that terminal velocity acquired by the ball = 9.8×10^-2 m/s,density of water =10^3 kg/m^3 , density of ball = 1.5×10^3 kg/m^3.

Answer 1

Answer:

the coefficient  of viscosity of water is

\eta = 10^{-7} Pa-sη=10−7Pa−s

Explanation:

As we know that when a ball is dropped into the viscous liquid then the terminal speed of the ball is given by the formula as

v_t = \frac{2r^2g}{9\eta}(\sigma - \rho)vt=9η2r2g(σ−ρ)

Now we know that

v_t = 9.8 \times 10^2 m/svt=9.8×102m/s

\sigma = 1.5 \times 10^3 kg/m^3σ=1.5×103kg/m3

\rho = 10^3 kg/m^3ρ=103kg/m3

r = 0.3 mm

now we will plug in all data in above equation

9.8 \times 10^2 = \frac{2(0.3\times 10^{-3})^2(9.8)}{9(\eta)}(1.5\times 10^3 - 10^3)9.8×102=9(η)2(0.3×10−3)2(9.8)(1.5×103−103)

100 = \frac{2 \times 10^{-8}}{\eta}(500)100=η2×10−8(500)

\eta = 10^{-7} Pa-sη=10−7Pa−s

#Learn

Topic : Terminal speed

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