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 x 102
m/s, density of water =10 Kg/mº, density of ball = 1.5 x 10 Kg/mº.​

Answer 1

Answer:

the coefficient  of viscosity of water is

$\eta = 10^{-7} Pa-s$

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

$v_t = \frac{2r^2g}{9\eta}(\sigma - \rho)$

Now we know that

$v_t = 9.8 \times 10^2 m/s$

$\sigma = 1.5 \times 10^3 kg/m^3$

$\rho = 10^3 kg/m^3$

r = 0.3 mm

now we will plug in all data in above equation

$9.8 \times 10^2 = \frac{2(0.3\times 10^{-3})^2(9.8)}{9(\eta)}(1.5\times 10^3 - 10^3)$

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

$\eta = 10^{-7} Pa-s$

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Topic : Terminal speed

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Answer 2

The coefficient  of viscosity of water is $1\times10^{-7}\ Pa-s$

Explanation:

Given that,

Radius = 0.3 mm

Terminal velocity $v_{t}=9.8\times10^2\ m/s$

Density of water$\rho=10^3\ kg/m^3$

Density of wall $\sigma=1.5\times10^3\ kg/m^3$

We need to calculate the coefficient  of viscosity of water

Using formula of terminal velocity

$v_{t}=\dfrac{2r^2g}{9\eta}(\sigma-\rho)$

Where, r = radius

$\eta$=coefficient  of viscosity

$\rho$= density of water

$\sigma$= density of wall

Put the value into the formula

$9.8\times10^2=\dfrac{2\times(0.3\times10^{-3})^2}{9\eta}(1.5\times10^{3}-10^3)$

$\eta=\dfrac{2\times(0.3\times10^{-3})^2}{9\times9.8\times10^2}(1.5\times10^{3}-10^3)$

$100=\dfrac{2\times10^{-8}}{\eta}\times500$

$\eta=1\times10^{-7}\ Pa-s$

Hence, The coefficient  of viscosity of water is $1\times10^{-7}\ Pa-s$

Learn more :

Topic : viscosity

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