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  • An experiment is conducted to determine the coefficient of viscosity () of a liquid using the measurement of terminal

An experiment is conducted to determine the coefficient of viscosity (\eta) of a liquid using the measurement of terminal velocity. A spherical body of radius r = 0.03 m and density ρ = 1200 kg/m3 is allowed to fall freely in the liquid. The terminal velocity of the sphere is observed to be vt = 0.08 m/s. Given that the acceleration due to gravity is g = 9.81 m/s2, calculate the coefficient of viscosity of the liquid.

Option: 1

1.38 * 10-3 Pa.s


Option: 2

2.38 * 10-3 Pa.s


Option: 3

1.38 * 10-2 Pa.s


Option: 4

1.38 * 10-4 Pa.s


Answers (1)

best_answer

Given values:

Radius of the spherical body (r) = 0.03 m
Density of the spherical body (ρ) = 1200 kg/m3
Terminal velocity (vt) = 0.08 m/s
Acceleration due to gravity (g) = 9.81 m/s2

The terminal velocity (vt) of a spherical body falling through a liquid is given by the formula:

                    v_{t} = \frac{2}{9} \frac{r^{2} (\rho - \rho_{liquid}) g}{\eta}

where \rho_{liquid} is the density of the liquid.

Solve for the coefficient of viscosity (\eta):

                    \eta = \frac{2}{9} \frac{r^{2} (\rho - \rho_{liquid}) g}{v_{t}}

Step 1: Calculate the density of the liquid (\rho_{liquid}):

                    \rho_{liquid} = \rho - \frac{4}{3} \pi r^{3}

Step 2: Substitute the values and calculate the coefficient of viscosity (\eta):

                    \eta = \frac{2}{9} \frac{(0.03m)^{2} (1200 kg/m^{3} - \rho_{liquid}) 9.81 m/s^{2}}{0.08 m/s}Step 3: Calculate the density of the liquid:

\rho_{liquid} = 1200 kg/m^{3} - \frac {4}{3} \pi (0.03m)^{3} = 1200 kg/m^{3} - 0.00113 kg/m^{3} = 1199.998 kg/m^{3}Step 4: Substitute the calculated value of \rho_{liquid} and the given values and calculate\eta:

                    \eta = \frac{2}{9} \frac{(0.03m)^{2} (1200 kg/m^{3} - 1199.998 kg/m^{3}) 9.81 m/s^{2}}{0.08 m/s}Step 5: Calculate \eta:

                    \eta = 1.38 *10-3 Pa.s

The coefficient of viscosity of the liquid is 1.38 *10-3 Pa.s.

Posted by

Suraj Bhandari

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