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  • A complex experiment is designed to determine the coefficient of viscosity (η) of a highly viscous liquid by measuring the terminal velocity of a spherical body falling through the liquid. The

A complex experiment is designed to determine the coefficient of viscosity (η) of a highly viscous liquid by measuring the terminal velocity of a spherical body falling through the liquid. The sphere used in the experiment has a radius (r) of 0.01 m and a density (ρ) of 1500 kg/m³. The density of the liquid (ρ0) is 1800 kg/m³, and the acceleration due to gravity (g) is 9.81 m/s². The measured terminal velocity of the sphere in the liquid is 0.02 m/s. The experiment is conducted at a temperature of 30°C, and the dynamic viscosity (µ) of the liquid at this temperature is 0.5 Ns/m². Calculate the coefficient of viscosity (η) of the given viscous liquid, taking into account the temperature-dependent dynamic viscosity relationship.

Option: 1

1.60Ns/m2


Option: 2

0.805Ns/m2


Option: 3

0.8Ns/m2


Option: 4

4.36Ns/m2


Answers (1)

best_answer

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

v_t=\frac{2 \cdot\left(\rho-\rho_0\right) \cdot g \cdot r^2}{9 \cdot \eta}

Where:

vt is the terminal velocity of the sphere

ρ is the density of the sphere

ρ0 is the density of the liquid

g is the acceleration due to gravity

r is the radius of the sphere

η is the coefficient of viscosity of the liquid

The dynamic viscosity (µ) of the liquid can be related to the coefficient of viscosity (η) using the temperature-dependent relationship:

\mu=\eta \cdot\left(\frac{T}{T_0}\right)^{\frac{3}{2}}

Where:

µ is the dynamic viscosity of the liquid

T is the absolute temperature of the liquid (in Kelvin)

T0 is a reference temperature (in Kelvin)

We are given:

Sphere radius (r) = 0.01 m

Sphere density (ρ) = 1500 kg/m³

Liquid density (ρ0) = 1800 kg/m³

Acceleration due to gravity (g) = 9.81 m/s²

Terminal velocity (vt) = 0.02 m/s

Dynamic viscosity (µ) = 0.5 Ns/m²

Reference temperature (T0) = 273.15 K (0°C)

Let’s start by calculating the absolute temperature of the liquid (in Kelvin):

T=T_0+30 \mathrm{~K}=273.15 \mathrm{~K}+30 \mathrm{~K}=303.15 \mathrm{~K}

Next, we can use the dynamic viscosity relationship to find the coefficient of viscosity (η):

\eta=\mu \cdot\left(\frac{T}{T_0}\right)^{\frac{3}{2}}

Substitute the values:

\begin{gathered} \eta=0.5 \mathrm{Ns} / \mathrm{m}^2 \cdot\left(\frac{303.15 \mathrm{~K}}{273.15 \mathrm{~K}}\right)^{\frac{3}{2}} \\ \eta \approx 0.805 \mathrm{Ns} / \mathrm{m}^2 \end{gathered}

The coefficient of viscosity of the given viscous liquid at the experimental temperature of 30°C is approximately 0.805 Ns/m².

Therefore, the correct option is B.

Posted by

vinayak

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