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  • An experiment is conducted to determine the coefficient of viscosity (η) of a viscous liquid by measuring the terminal velocity of a spherical body falling through the liquid. The sphere used i

An experiment is conducted to determine the coefficient of viscosity (η) of a 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 0.03 m and a density (ρ) of 800 kg/m³. The density of the liquid (ρ0) is 1000 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 0.08 m/s. Calculate the coefficient of viscosity (η) of the given viscous liquid.

Option: 1

4.90Ns/m2


Option: 2

8.70Ns/m2


Option: 3

0.8Ns/m2


Option: 4

4.6Ns/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

We are given:

Sphere radius (r) = 0.03 m

Sphere density (ρ) = 800 kg/m³

Liquid density (ρ0) = 1000 kg/m³

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

Terminal velocity (vt) = 0.08 m/s

Let’s rearrange the equation to solve for the coefficient of viscosity (η):

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

Now, substitute the given values:

\begin{gathered} \eta=\frac{2 \cdot\left(800 \mathrm{~kg} / \mathrm{m}^3-1000 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot(0.03 \mathrm{~m})^2}{9 \cdot 0.08 \mathrm{~m} / \mathrm{s}} \\ \eta=\frac{2 \cdot\left(-200 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot 0.0009 \mathrm{~m}^2}{0.72 \mathrm{~m} / \mathrm{s}} \\ \eta=\frac{-3.5286 \mathrm{Ns} / \mathrm{m}^2}{0.72 \mathrm{~kg} / \mathrm{ms})} \\ \eta \approx-4.90 \mathrm{Ns} / \mathrm{m}^2 \end{gathered}

Since viscosity cannot be negative, it seems there might be an error or discrepancy in the given values or measurements. Please double-check the data provided for accuracy. Therefore, the correct option is A. article

Posted by

Ajit Kumar Dubey

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