Question

An iron ball of radius 0.3 cm falls through a column of oil of density 0.94g/cm3. It is found to attain a terminal velocity of 0.54 m/s. Determine the viscosity of the oil. Given that the density of iron is 7.8g/cm3.

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Answer 1

Answer:

Given:

Radius of ball = 0.3 cm

Density of oil = 0.94 g/cm³

Density of iron = 7.8 g/cm³

Terminal velocity = 0.54 m/s

To find:

Coefficient of viscosity of oil

Formulas used:

$terminal \: velocity \: = \frac{2g(σ-ρ) {r}^{2} }{9η}$

Conversion:

1. Radius of ball = 0.3 cm = 0.003 m

2. Density (oil) = 940 kg/m³

3. Density (iron) = 7800 kg/m³

Calculation:

Putting all the necessary values for the given Equation :

Vel. = 2g (7800 - 940) × {3×10^(-3)}²/9η

=> 0.54 = 20 ×6860 × {9 × 10^(-6)}/9η

=> η = 2540.74 × {10^(-4)}

=> η = 0.254074 poise.

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\tt Given \begin{cases} \sf{Radius \: of \: Iron \: ball \: (r) \: = \: 0.3 \: cm \: = \: 0.03 \: m} \\ \sf{Terminal \: Velocity \: (v) \: = \: 0.54 \: m/s} \\ \sf{Density \: of \: Iron (\rho) \: = \: 7.8 \: g/cm^3 \: = \: 7800 \: g/m^3} \\ \sf{Density \: of \: Oil \: (\sigma) \: = \: 0.94 \: g/cm^3 \: = \: 940 \: kg/m^3} \end{cases}$

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To Find :

Coefficient of Viscosity of Ball

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Solution :

Use formula for Terminal Velocity :

$\LARGE \implies \star {\underline{\boxed{\sf{v \: = \: \frac{2(\rho \: - \: \sigma) r^2g}{9 \eta}}}}}$

$\Large \leadsto {\sf{\eta \: = \: \frac{2(\rho \: - \: \sigma) r^2g}{9v}}}$

Put the given Values

$\large \leadsto {\sf{\eta \: = \: \frac{2(7800 \: - \: 940)(0.03)^2(10)}{9(0.54)}}}$

$\large \leadsto {\sf{\eta \: = \: \frac{2(6860)(0.000009)(10)}{9(0.54)}}}$

$\large \leadsto {\sf{\eta \: = \: \frac{13720(0.000009)(10)}{4.86}}}$

$\large \leadsto {\sf{\eta \: = \: \frac{(137200)(0.000009}{4.86}}}$

$\large \leadsto {\sf{\eta \: = \: \frac{137200)(9)*10^{-6}}{4.86}}}$

$\LARGE \implies {\boxed{\boxed{\sf{\eta \: = \: 0.254 \: Poise}}}}$

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