Question

) The terminal velocity of a copper ball of radius 2 mm falling through a
tank of oil at 20°C is 6.5 cm/s. Compute the viscosity of the oil at
20°C. Given poil = 1.5*103 kg m-3, pcopper = 8.9x10²kg m”.
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Answer 1

Radius of copper ball, \sf{r=2\times10^{-3}\ m.}r=2×10 −3m.

Terminal velocity of copper ball, \sf{v=6.5\times10^{-2}\ m\,s^{-1}.}v=6.5×10

−2 ms −1 .

Density of copper, \sf{\rho=8.9\times10^3\ kg\,m^{-3}.}ρ=8.9×10

3 kgm −3

Density of oil, \sf{\sigma=1.5\times10^3\ kg\,m^{-3}.}σ=1.5×10

3kgm −3 .

Let \sf{g=10\ m\,s^{-2}.}g=10 ms

−2 .

The expression for terminal velocity,

$\longrightarrow\sf{v=\dfrac{2gr^2(\rho-\sigma)}{9\eta}}⟶v=$

9η2gr 2 (ρ−σ)

Then, coefficient of viscosity,

$\longrightarrow\sf{\eta=\dfrac{2gr^2(\rho-\sigma)}{9v}}⟶η=$

9v2gr 2

(ρ−σ)

$\longrightarrow\sf{\eta=\dfrac{2\times10\times(2\times10^{-3})^2\times10^3(8.9-1.5)}{9\times6.5\times10^{-2}}}⟶η=$

9×6.5×10 −2

2×10×(2×10 −3 ) 2×10 3 (8.9−1.5)

$\longrightarrow\sf{\eta=\dfrac{2\times10\times4\times10^{-6}\times10^3\times7.4}{9\times6.5\times10^{-2}}}⟶η=$

9×6.5×10 −22×10×4×10 −6×10 3 ×7.4

$\longrightarrow\sf{\underline{\underline{\eta=1.01\ kg\,m^{-1}\,s^{-1}}}}⟶$

η=1.01 kgm−1s−1

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Answer 2

Answer:

Radius of copper ball, \sf{r=2\times10^{-3}\ m.}r=2×10 −3m.

Terminal velocity of copper ball, \sf{v=6.5\times10^{-2}\ m\,s^{-1}.}v=6.5×10

−2 ms −1 .

Density of copper, \sf{\rho=8.9\times10^3\ kg\,m^{-3}.}ρ=8.9×10

3 kgm −3

Density of oil, \sf{\sigma=1.5\times10^3\ kg\,m^{-3}.}σ=1.5×10

3kgm −3 .

Let \sf{g=10\ m\,s^{-2}.}g=10 ms

−2 .

The expression for terminal velocity,

\longrightarrow\sf{v=\dfrac{2gr^2(\rho-\sigma)}{9\eta}}⟶v=⟶v=

9η

2gr

2

(ρ−σ)

⟶v=

9η2gr 2 (ρ−σ)

Then, coefficient of viscosity,

\longrightarrow\sf{\eta=\dfrac{2gr^2(\rho-\sigma)}{9v}}⟶η=⟶η=

9v

2gr

2

(ρ−σ)

⟶η=

9v2gr 2

(ρ−σ)

\longrightarrow\sf{\eta=\dfrac{2\times10\times(2\times10^{-3})^2\times10^3(8.9-1.5)}{9\times6.5\times10^{-2}}}⟶η=⟶η=

9×6.5×10

−2

2×10×(2×10

−3

)

2

×10

3

(8.9−1.5)

⟶η=

9×6.5×10 −2

2×10×(2×10 −3 ) 2×10 3 (8.9−1.5)

\longrightarrow\sf{\eta=\dfrac{2\times10\times4\times10^{-6}\times10^3\times7.4}{9\times6.5\times10^{-2}}}⟶η=⟶η=

9×6.5×10

−2

2×10×4×10

−6

×10

3

×7.4

⟶η=

9×6.5×10 −22×10×4×10 −6×10 3 ×7.4

\longrightarrow\sf{\underline{\underline{\eta=1.01\ kg\,m^{-1}\,s^{-1}}}}⟶⟶

η=1.01 kgm

−1

s

−1

⟶

η=1.01 kgm−1s−1

Explanation:

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