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  • A cylindrical glass tube with an inner radius of 0.2 mm is dipped into a beaker of mercury. Given the surface tension of mercury is 0.465 N/m and the density is 13546 kg/m3 , calculate t

A cylindrical glass tube with an inner radius of 0.2 mm is dipped into a beaker of mercury. Given the surface tension of mercury is 0.465 N/m and the density is 13546 kg/m3 , calculate the height to which the mercury rises in the tube. Assume standard gravity (9.81 m/s2 ).

 

Option: 1

160Pa · s


Option: 2

170Pa · s


Option: 3

10.44Pa · s


Option: 4

20Pa · s


Answers (1)

best_answer

Given:

r = 0.0002 m,

σ = 0.465 N/m,

ρ = 13546 kg/m3 ,

g = 9.81 m/s2 .

Using the capillary rise equation:

h=\frac{2 \sigma}{r \rho g}=\frac{2 \times 0.465}{0.0002 \times 13546 \times 9.81} \approx 0.035 \mathrm{~m} .

Given data:

Diameter of the sphere (d) = 0.02 meters

Mass of the sphere (m) = 0.1 kilograms

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

Density of the viscous liquid (ρ) = 800 kg/m3

Terminal velocity of the sphere (v) = 0.05 m/s

Step-by-step calculations:

1. Calculate the radius of the sphere:

r=\frac{d}{2}=\frac{0.02}{2}=0.01 \mathrm{~m}

2. Calculate the gravitational force acting on the sphere: Fgravity = mg = 0.1 kg × 9.81 m/s2 = 0.981 N

3. Calculate the drag force: We can rearrange the Stokes’ law equation to solve for the coefficient of viscosity (η):

\eta=\frac{F_{\mathrm{drag}}}{6 \pi r v}

Plugging in the values:

\eta=\frac{0.981}{6 \pi \times 0.01 \times 0.05} \approx 10.44 \mathrm{~Pa} \cdot \mathrm{s}

So, the coefficient of viscosity (η) of the given viscous liquid is approximately 10.44 Pa · s.

Therefore, the correct option is A.

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