Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^{–3} kg s^{–1}, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 10^{3} kg m^{–3} and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Asked by Pragya Singh | 1 year ago | 62

Given:

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02 m

Glycerine is flowing at the rate of 4.0 × 10^{-3} kg/s

M = 4.0 × 10^{-3} kg/s

Density of glycerine, ρ = 1.3 × 10^{3} kg m^{-3}

Viscosity of glycerine, η = 0.83 Pa s

We know, volume of glycerine flowing per sec:

V =\( \dfrac{ M }{ density }= \dfrac{ 4\; \times \;10^{ -3 } }{ 1.3\; \times \;10^{ 3 } }\)

= 3.08 x 10 ^{-6} m^{3} /s

Using Poiseville’s formula, we get:

V = \( \dfrac{ \pi p^{‘} r^{ 4 }}{ 8\; \eta \;l}p^{‘} = \dfrac{ V \;8\; \eta \;l}{ \pi r^{ 4 }}\)

Where p’ is the pressure difference between the two ends of the pipe.

p’ = \( \dfrac{3.08\; \times \;10^{-6} \;\times \;8\; \times \;0.83\; \times \;2 }{\pi \;\times \;(0.01)^{4} }\)

= 9.8 x 10^{2 }Pa

And, we know:

R = \( \dfrac{ 4 \;\rho \;V }{ \pi \;d\; \eta }\), [Where R = Reynolds’s number]

R = \( \dfrac{4 \times 1.3 \times 10^{3} \times 3.08\times 10^{-6} }{\pi \times 0.83 \times 0.02 }= 0.3\)

Since the Reynolds’s number is 0.3 which is way smaller than 2000 , the flow of glycerine in the pipe is laminar.

Answered by Abhisek | 1 year ago**Answer the following questions:-**

**(a)** According to the law of atmospheres density of air decreases with increase in height y as \( ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}\). Where ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

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Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m^{–1}. Density of mercury = 13.6 × 10^{3} kg m^{–3}

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius \( 2.0 × 10^{–5} m\) and density \( 1.2 × 10^3 kg m^{–3}\). Take the viscosity of air at the temperature of the experiment to be \( 1.8 × 10^{–5} Pa \). How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

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