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.

Asked by Abhisek | 1 year ago | 188

Acceleration due to gravity, g = 9.8 m/s^{2}

Radius of the uncharged drop, r = 2.0 × 10^{-5} m

Density of the uncharged drop, ρ = 1.2 × 10^{3} kg m^{-3}

Viscosity of air, η = 1.8 × 10^{-5} Pa s

We consider the density of air to be zero in order to neglect the buoyancy of air.

Therefore terminal velocity (v) is :

v = \( \dfrac{ 2 r^{ 2 } g \rho }{ 9 \eta }\)

= \( \dfrac{ 2 \left ( 2.0 \times 10^{ -5 } \right )^{ 2 } \times 9.8 \times 1.2 \times 10^{ 3 } }{ 9 \times 1.8 \times 10^{ -5 } }\)

And the viscous force on the drop is :

F = 6πηrv

= 6 x 3.14 x 1.8 × 10^{-5}x2 x 10^{-5} x 5.8 10^{-2}

= 3.91 x 10^{-10} N

**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}].

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10^{–2} N m^{–1}. Take the angle of contact to be zero and density of water to be 1.0 × 10^{3} kg m^{–3} (g = 9.8 m s^{–2}).

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}

A plane is in level flight at constant speed and each of its wings has an area of \( 25 m^2\). If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be \( 1 kg/m^3\)), g = \( 9.8 m/s^2\)

Answe the following question:-

**(a) **What is the largest average velocity of blood flow in an artery of radius 2×10^{–3}m if the flow must remain lanimar?

**(b)** What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10^{–3} Pa s).