MHT-CET Physics Crash Course “Terminal Velocity” Class 12

“Terminal Velocity” MHT-CET Class 12 Physics Crash Course | Chapter Mechanical Properties of Fluids

In this session G.S. Khairnar discussed the theory and MCQ’s of “Terminal Velocity”. The session will be useful for all the aspirants of MHT-CET 2022. The session will be in Marathi and English.

Terminal Velocity:-

Dear students, we studied about the viscous force acting between the layers of streamline flow, viscous force acting on the spherical body falling freely in viscous fluid. Recalling it we can write as, viscous force is acting between layers of liquid in streamline flow is,

1. Directly proportional to area (A) of layer
2. Directly proportional to velocity gradient of liquid

Then from the above conditions we can write that,

Where ‘η’ is constant of proportionality known as coefficient of viscosity. From above equation we have,

Also by Stoke’s law, the formula for viscous force acting on the spherical body falling freely in viscous fluid is,

∴ F=6π∙ r∙ v ∙η

In this article we are going to discuss about the concept of terminal velocity.

Let’s derive the expression for terminal velocity……………!

Terminal velocity is the constant velocity attained by the body when it falls freely in viscous fluid.

Consider the spherical object of mass ‘m’ radius ‘R’ and density ‘ρ’ is falling freely in viscous fluid of density ‘σ’ and coefficient of viscosity ‘η’ as shown in fig. below.

When the spherical body is falling freely in viscous fluid, the forces acting on the object are,

1. Downward force: Gravitational force as the weight of body
2. Upward force: It is provided by viscous force acting on body due to different layers of liquid and buoyant force acting on the surface of object by fluid.

Therefore the upward force acting on the spherical object is given as,

Upward force =Viscous force + Buoyant force

∴ Upward force=6π R v η+weight of liquid that spills out of vessel

∴Upward force=6π R v η+m’ g (m’=mass of liquid)

But density of liquid can be give as,

Now the downward force is due to gravitation

∴ 𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒=weight of object

∴ 𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒=mg

∴ 𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒=ρVg

This is the formula for terminal velocity of object falling freely in viscous fluid.

Some important points to note…!

• Terminal velocity of object is independent of mass of object.
• Terminal velocity is directly proportional to square of radius of object.
• Terminal velocity is inversely proportional to inversely proportional to coefficient of viscosity of fluid.

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Terminal Velocity:-

Let’s solve following MCQ’s on terminal velocity…..!

Q.1) The correct equation for terminal velocity from the given below is….

Q.2) Two metal balls of same material having radii of 1.2 cm and 1.8 cm falling freely in viscous fluid. If the smaller ball has terminal velocity of 3.2 cm/s, terminal velocity of bigger ball is…

a) 1.8 cm/s

b) 3.6 cm/s

c) 6.4 cm/s

d) 7.2 cm/s

Answer

d) 7.2 cm/s

Q.3) Which among the following is not correct in terms of terminal velocity?

a) Terminal velocity of object is independent of mass of object.

b) Terminal velocity is directly proportional to square of radius of object.

c) Terminal velocity is independent of acceleration due to gravity.

d) Terminal velocity is inversely proportional to coefficient of viscosity of fluid.

Answer

c) Terminal velocity is independent of acceleration due to gravity.

Q.4) Mercury drop of radius 0.02 mm falling freely in water of having coefficient of viscosity 8.5 ×10^-5 N.s/m² (ρ =13600 kg/m³, σ =1000 kg/m³)…

a) 13.17 m/s

b) 13.17 cm/s

c) 13.17 mm/s

d) 13.17 μm/s

Answer

b) 13.17 cm/s

Q.5) Two metal balls of same material falling with same terminal velocity in mercury and in water whose coefficient of viscosity is in ratio 5:2. Ratio of radius of metal balls should be…

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