Question

if temperature of liquid is increased how the terminal velocity will affect​

Answer 1

Answer:

The density is a function of the air pressure and temperature, which are themselves functions of altitude. So, the terminal velocity decreases slightly as the object falls to earth; it is not truly a constant value.

Answer 2

Answer:

Yes. The terminal velocity of a small , smooth spherical body falling in vast viscous medium is given by

vt=(2/9)(r^2g/n)(d-do)……………….(1)

Here, r is radius of falling body.

n is coefficient of viscosity of the viscous medium.

d is density of the falling body.

do is density of viscous material.

The coefficient of viscosity depends on the type of the fluid and temperature of the fluid.

For liquids the coefficient of viscosity decreases with increase in temperature ,but , for gases the coefficient of viscosity increases with increase in temperature .As an example, coefficient of viscosity of engine oil at 16 degree Celsius is 1.13X10^-1N s m^-2 and at 38 degree Celsius it is 3.4X10^-2 Nsm^-2.

For air at 0 degree Celsius the coefficient of viscosity is 1.71X10^-5Nsm^-2 and at 40 degree Celsius it is 1.9X10^-5 Nsm^-2.

Also, with increase in temperature the density of fluid and that of falling object decreases.

For liquids, nt=no/(1+at+bt^2), a and b are constant. This is called Slotte's formula.

There are more number of formulae.

n=Ae^(C/T) is Andrade relation.

nV^(1/3)=Ae^(c/VT): this is more accurate relation. V is specific volume.

For gases there is Sutherland formula: n=no[(k sqrt of T)/(1+C/T

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