Question

A small spherical ball of radius R Falls with viscosity V through a liquid having coefficient of viscosity. Make a suitable equation to find the relation among these.​

Answer 1

Answer:

We can write the formula of the viscous fore V =6πηrv.

Where η is coefficient of viscosity of liquid.

As, the spherical ball is falling with viscosity V hence will have a constant velocity while it is falling.

This viscous force is mainly downwards and is always counteracted by an opposite force or the buoyant force that is against the weight of the body in downward direction.

Answer 2

Answer:

Let, F  ∝ (R)a

      F  ∝ (V)b

      F  ∝ (n)c

Or,  F ∝ (R)a(V)b(n)c  

      F= k(R)a(V)b(n)c………………………………..(1)

Here, K is constant of proportionality.

Substitute the dimensional formula of each quantity in eq (1)

[M1L1T-2] = [M0 L1T0]a [M0L1T-1]b [M1L-1T-1]c

Or, [M1L1T-2]= [Mc(L)a+b-c(T)-b-c]

Equating the power of M,L,T on both sides,

C=1

a+ b – c=1

-b –c= -2

On solving, a=1, b=1, c=1

Put these values in eq (1)

F= krvn