Eight equal drops of oil are falling through the air with a steady velocity of 0.1 m s⁻¹. The drops recombine to form a single drop, what should be the new terminal velocity. (0.4 m/s) 


Given:

Terminal velocity of the drops = vt = 0.1 m s⁻¹

To Find:

Terminal velocity of new big drops = vt' = ?

Solution:

Let the radius of each 8 drops is r and the combined new big drop be r'.  Then the terminal velocities of each small 8 drops and the new big drop are given below respectively:


vt = `\frac {2⍴gr^2}{9η}` ----------(1)

and

vt' = `\frac {2⍴g(r')^2}{9η}` ------------(2)

In order to solve this question we have to find the relation between r' and r


So the volume of each 8 spherical drops is

V = `\frac {4}{3}` 𝜋r³

and volume for the big (new) drop is

V' = `\frac {4}{3}` 𝜋 (r')³

As the new big drop is made of 8 drops so the volume of the big drop will be

V' = 8 V

or

`\frac {4}{3}` 𝜋 (r')³ = 8 `\frac {4}{3}` 𝜋r³

(r')³ = 8 r³

(r')³ = (2 r)³         [ 2³ = 8]

Multiplying 1/3 in exponent on both sides, we get

r' = 2 r 

Now replacing r' of the equation (2) with 2r 

vt' = `\frac {2⍴g(2r)^2}{9η}`

vt' = `\frac {2⍴g(4r^2)}{9η}`

or

vt' = 4 `\frac {2⍴gr^2}{9η}`

using equation (1)

vt' = 4 vt

putting value of vt = 0.1 m s⁻¹(given)

vt' = 4 ｘ0.1 m s⁻¹

vt' = 0.4 m s⁻¹ -----------Ans.

Thus, the new big drop will fall with a terminal velocity of 0.4 m s⁻¹



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