Question

Two equal drops of water falling through air with a steady velocity 5 cm//s. If the drops combire to from a single drop, what will be new terminal velocity?

Answer 1

here is ur answer

When the two equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say r is radius of drop before these combine and R is radius after these combine. 

We have,

2×34πr3=34πR3⇒R=21/3r

and we know that terminal velocity VT ∝ radius2, so we have

VTVT′=r2R2

⇒VT′=(rR)2VT 

=(21/3)2v

=22/3v m/s

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Answer 2

Given:

The steady velocity of the two drops of water = 5 cm/s

To find:

If the two drops combine to form a single drop then what will be the new terminal velocity.

Solution:

When the two drops would combine their volume would remain constant.

Let the radius of the two drops be r

And the radius of the final drop formed be R.

Since the volume is conserved:

2*4/3* πr³ = 4/3* πR³

R = ∛2 r

Also terminal velocity ∝ radius²

Let the velocity of the final drop = V₁

And velocity of the two small drop = V₂

Therefore, V₁/ V₂ = R²/ r²

⇒ V₁ = R² V₂/ r²

Putting the values we get:

V₁ =  (∛2 r)² (5 cm/s)/ r²

⇒V₁ = 2^2/3 * 5 cm/s

⇒V₁ = ∛4 * 5 cm/s = 1.587*5 cm/s = 7.937 cm/s ≈ 7.94 cm/s

Therefore, the new terminal velocity is 7.94 cm/s.