Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced what will be the new velocity?

Question

Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced what will be the new velocity? (A)  (2)(1/3)v (B) (2)(3/2)v  (C) (2)(2/3)v  (D) (2)(1/4)v

Tardigrade Admin · Accepted Answer

Let r be the radius of the each drop. The terminal velocity of drop will be given by

v = \frac{2}{9} \frac{r ^{2} ( ρ - σ ) g}{η}

....(i)
where

ρ

is density of drop and a is density of viscous medium of coefficient of viscosity

η

.
When two drops each of radius

r

coalesce to form a new drop, then the radius of coalesced drop will be

R = (2)^{1/3} r

Hence, new terminal velocity of coalesced drop will be

v^{'} = \frac{2}{9} [\frac{( 2 ^{1/3} r ) ^{2} ( ρ - σ ) g}{η}]

....(ii)
From Eqs. (i) and (ii), we get

\frac{v ^{'}}{v} = (2)^{2/3}

or

v^{'} = (2)^{2/3} v

Q. Two equal drops of water are falling through air with a steady velocity $v$ . If the drops coalesced what will be the new velocity?

$(2)^{\frac{1}{3}} v$

$(2)^{\frac{3}{2}} v$

$(2)^{\frac{2}{3}} v$

$(2)^{\frac{1}{4}} v$

Q. Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced what will be the new velocity?

(2)31​v

(2)23​v

(2)32​v

(2)41​v

Q. Two equal drops of water are falling through air with a steady velocity $v$ . If the drops coalesced what will be the new velocity?

$(2)^{\frac{1}{3}} v$

$(2)^{\frac{3}{2}} v$

$(2)^{\frac{2}{3}} v$

$(2)^{\frac{1}{4}} v$