Question

Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be(a) 10 cm per sec(b) 2.5 cm per sec(c) $5\times (4)^{\frac{1}{3}}$ cm per sec(d) 5 × √3 cm per sec

Answer 1

The ans will be - 5×(4) to the power 1/3m/s

If two drops of same radius r coalesce then the radius of new drop will be given by R

R = 4/3πR³ =4/3πr³+4/3πr³

R³ =2r³ = R=2 1/3r  

Thus, if a drop of radius r is falling in viscous medium then it will acquire a critical velocity v and v∝r² v2/v1

=(R/r)²=(21/3r/r)²

=  v2=2 to the power 2/3×v1 =2 to the power 2/3×(5)

=5×(4) to the power 1/3m/s