Two drops of the same radius are falling through air with a steady velcoity of `5 cm s^(-1).` If the two drops coalesce, the terminal velocity would be

Options

A. 10 cm per sec
B. `2.5` cm per sec
C. `5 xx(4) ^(1//3)` cm per sec
D. `5 xx sqrt3 ` cm per sec

Answer

C

Explanation

If R is radius of bigger drop formed, then
` 4/3 pi R ^(3) = 2 xx 4/3 pi r ^(3) or R = 2 ^(1//3)r`
As `v _(0) prop r ^(2) therefore (v _(01))/(v _(0)) = (R ^(2))/( r ^(2))= ((2 ^(1//3)r ) ^(2))/( r ^(2)) = 2^(2//3)`
or `v _(01) = v_(0) xx 3 ^(2//3) = 5 xx (4) ^(1//3)`