Question

Two equal drops are falling through air with a steady velocity of $5cm{s^{ - 1}}$. If the drop coalesces the new terminal velocity will become:
$\left( A \right)5 \times 2cm{s^{ - 1}}$
$\left( B \right)5 \times \sqrt 2 cm{s^{ - 1}}$
$\left( C \right)5 \times {4^{\dfrac{1}{3}}}cm{s^{ - 1}}$
$\left( D \right)\dfrac{5}{{\sqrt 2 }}cm{s^{ - 1}}$

Answer 1

Velocity of the two equal drops is given to use. When these two drops combine to form a bigger drop then the volume of the two drops is conserved in the form of a bigger drop. With the help of this we can find the relation between the better drop radius and the smaller drop radius. Now we know that terminal velocity is directly proportional to square of radius with this we can find the new terminal velocity of the bigger drop.

Complete Step By Step Answer:
As per the given problem two equal drops are falling through air with a steady velocity of $5cm{s^{ - 1}}$.
Here we need to calculate the new terminal velocity of the combined drop or we can say bigger drop.
Let say two drop as drop 1 and drop 2 respectively,
Bigger drop = drop 1 + drop 2

Let ${v_1},{v_2}$ be the which is given in the question and equals to $5cm{s^{ - 1}}$ .
Hence we can say,
${v_1} = {v_2} = v = 5cm{s^{ - 1}}$
As given the two small drops are equals hence, let r be the radius of each of the drops.
Now we know the formula for the terminal velocity as,
$V = \dfrac{{2{R^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g$
Except r all values will remain constant as the bigger drop and the small drop is made from the same matter.
Hence we can write,
$V\alpha {R^2}$
For small drop,
$v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g = 5cm{s^{ - 1}}$
When two drops combine the bigger drops form let the radius of the bigger drop be R.
Since the bigger drop volume will be the same as the sum of two equal drops, the volume remains constant.
Now we can write,
Volume of the bigger drop = volume of drop 1 + volume of drop 1
As both the drops are equal to each other we can write the above statement as,
Volume of the bigger drop = 2 (volume of small drop)
As the drop are in the shape of sphere
Volume of the sphere $= \dfrac{4}{3}\pi {r^3}$
Now according to the problem statement we can write,
$\dfrac{4}{3}\pi {R^3} = 2 \times \dfrac{4}{3}\pi {r^3}$
Cancelling the common terms we get,
${R^3} = 2{r^3}$
$\Rightarrow R = {2^{\dfrac{1}{3}}}r$
Hence terminal velocity of the bigger drop will be,
$V\alpha {R^2}$
$\Rightarrow V = \dfrac{{2{R^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g$
In place of R write ${2^{\dfrac{1}{3}}}r$ ,we get
$V = \dfrac{{2 \times {{\left( {{2^{\dfrac{1}{3}}}r} \right)}^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g$
$\Rightarrow V = \dfrac{{2{r^2} \times \left( {{2^{\dfrac{2}{3}}}} \right)\left( {\rho - \sigma } \right)}}{{9\eta }}g$
$\Rightarrow V = \left( {{2^{\dfrac{2}{3}}}} \right) \times \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g$
We know that $v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}g$ we get
$\Rightarrow V = \left( {{2^{\dfrac{2}{3}}}} \right) \times v$
Putting $v = 5cm{s^{ - 1}}$ we get,
$V = \left( {{2^{\dfrac{2}{3}}}} \right) \times 5cm{s^{ - 1}}$
Further simplifying we get,
$V = 5cm{s^{ - 1}} \times {\left( 4 \right)^{\dfrac{1}{3}}}$
Therefore the correct option is $\left( C \right)$ .

Note:
No need to convert the velocity into meters per second term as our option is given in centimeters per second. Keep in mind, except for radius all the other terms remain constant and same for both bigger and smaller drops because all the drops are the same matter either water or any other matter but all are of the same kind.