Answer:
Time taken by the
diameter droplet is 60 ns
Solution:
As per the question:
Diameter of the droplet, d = 1 mm = 0.001 m
Radius of the droplet, R = 0.0005 m
Time taken for complete evaporation, t = 1 min = 60 s
Diameter of the smaller droplet, d' =
![1* 10^(- 6) m](https://img.qammunity.org/2020/formulas/engineering/college/5suke4yp6pl183glguaugzyj68lxnj5voy.png)
Diameter of the smaller droplet, R' =
![0.5* 10^(- 6) m](https://img.qammunity.org/2020/formulas/engineering/college/rod1fcjaxhgtnvoz3ivn6s0yft9sjbbekm.png)
Now,
Volume of the droplet, V =
![(4)/(3)\pi R^(3)](https://img.qammunity.org/2020/formulas/engineering/college/os3dkvx9hdfo404s4wa7gebvvq3telzzcb.png)
Volume of the smaller droplet, V' =
![(4)/(3)\pi R'^(3)](https://img.qammunity.org/2020/formulas/engineering/college/uj5ybhud7n4hv12dymf3hxufhwv9sbqijc.png)
Volume of the droplet ∝ Time taken for complete evaporation
Thus
![(V)/(V') = (t)/(t')](https://img.qammunity.org/2020/formulas/engineering/college/an83ytiqqnwj4njs48lb9eqfc6pbn7l69v.png)
where
t' = taken taken by smaller droplet
![((4)/(3)\pi R^(3))/((4)/(3)\pi R'^(3)) = (60)/(t')](https://img.qammunity.org/2020/formulas/engineering/college/dd69a6n1pvza69gtvy4n4j3hira308b7zd.png)
![((4)/(3)\pi 0.0005^(3))/((4)/(3)\pi (0.5* 10^(- 6))^(3)) = (60)/(t')](https://img.qammunity.org/2020/formulas/engineering/college/nknh61skssoaoj2cbk4rwg2h8ymg1usll8.png)
t' =
![60* 10^(- 9) s = 60 ns](https://img.qammunity.org/2020/formulas/engineering/college/zo80q84mt52rckp3iwdl3edigtyt26x32z.png)