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\times 10^{- 6} m](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-%206%7D%20m)
Diameter of the smaller droplet, R' = ![0.5\times 10^{- 6} m](https://tex.z-dn.net/?f=0.5%5Ctimes%2010%5E%7B-%206%7D%20m)
Now,
Volume of the droplet, V = ![\frac{4}{3}\pi R^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Cpi%20R%5E%7B3%7D)
Volume of the smaller droplet, V' = ![\frac{4}{3}\pi R'^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Cpi%20R%27%5E%7B3%7D)
Volume of the droplet ∝ Time taken for complete evaporation
Thus
![\frac{V}{V'} = \frac{t}{t'}](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7BV%27%7D%20%3D%20%5Cfrac%7Bt%7D%7Bt%27%7D)
where
t' = taken taken by smaller droplet
![\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20R%5E%7B3%7D%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20R%27%5E%7B3%7D%7D%20%3D%20%5Cfrac%7B60%7D%7Bt%27%7D)
![\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%200.0005%5E%7B3%7D%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%280.5%5Ctimes%2010%5E%7B-%206%7D%29%5E%7B3%7D%7D%20%3D%20%5Cfrac%7B60%7D%7Bt%27%7D)
t' = ![60\times 10^{- 9} s = 60 ns](https://tex.z-dn.net/?f=60%5Ctimes%2010%5E%7B-%209%7D%20s%20%3D%2060%20ns)