Question

show that if 'n ' equal raindrops falling through
air with equal steady velocity 10 cm/s
combine. The resultant drop attends a new turminal velocity of 10n ^2/3 cm/sec
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Answer 1

Given:

'n ' equal raindrops falling through

air with equal steady velocity 10 cm/s

combine

To Show:

The resultant drop attends a new terminal velocity of 10n ^2/3 cm/sec

​

Solution:

Total volume of small droplets = Volume of larger drop

• n x $\frac{4}{3}$ πr³ = $\frac{4}{3}$ πR³
• R = r ∛n .

Terminal velocity of droplet is given by ,

• V = $\frac{2r^{2}(\rho - \sigma) }{9g}$  

Now terminal velocity of smaller drop = >

• v = $\frac{2r^{2}(\rho - \sigma) }{9g}$  

Terminal velocity of larger drop = >

• V = $\frac{2R^{2}(\rho - \sigma) }{9g}$

Dividing both the equations,

• v/V = r²/R²
• V = vR²/r² = v $n^{2/3}$ = 10$n^{2/3}$ cm/s  

Thus showed that resultant drop attends a new terminal velocity of 10$n^{2/3}$ cm/sec.