Question

दो गोल कार बूंदों की त्रिज्या ओं का अनुपात 1:3 है उनके सीमांग विवो का अनुवाद क्या होगा​

Answer 1

Given:

दो गोल कार बूंदों की त्रिज्या ओं का अनुपात 1:3 है

To find:

उनके सीमांग विवो का अनुवाद क्या होगा​

Solution:

Let,

"v" → the terminal velocity of the first spherical raindrop

"r" → the radius of the first spherical raindrop

"V" → the terminal velocity of the second spherical raindrop.

"R" → the radius of the second spherical raindrop

Since the ratio of the radii of the two raindrops is 1:3, then, let,

r be "x"

and

R be "3x"

We know the formula of terminal velocity is:

$\boxed{\bold{v = \frac{2}{9} \times \frac{r^2\:(\rho \:-\:\sigma )g}{\eta} }}$

where

ρ = mass density of the particles

σ = mass density of the fluid

η = dynamic viscosity

g = gravitational acceleration

By using the above formula, we get

The terminal velocity of the first raindrop will be,

$v = \frac{2}{9} \times \frac{r^2\:(\rho \:-\:\sigma )g}{\eta}=\frac{2}{9} \times \frac{x^2\:(\rho \:-\:\sigma )g}{\eta}$ . . . .  Equation 1

The terminal velocity of the second raindrop will be,

$V = \frac{2}{9} \times \frac{R^2\:(\rho \:-\:\sigma )g}{\eta}= \frac{2}{9} \times \frac{(3x)^2\:(\rho \:-\:\sigma )g}{\eta}$ . . . . Equation 2

Now, from equation 1 and equation 2, we get

The ratio of their terminal velocity will be,

$\frac{v}{V} = \frac{ \frac{2}{9} \times \frac{x^2\:(\rho \:-\:\sigma )g}{\eta} }{\frac{2}{9} \times \frac{(3x)^2\:(\rho \:-\:\sigma )g}{\eta} }$

on cancelling all the similar terms, we get

$\implies \frac{v}{V} = \frac{x^2}{(3x)^2 }$

$\implies \frac{v}{V} = \frac{x^2}{9x^2 }$

$\implies \bold{\frac{v}{V} = \frac{1}{9 }}$  

Thus, the ratio of the terminal velocities of the two raindrops is → 1 : 9.

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