Question

Two rain drops reach the earth with their terminal velocities in the ratio 25:64.The ratio of their radii is​

Answer 1

Given:

The ratio of the terminal velocities of two rain drops reaching the earth is 25:64.

To find:

The ratio of their radii

Solution:

Let' s assume,

"v₁" → the terminal velocity of first rain drop

"v₂" → the terminal velocity of the second rain drop.

"r₁" → the radius of the first rain drop

"r₂" → the radius of the second rain drop

We have

The formula of terminal velocity as,

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

where

ρ = mass density of the particles

σ = mass density of the fluid

η = dynamic viscosity

g = gravitational acceleration

Therefore,

The terminal velocity of the first rain drop,

$v_1\:=\:\frac{2}{9} \times\:\frac{r_1^2(\rho\:-\:\sigma)g}{\eta}$ ......... (i)

The terminal velocity of the second rain drop,

$v_2\:=\:\frac{2}{9} \times\:\frac{r_2^2(\rho\:-\:\sigma)g}{\eta}$ ........... (ii)

The ratio of the terminal velocity is given as, $\frac{v_1}{v_2} = \frac{25}{64}$  ...... (iii)

Now, from (i), (ii) & (iii), we get

$\frac{v_1}{v_2} = \:\frac{\frac{2}{9} \times\:\frac{r_1^2(\rho\:-\:\sigma)g}{\eta}}{\:\frac{2}{9} \times\:\frac{r_2^2(\rho\:-\:\sigma)g}{\eta}}$

⇒ $\frac{25}{64} = \:\frac{{r_1^2}}{{r_2^2}{}}$

⇒ $\frac{r_1}{r_2} = \frac{\sqrt{25}}{\sqrt{64}}$

⇒ $\frac{r_1}{r_2} = \bold{\frac{5}{8}}$

Thus, the ratio of the radii of the two rain drops reaching the earth is 5:8.

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Answer 2

The ratio of their radii is​ 5 : 8

Explanation:

The terminal velocity is given by the formula:

v = 2/9 × (r² (ρ - σ) g)/(μ)

The terminal velocity are:

v₁ = 2/9 × (r₁² (ρ - σ) g)/(μ)

v₂ = 2/9 × (r₂² (ρ - σ) g)/(μ)

Now,

v₁/v₂ = (2/9 × (r₁² (ρ - σ) g)/(μ))/(2/9 × (r₂² (ρ - σ) g)/(μ))

v₁/v₂ = (r₁²)/(r₂²)

On substituting the value of terminal velocity, we get,

25/64 = (r₁²)/(r₂²)

On taking square root, we get,

5/8 = (r₁)/(r₂)

∴ r₁ : r₂ = 5 : 8