Question

two exactly similer rain drops falling with terminal velocity of 2 1/3 m/s joined to form a bigger drop. Find the new terminal velocity of the bigger drop.

Answer 1

The terminal velocity of the bigger drop is 2m/s

Explanation:

Let r be the radius of small drops and R be the radius of Big drop

Then According to question small drops joined and from new big drop i.e

Then the volume of small drops = volume of big drop

[(4π/3)r³ + (4π/3)r³] = (4/3π)R³

2r³ = R³

R = (2)¹/³r

Terminal velocity is given by V = $\frac{2}{9}\frac{r^{2} }{n}[ Density of body - Density of Liquid]g$

i.e v∝r² then initial terminal velocity V₁ =  2¹/³ m/sec

Terminal Velocity final÷ Terminal velocity initial = R²/r² = [(2²/3)r²}÷r²

• V₂/V₁ = R²/r² = [(2²/3)r²}÷r²

• V₂ = 2²/3V₁

• V₂ = (2¹/3)ₓ(2¹/³)

• V₂ = 2m/sec final terminal velocity

Answer 2

The terminal velocity of the bigger drop is 2m/s

Explanation:

Let r be the radius of small drops and R be the radius of Big drop

Then According to question small drops joined and from new big drop i.e

Then the volume of small drops = volume of big drop

[(4π/3)r³ + (4π/3)r³] = (4/3π)R³

2r³ = R³

R = (2)¹/³r

Terminal velocity is given by V =  

i.e v∝r² then initial terminal velocity V₁ =  2¹/³ m/sec

Terminal Velocity final÷ Terminal velocity initial = R²/r² = [(2²/3)r²}÷r²

V₂/V₁ = R²/r² = [(2²/3)r²}÷r²

V₂ = 2²/3V₁

V₂ = (2¹/3)ₓ(2¹/³)

V₂ = 2m/sec final terminal velocity

ANSWER BY MAD210216