Question

N identical liquid drops are falling with terminal velocity V0.if they combine to form a single drop then new terminal velocity of the big drop will be

Answer 1

Answer:

The terminal velocity of an object is directly proportional to the square of the radius of the drop The volume of the small drop is 4/3 πr^3 The volume of the bigger drop formed by collection of N drops is N(4/3πR^3) Thus the radius of the bigger drop is R=N^1/3×r (v2/v1)=(R/r)^2 (v2/v1)=[(r.N^1/3)/r]^2 v2/v1=N^2/3 4v=N^2/3.v N=8 , please mark my answer brainliest

Answer 2

$\therefore v_o=\sqrt{(2N.m.g)\div(\rho.A'.C_D)}$ is the new terminal velocity with A' as new area normal to the velocity.

Explanation:

Any body that falls through a fluid medium experiences a drag force in the opposite direction which depends upon the mass of the object and the area of the object normal to the velocity.

The drag force is approximately equal to the square of the velocity of the falling object. This force goes on increasing until it balances the (weight) force of gravity on the falling body and the body starts to fall with a constant velocity called terminal velocity.

Drag force on the drop can be given by the equation:

$F_D=\frac{1}{2} \rho.A.v^2.C_D$

where

$\rho=$ density of (air) the fluid medium in which the drop is falling

$A=$ area of the falling object subjected normal to the velocity

$v=$ velocity of the falling object

$C_D=$ coefficient of drag

In case of terminal velocity, drag force becomes equal to the weight of the body:

$m.g=\frac{1}{2} \rho.A.v_o^2.C_D$ ................................(1)

where:

$v_o=$ terminal velocity

when N drops combine then the eq. (1) become:

$N\times m.g=\frac{1}{2} \rho.A'.v_o^2.C_D$

here

$A'=$ new normal area subjected to the velocity after the combination of N drops resulting in N.m.g weight.

$\therefore v_o=\sqrt{(2N.m.g)\div(\rho.A'.C_D)}$

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TOPIC: terminal velocity

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