# Two hail stones whose radii are in the ratio of 1 : 2 fall from a heightof 50 km. Their terminal velocities are in the ratio of(a) 1 : 9 (b) 9 : 1(c) 4 : 1

(d) 1 : 4

1
by mansaridevabar 02.10.2015

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(d) 1 : 4

1
by mansaridevabar 02.10.2015

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As per Stokes drag formula, the terminal velocity is proportional to square of radius. so it is (d) 1 : 4

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__There is also another method for terminal velocity in a fluid. But it gives a different result.__

Here we can apply the equations of dynamics involving gravity force, drag force due to the air , and the buoyancy force due to air. We assume that the densities of both hail stones are same. The hail stones are falling from a large height. So we can assume that they reach some sort of terminal velocity. That is the velocity remains same. The gravitational force is equal to the drag force and buoyancy in opposite directions.

V = volume of stones = 4/3 * πR³

d = relative density of hail stone wrt water

d_a = density of air layer near the surface of Earth

*v = terminal velocity of a solid travelling in a viscous fluid*

μ = viscosity constant

Cd = drag coefficient

R = radius of the solid falling

A = surface area of object, that is projected on a plane = πR²

Fg = m g = gravity = d * 4/3π R³ * g

Fb = buoyancy = d_a * 4/3 π R³ * g

1)*Stokes law and stokes drag*

*Fd = drag force = 6 π μ R v * : this is for Viscous flow of a fluid around a solid spherical body.

So*Fd = Fb + Fg * => * v = 2/9 (d - d_w) g R² / μ*

2) A formula is based on the formulation that the drag in air in the atmosphere is

* Fd =* 1/2 * Cd * d_a * A * v² =* 1/2 Cd d_a πR² v²*

Then we get :** v² = (8π/3) R (d/d_a - 1) / Cd**

Here*in this case, v is proportional to the square root of Radius.*

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Here we can apply the equations of dynamics involving gravity force, drag force due to the air , and the buoyancy force due to air. We assume that the densities of both hail stones are same. The hail stones are falling from a large height. So we can assume that they reach some sort of terminal velocity. That is the velocity remains same. The gravitational force is equal to the drag force and buoyancy in opposite directions.

V = volume of stones = 4/3 * πR³

d = relative density of hail stone wrt water

d_a = density of air layer near the surface of Earth

μ = viscosity constant

Cd = drag coefficient

R = radius of the solid falling

A = surface area of object, that is projected on a plane = πR²

Fg = m g = gravity = d * 4/3π R³ * g

Fb = buoyancy = d_a * 4/3 π R³ * g

1)

So

2) A formula is based on the formulation that the drag in air in the atmosphere is

Then we get :

Here