A spherical solid ball of volume V is made of a material of density ρ1 It is falling through a liquid of density ρ2(ρ2

Question

A spherical solid ball of volume V is made of a material of density ρ1 It is falling through a liquid of density ρ2(ρ2 < ρ1). Assume that the liquid applies a viscous force on the ball that is F = krv where r is the radius of the ball and v is the speed and k is the constant of proportionality. The terminal speed of the ball is (A) (4π r2 g(ρ1 -ρ2)/3k) (B) (π r2 g(ρ1 -ρ2)/3k) (C) (π r2 g(ρ1 -ρ2)/k) (D) (π r2 g(ρ1 -ρ2)/4k)

Tardigrade Admin · Accepted Answer

When the falling ball attains the terminal velocity

υ_{T}

, then the acceleration of the ball is zero. In this situation

m g = F_{B} + F_{V}

V ρ_{1} g = V ρ_{2} g + k r υ_{T}

υ_{T} = \frac{V ( ρ _{1} - ρ _{2} ) g}{k r}

υ_{T} = \frac{\frac{4}{3} π r ^{3} ( ρ _{1} - ρ _{2} ) g}{k r}

= \frac{4 π ^{2} ( ρ _{1} - ρ _{2} ) g}{3 k}

$\frac{4 π r ^{2} g ( ρ _{1} - ρ _{2} )}{3 k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{3 k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{4 k}$

3k4πr2g(ρ1​−ρ2​)​

3kπr2g(ρ1​−ρ2​)​

kπr2g(ρ1​−ρ2​)​

4kπr2g(ρ1​−ρ2​)​

When the falling ball attains the terminal velocity υT​, then the acceleration of the ball is zero. In this situation mg=FB​+FV​ Vρ1​g=Vρ2​g+krυT​ υT​=krV(ρ1​−ρ2​)g​ υT​=kr34​πr3(ρ1​−ρ2​)g​ =3k4π2(ρ1​−ρ2​)g​

$\frac{4 π r ^{2} g ( ρ _{1} - ρ _{2} )}{3 k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{3 k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{k}$

$\frac{π r ^{2} g ( ρ _{1} - ρ _{2} )}{4 k}$