A small spherical ball of steel falls through a viscous medium with terminal velocity v. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

Question

A small spherical ball of steel falls through a viscous medium with terminal velocity v. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy) (A) (v/2) (B) (v/√2) (C) v (D) 2 v

Tardigrade Admin · Accepted Answer

Given,v =(2 r2 ρ g/9 η) ...(i) Mass =(4/3) π r3 ρ=(4/3) π(2 r)3 ρ1 or ρ1 =ρ / 8 Terminal velocity of second ball is v1=(2(2 r)2(ρ / 8) g/8 η)=(v/2)

Q. A small spherical ball of steel falls through a viscous medium with terminal velocity $v$ . If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

$\frac{v}{2}$

$\frac{v}{2}$

v

2 v

Given, $v = \frac{2 r ^{2} ρ g}{9 η} ... (i)$
Mass $= \frac{4}{3} π r^{3} ρ = \frac{4}{3} π (2 r)^{3} ρ_{1}$
or $ρ_{1} = ρ /8$
Terminal velocity of second ball is
$v_{1} = \frac{2 ( 2 r ) ^{2} ( ρ /8 ) g}{8 η} = \frac{v}{2}$

Q. A small spherical ball of steel falls through a viscous medium with terminal velocity v. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

2v​

2​v​

v

2 v

Given,v=9η2r2ρg​...(i) Mass =34​πr3ρ=34​π(2r)3ρ1​ or ρ1​=ρ/8 Terminal velocity of second ball is v1​=8η2(2r)2(ρ/8)g​=2v​

Q. A small spherical ball of steel falls through a viscous medium with terminal velocity $v$ . If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminal velocity (neglect buoyancy)

$\frac{v}{2}$

$\frac{v}{2}$

Given, $v = \frac{2 r ^{2} ρ g}{9 η} ... (i)$
Mass $= \frac{4}{3} π r^{3} ρ = \frac{4}{3} π (2 r)^{3} ρ_{1}$
or $ρ_{1} = ρ /8$
Terminal velocity of second ball is
$v_{1} = \frac{2 ( 2 r ) ^{2} ( ρ /8 ) g}{8 η} = \frac{v}{2}$