A small spherical ball falling through a viscous medium of negligible density has terminal velocity v. Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity

Question

A small spherical ball falling through a viscous medium of negligible density has terminal velocity v. Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity (A) v (B) v / 4 (C) v / 2 (D) 2 v

Tardigrade Admin · Accepted Answer

If ρ is the density of the ball and ρ prime that of the another ball, m for the balls are the same, but r prime=2 r ∴ m g=6 π r η v (by Stoke's law) or, 6 π r η v=6 π 2 m v prime So, v prime=(v/2)

Q. A small spherical ball falling through a viscous medium of negligible density has terminal velocity $v$ . Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity

$v$

$v /4$

$v /2$

$2 v$

If $ρ$ is the density of the ball and $ρ^{'}$ that of the another ball, $m$ for the balls are the same, but $r^{'} = 2 r$
$∴ m g = 6 π rη v$ (by Stoke's law)
or, $6 π rη v = 6 π 2 m v^{'}$ So, $v^{'} = \frac{v}{2}$

Q. A small spherical ball falling through a viscous medium of negligible density has terminal velocity v. Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity

v

v/4

v/2

2v

If ρ is the density of the ball and ρ′ that of the another ball, m for the balls are the same, but r′=2r ∴mg=6πrηv (by Stoke's law) or, 6πrηv=6π2mv′ So, v′=2v​

Q. A small spherical ball falling through a viscous medium of negligible density has terminal velocity $v$ . Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity

$v$

$v /4$

$v /2$

$2 v$

If $ρ$ is the density of the ball and $ρ^{'}$ that of the another ball, $m$ for the balls are the same, but $r^{'} = 2 r$
$∴ m g = 6 π rη v$ (by Stoke's law)
or, $6 π rη v = 6 π 2 m v^{'}$ So, $v^{'} = \frac{v}{2}$