A small metal ball of mass m is dropped in a liquid contained in a vessel , attains a terminal velocity v . If a metal ball of same material but of mass 8 m is droped is same liquid then the terminal velocity will be

Options

A. V
B. 2V
C. 4 V
D. 8 V

Answer

C

Explanation

Thermal velocity
`v = (2 )/(g) (r^(2) (rho - rho_(0)) g)/(eta)`
In both cases all the variables will remain same except radius , density remains same as material is same
`m_(1) = m` or `m_(2) = 8`m
Moreover m = volume `xx` density = `(4)/(3) pi r^(3) xx d`
`therefore " " m_(1) = (4)/(3) pi r_(1)^(3) d implies m_(2) = (4)/(3) pi r_(2)^(3)` d
`(m_(1))/(m_(2)) = (m)/(8m) = ((4)/(3) pi r_(1)^(3) d)/((4)/(3) pi r_(2)^(3) d)) implies (1)/(8) = (r_(1)^(3))/(r_(2)^(3))`
`(r_(1))/(r_(2)) = (1)/(r) implies r_(2) = 2r_(1)`
`(v_(1))/(v_(2)) = (r_(1)^(2))/(r_(2)^(2))`
`implies (v)/(v_(2)) = ((r_(1))/(2r_(1)))^(2) or v_(2) = 4v`