Question

A metal ball of radius $r$ and density $d$ travels with a terminal velocity $v$ in a liquid of density $\dfrac{d}{4}$ . the terminal velocity of another ball of radius $2r$ and density $3d$ in the same liquid is
(A) $\dfrac{44v}{3}$
(B) $\dfrac{22v}{3}$
(C) $\dfrac{11v}{3}$
(D) $\dfrac{3v}{44}$

Answer 1

Hint : Use the equation to find the terminal velocity of a sphere in a fluid and find the ratio of the terminal velocity equations of both the metal balls and find an equation for the terminal velocity of the second ball in terms of the terminal velocity of the first ball.

Complete step by step answer
Let the terminal velocity of the first ball be $v$ and that of the second ball be $v'$ . Let the acceleration due to gravity be denoted by the variable $g$ . Let the radius of the first ball be $r$ and therefore, the radius of the second ball becomes $2r$ . Let the density of the first ball be $d$ and therefore, the density of the second ball becomes $3d$ as per the question. Let ${{d}_{l}}$ be the density of the liquid.
Terminal velocity of the first ball is given by
$\Rightarrow v=\dfrac{2g{{r}^{2}}}{9\eta }\left( d-{{d}_{l}} \right)$
Terminal velocity of the second ball is given by
$\Rightarrow {{v}^{'}}=\dfrac{2g{{\left( 2r \right)}^{2}}}{9\eta }\left( 3d-{{d}_{l}} \right)$
Therefore, taking ratios of both the terminal velocity and substituting the given value of the density of liquid gives us
$\Rightarrow \dfrac{v'}{v}=\dfrac{\left( \dfrac{2g\times 4{{r}^{2}}}{9\eta } \right)\left( 3d-\dfrac{d}{4} \right)}{\left( \dfrac{2g{{r}^{2}}}{9\eta } \right)\left( d-\dfrac{d}{4} \right)}$
Canceling all common terms and simplifying gives us
$\Rightarrow \dfrac{v'}{v}=4\times \dfrac{11}{4}\times \dfrac{4}{3}$
Therefore we get an equation for the terminal velocity of the second metal ball as
$\Rightarrow v'=\dfrac{44v}{3}$
Hence, the terminal velocity of the second metal ball in terms of the terminal velocity of the first ball is given by $\dfrac{44v}{3}$ .

Note
Terminal velocity is achieved by an object in a fluid, when the forces acting on the object balances, that is, the weight of the object will balance the force due to buoyancy and the force due to the viscosity of the liquid. Viscous force occurs due to the resistance offered by the fluid when the object is moving. The buoyant force acts on the body due to the pressure difference with an increase in depth of the liquid.