Question

A small metal ball of diameter 4mm and density $10.5 g/cm^3$ is dropped in glycerine of density $1.5 g/cm^3$. The ball attains a terminal velocity $8 cm/sec$. The coefficient of viscosity of Glycerine is :

$\begin{aligned} & A.4.9poise \\\ & B.9.8poise \\\ & C.98poise \\\ & D.980poise \\\ \end{aligned}$

Answer 1

Hint: Use the expression for the terminal velocity of a body dropped in viscous liquid to calculate the coefficient of viscosity of the glycerine by manipulating the LHS and RHS

Step by step solution:
We are given a ball of diameter 4mm and density $10.5 g/cm^3$ . The density of glycerine is given $1.5 g/cm^3$.
We know the terminal velocity of a ball is given by the expression:
$V_t = \dfrac{2}{9} \dfrac{r^2(\rho-\rho^I)g}{\eta}$
Where the density of liquid where the ball is dropped is denoted by $\rho ^I$ and the $\rho$ denotes the density of the ball dropped. $\eta$ is the viscous coefficient to be found
r is the radius of the ball and the g is the acceleration due to gravity.
We have r = 2mm and $V_t = 8 cm/sec$
Manipulating the LHS and RHS such that RHS is $\eta$ we get :
$\eta = \dfrac{2}{9} \dfrac{r^2(\rho-\rho^I)g}{V_t}$
Substituting the values we get as:
$\eta = \dfrac{2}{9} \dfrac{0.2^2(10.5-1.5)980}{8} = 9.8 poise$

Thus, we have used the expression for the terminal velocity of a body in a viscous liquid and found out the unknown constant it the viscosity of the liquid by manipulating the LHS and RHS of the equation and we got the viscosity of the liquid as 9.8 poise.

Note: One of the possible mistakes that we may make is that we confuse between the terms $rho$ and $\rho^I$ . We often confuse which one refers to the density of the viscous liquid and the body dropped in the liquid, So we need to be careful about that.