Question

Water passes through a viscometer in 30 s. The same
volume of oil required 2263.7 s. If the density of oil & water
are 1100 kg/m3 & 998 kg/m3, calculate the viscosity of oil, if
the viscosity of water = 0.00101 kg/(m-s) (Kerala, 2011)​

Answer 1

Given:

$t_w$ = 30 s

$t_o$ = 2263.7 s

$d_w$ = 998 kg/m3

$d_o$ = 1100 kg/m3

$V_w$ = 0.00101 kg/(m-s)

To find:

the viscosity of oil $V_o$ =?

Explanation:

Viscosity is a count of a fluid's resistance to flow.

Viscosity is directly proportional to the density of the fluid.

i.e.  V ∝ d

Viscosity is inversely proportional to the runoff timing of fluid.

i.e. V ∝$\frac{1}{t}$

∴ The ratio of the viscosity of water to the viscosity of oil is given by

$\displaystyle \frac{V_w}{V_o} = \frac{d_w \times t_o}{d_o \times t_w}$

Put given values in above eq

$\displaystyle \frac{0.00101}{V_o} = \frac{998\times2263.7}{1100\times30}$

Solving above eq we get

$\displaystylef{ V_o}$ = 1.4753 $\times 10^{-5$

$\displaystyle\mathbf{ V_o}$ = 0.000014  kg/(m-s)