Question 23: Properties of Matter

Question

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$ while falling under gravity through an oil of density $925 \mathrm{~kg} \mathrm{~m}^{-3}$. Take density of steel as $7825 \mathrm{~kg} \mathrm{~m}^{-3}$ and g as $9.8 \mathrm{~m} / \mathrm{s}^2$. The viscosity of the oil in SI unit is

Accepted Answer

To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid:
$ v_T = \frac{2}{9} \frac{(ho_s - ho_f) r^2 g}{\eta} $
Where:
$ v_T $ is the terminal velocity,
$ ho_s $ is the density of the steel ball,
$ ho_f $ is the density of the fluid,
$ r $ is the radius of the ball,
$ g $ is the acceleration due to gravity,
$ \eta $ is the viscosity of the fluid.
Given:
Diameter of the steel ball = 3.6 mm, therefore radius $ r = 1.8 $ mm = $ 1.8 	imes 10^{-3} $ m,
Terminal velocity $ v_T = 2.45 	imes 10^{-2} $ m/s,
Density of oil $ ho_f = 925 $ kg/m$^3$,
Density of steel $ ho_s = 7825 $ kg/m$^3$,
Acceleration due to gravity $ g = 9.8 $ m/s$^2$.
Rearranging the formula to solve for $\eta$:
$ \eta = \frac{2}{9} \cdot \frac{(ho_s - ho_f) \cdot r^2 \cdot g}{v_T} $
Substituting the known values into the equation:
$ \eta = \frac{2}{9} \cdot \frac{(7825 - 925) \cdot (1.8 	imes 10^{-3})^2 \cdot 9.8}{2.45 	imes 10^{-2}} $
Calculating this, we approximate the viscosity $\eta$ to be:
$ \eta \approx 1.99 \, 	ext{Pa}\cdot	ext{s} $
Thus, the viscosity of the oil is approximately 1.99 PaÃÂ·s.

Answer

2.18

Answer

1.68

Answer

2.38

Answer

1.99

Properties of Matter

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