The terminal velocity of a copper ball of radius $2.0 \;mm$ falling through a tank of oll at $20\,^{\circ} C$ is $6.5 \;cm s ^{-1} .$ Compute the viscosity of the oil at $20\,^{\circ} C .$ Density of oil is $1.5 \times 10^{3} \;kg m ^{-3},$ density of copper is $8.9 \times 10^{3} \;kg m ^{-3}$
$1.1 \times 10^{-1} kg m ^{-1} s ^{-1}$
$9.9 \times 10^{-1} kg m ^{-1} s ^{-1}$
$6.37 \times 10^{-2} kg m ^{-1} s ^{-1}$
$5.98 \times 10^{-1} kg m ^{-3} s ^{-1}$
Write the equation of terminal velocity.
If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly
Which of the following is the incorrect graph for a sphere falling in a viscous liquid? (Given at $t = 0$, velocity $v = 0$ and displacement $x = 0$.)
State stokes’ law. By using it deduce the expression for :
$(i)$ initial acceleration of smooth sphere and
$(ii)$ equation of terminal velocity of sphere falling freely through the viscous medium.
$(iii)$ Explain : Upward motion of bubbles produced in fluid.
A thin square plate of side $2\ m$ is moving at the interface of two very viscous liquids of viscosities ${\eta _1} = 1$ poise and ${\eta _2} = 4$ poise respectively as shown in the figure. Assume a linear velocity distribution in each fluid. The liquids are contained between two fixed plates. $h_1 + h_2 = 3\ m$ . A force $F$ is required to move the square plate with uniform velocity $10\ m/s$ horizontally then the value of minimum applied force will be ........ $N$