Class 12 Maharashtra State Board

With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m² and specific gravity 0.9? Density of air is 1.29 kg/m³ .

As per the question we can put some of the given data as,

diameter D= 0.4mm

radius r = 0.2mm = 0.0002 m

coefficient of viscosity η = 0.1 Ns/m²

density of liquid d = 0.9 x 103 kg/m3 = 900kg/m³

density of air ρ= 1.29 kg/m³

To find - terminal velocity v = ?

Solution -

The terminal velocity can be given as,

$https://latex.codecogs.com/svg.image?\mathrm{v = \frac{2r^{2}g(\rho -d)}{9\eta }}$

$https://latex.codecogs.com/svg.image?\mathrm{v = \frac{2 x 0.0002^{2} x 9.8(1.29-900)}{9x 0.1}} = \mathrm{\frac{-19.6x 0.00000004 x 898.71 }{0.9}}$

$https://latex.codecogs.com/svg.image?\mathrm{v=\mathrm{\frac{-196x4 x 10^{-8} x 898.71 }{9}} = \mathrm{\frac{-784 x 898.71 x 10^{-8}}{9}}$

$https://latex.codecogs.com/svg.image?\mathrm{v= - 87.11 x 898.71 x 10^{-8}}$

v = -Al ( log 87.11 + log 898.71 ) x 10^-8

v = -Al( 1.9400 + 2.9536 ) x 10^-8

v = -Al(4.8936) x 10^-8

v = - 78270 x 10^-8

$https://latex.codecogs.com/svg.image?\mathrm{v= - 7.287 x 10^{-4} m/s}$

The mass of the earth is 6 x 10²⁴ kg. The distance between the earth and the Sun is 1.5x 10¹¹ m. If the gravitational force between the two is 3.5 x 10²² N, what is the mass of the Sun?