Class 12 Maharashtra State Board

Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is 1.8 × 10^-5 Ns/m².

As per the question we can list some data as,

diameter d = 1mm

$https://latex.codecogs.com/svg.image? r = 0.5mm = 0.5 \mathrm{x} 10^{-3}m = 5\mathrm{x} 10^{-4}m$

velocity v= 2m/s

$https://latex.codecogs.com/svg.image?\eta = 1.8 \mathrm{x}10^{-5} \mathrm{Ns/m^{2}}$

To find , viscous force F= ?

Solution -

Viscous force acting on spherical object can be given by Stoke's Law

$https://latex.codecogs.com/svg.image?\mathrm{F = 6\pi \eta rv}$

$https://latex.codecogs.com/svg.image?\mathrm{F = 6 x 3.142 x 1.8 x 10^{-5} x 5x 10^{-4}x 2}$

$https://latex.codecogs.com/svg.image?\mathrm{F = 6 x 3.142 x 18 x 10^{-9}}$

$https://latex.codecogs.com/svg.image?\mathrm{F = 108 x 3.142 x 10^{-9}}$

$https://latex.codecogs.com/svg.image?\mathrm{F = Al(log 108 + log 3.142) x 10^{-9}}$

$https://latex.codecogs.com/svg.image?\mathrm{F = Al(2.0334+ 0.4972) x 10^{-9}}$

$https://latex.codecogs.com/svg.image?\mathrm{F = Al(2.5306) x 10^{-9}} = \mathrm{339.3 x 10^{-9}}$

$https://latex.codecogs.com/svg.image?\mathrm{F = 3.393 x 10^{-7} N}$

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?