The mean pressure is $\frac{6K}{5}$ in SI unit, which rain renders to vertical windshield of automobile, moving with constant velocity of magnitude $v=12 m/s.$ Consider that raindrops fall vertically with speed  $u=5 m/s.$ The intensity of rainfall deposits $h=2cm$ of sediments in time $\tau =1\text{\hspace{0.17em}\hspace{0.17em}minute}$. [$\rho ={10}^{3}kg/m^3$ is the density of liquid] (Assume collisions are perfectly inelastic). Calculate K.

Let in time t, m mass of liquid collides with windshield. So
$F=\left|\frac{p_f-p_i}{t}\right|=\left|\frac{0-mv}{t}\right|=\frac{mv}{t}$
$\lambda \Rightarrow$density of liquid molecules in air so
$m=Svt\times \lambda =Sv\lambda t$
Where $S\Rightarrow$ surface area of the windshield
$F=S{v}^2\lambda \Rightarrow Meanpressure=\frac{F}{S}=\lambda v^2\mathrm{.....}\left(1\right)$
Now as we assumed that $\lambda$ is the density of liquid molecules of drops in air so in $\tau$ time $U\tau S_0$ volume of liquid will strikes on the surface of earth so mass of liquid strikes on the earth in time interval $\tau$ will be
$m_0=S_{0}u\tau \lambda \Rightarrow S_0\rho h=S_{0}u\tau \lambda \Rightarrow \lambda =\frac{\rho h}{u\tau }$
Putting this value in equation (1), we have
Mean pressure $=\frac{\rho h{v}^2}{u\tau }=\frac{48}{5}$