A tank, closed tightly with a lid, is completely filled with water and is rigidly fixed on a car. The car moves with a constant acceleration of `20cm*s^(-2)`. Determine the pressure at a point 10 cm below the lid and at a distance of 10 cm from the front wall of the tank.

Explanation

Let us consider a vertical disc of unit area kept at a depth h from the upper surface of the tank. The disc is at a distance x from the front wall of the car and its width is dx. Let the pressures at distances x and (x + dx) be p and (p + dp) respectively.

`therefore" "p+dp-p=rho*dx*a`
[`rho` = density of the liquid, `rhodx` = mass of the liquid displaced by the disc, and a = acceleration of the car]
or, `dp=arhodx`
`therefore" "intdp=arhointdx`
or, `p=arhox+c" "`[c = integration constant]
Now, if x = 0, then p = `hrhog`.
`therefore" "hrhog=c`
`therefore" "p=arhox+hrhog=rho(ax+hg)`
`therefore` The pressure at a point 10 cm below the lid and at a distances of 10 cm from the front wall of the tank,
`p=1(20xx10+10xx980)=200+9800`
`= 10000dyn*cm^(-2)`