A closed container shown in figure is filled with water `(rho=10^(3) kg//M^(3))`

This is accelerated in horizonal direction with an acceleration, `a=2m//s^(2)`.Find (a)`p_(C)-p_(D)` and (b) `p_(A)-p_(D)`.

Explanation

(a) In horizontal direction, pressure decreases in the direction of acceleration.
Thus, `p_(C) gt p_(D)`
or, `p_(C)-p_(D)=+ rhoax`
Substituting the values, we have
`p_(C)-p_(D)=(10^(3))(2)(2)`
or,`p_(C)-p_(D)=4.0xx10^(3) N//m^(2)`
(b) In vertical direction, pressure increses with depth.
`:. p_(C) gtp_(A)`
or, `p_(A)-p_(C)=-rho gh`
`=-(10^(3))(10)(6) (g_(e) =g)`
`=-60xx10^(3) N//m^(2)`
(b) Now, `p_(A)-p_(D)=(p_(A)-p_(C))+(p_(C)-p_(D))`
`=(-60xx10^(3))+(4.0xx10^(3))`
`=-56xx10^(3) N//m^(2)`
`=-56xx10^(4) N//m^(2)`.