Question

A beaker containing a liquid of density rho moves down with an acceleration 'a'. The pressure due to the liquid at a depth h below free surface of the liquid is

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Due}\mathrm{to}\mathrm{upward}\mathrm{acceleration}\mathrm{pseudo}\mathrm{force}\mathrm{will}\mathrm{act}\mathrm{downwards}\mathrm{so}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity} \\ \mathrm{will}\mathrm{increases}\mathrm{by}\text{'}\mathrm{a}\text{'}. \\ \therefore \mathrm{g}\text{'}=\left(\mathrm{g}+\mathrm{a}\right)........\left(1\right) \\ \mathrm{P}=\mathrm{\rho g}\text{'}\mathrm{h} \\ \mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}, \\ \mathrm{P}=\mathrm{\rho }\left(\mathrm{g}+\mathrm{a}\right)\mathrm{h} \end{gathered}$$