• # question_answer When at rest, a liquid stands at the same level in the tubes shown in the figure. But as indicated a height difference h occurs when the system is given an acceleration a towards the right. Here h is equal to: A)  $\frac{aL}{2g}$                               B) $\frac{gL}{2a}$C) $\frac{gL}{a}$                                  D) $\frac{aL}{g}$

Let ${{P}_{1}}$and ${{P}_{2}}$be the pressures at the bottom of the left and right end of the tube respectively. Then $F=({{P}_{1}}-{{P}_{2}})A=\rho ghA$ Where A is the cross-section of the tube. The mass of the liquid in the horizontal position is $m=\rho LA$ Now,$F=ma,$so $\rho gh\,A=\rho LAa,$$\therefore$$h=\frac{aL}{g}$ Hence, the correction option is [d].