question_answer 48)

                  The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle \[\theta \]. If the acceleration is \[a\,m\,{{s}^{-2}},\] what will be the slope of the free surface?                

Answer:

                  Consider two points A and B in liquid separated by a distance dx as shown in figure.                 Let P = presure at A                 and P + Pd = pressure at B                 \ Force at A ? force at B = ma                 or \[PA-(P+dP)A=ma,\]where A is the area of cross-section of a cylinder AB or \[(-dP)A=ma\]                 But         \[m=(Adx)\rho \]                 \[\therefore \] \[(-dP)A=Ad\times \rho a\]                 or            \[dP=-\rho a\,dx\]                    ?. (i)                 But         \[dP=\rho g({{h}_{2}}-{{h}_{1}})=-\rho g({{h}_{1}}-{{h}_{2}})\]                 \[\therefore \] eqn. (i),                 \[g({{h}_{1}}-{{h}_{2}})=a\,dx\]                 or  \[\frac{({{h}_{1}}-\,{{h}_{2}})}{dx}\,=\,\frac{a}{g}\]                 Since \[\frac{{{h}_{ & 1}}-{{h}_{2}}}{dx}\,=\,\tan \,\theta \] \[\therefore \]\[\tan \,\theta =\,\frac{a}{g}\]