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The minimum horizontal acceleration of the container so that the pressure at point A of the container becomes atmospheric is (the tank is of sufficient height) 

                                                           

Option: 1

\frac{2g}{3}


Option: 2

\frac{4g}{3}


Option: 3

\frac{8g}{3}


Option: 4

None \ of \ the \ above


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                                                    \begin{array}{l}{\text { Volume equality gives }} \\ \\ {2 \times 3=\frac{1}{2} \times h \times 3 \Rightarrow h=4 m} \\ \\ {\therefore \tan \theta=\frac{4}{3}=\frac{a}{g} \Rightarrow a=\frac{4}{3} g}\end{array}

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