Q: 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)

(a) 3g/2

(b) 4g/3

(c) 4g/2

(d) 3g/4

Ans: (b)

Sol: Let height of liquid becomes h .

Now equating volume ;

$\large 2\times 3 = \frac{1}{2}\times h \times 3$

h = 4 m

$\large tan\theta = \frac{h}{3} = \frac{4}{3}$

$\large \frac{a}{g} = \frac{4}{3}$

$\large a = \frac{4}{3} g$