The free surface of oil in a tanker, at rest, is horizontal if the tanker starts ac celerating the free surface will be titled by an angle `theta.` If the ac celeration is a ` ms^(-2)` what will be the slope of the free surface ?

Explanation

If the tanker ac celerates towareds right hand side, then the water will bulge at the back of tanker as shown in

Consider a horizontal liquid cylinder in the liquid, of length L, area of cross -section A. Let `h_1 and h_2` be the heights of liquid coluum at the ends of cylinder from the surface of liquid as shown in fig. 7 (EP(19. For translatory motion of cylinder with ac celeration a, `F_1 - F_2 =ma or P_1 A - P_2 A = AL rho a`
or `(h_1 rho g - h_2 rho g) A = AL rho a '[:. P = h rho g]`
or `(h_1 - h_2) g =L a or (h_1 - h_2)/(L) = (a)/(g)`
From fig. `7 (EP) 19. tan theta = (h_1 - h_2)/(L) = (a)/(g)` , so `theta = tan^(-1) (a//g)`