Question

an open tank 10m long and 2m deep is filled upto height with oil of specific gravity 0.82. the tank is uniformly accelerated along its length from rest to a speed of 20m/s horizontally. the shortest time in which the speed may be attained without spilling any oil is :[g=10m/s
${10}^{2}$
]

Answer 1

It is given that when the tank is at rest, the level of oil is up to h=1.5 tn.
If the tank is uniformly accelerated, the height of liquid will increase on the back side and decreases on the front side.
As an increase in acceleration height difference will also increase and after the certain limit the liquid will spill out.
According to the variation of pressure formula, The pressure at point q is,
Pq = Pa + Pgh = Pa + pg x 1= Pa + pg
Here,
Pa is the atmospheric pressure,
p is the density of the oil.
The pressure at point p is,
Pp = Pa + pgh = Pa + pg x 2 = Pa + 2pg

Now, the pressure difference at two points p & q in terms of acceleration is, Pp = Pq + paL = Pq + pa x 10
so, Pp — Pq = 10 pa
Then,
Pp — Pq = 10
 = pa + 2pg — Pa—P9 = 10 pa
a = g/10 = 1m/s²30
Now, the minimum time required to change the velocity of tank from zero to 10 m/s is,
V = U + at
t = (V-U)/a
= 10-0/1
 t = 1 sec