plz solve this question as given in the image pl
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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/s²30Now, the minimum time required to change the velocity of tank from zero to 10 m/s is, V = U + att = (V-U)/a= 10-0/1 t = 1 sec
. 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/s²30Now, the minimum time required to change the velocity of tank from zero to 10 m/s is, V = U + att = (V-U)/a= 10-0/1 t = 1 sec
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