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Asked by Prashant DIGHE 9th October 2019, 10:05 PM
Figure shows the status of liquid level, when uniform acceleration is given along the direction of length L.

when tha tank is accelerated front side level of liquid is lower than the backside.

Height of container 2 m.  Initially level of liquid is at height 1.5 m .  when it is accelerated, backside level just before spilling is 2 m
and front side level at same time is 1 m

Pressure at points A and B , when the liquid about to spill is given as follows

P= ρg (2) and PB = ρ g (1) and pressure difference  is PA - PB = ρ g ( 2 - 1 ) = ρ g (1) ....................(1)
where ρ is density

Pressure difference and acceleration are related as,  PA - PB = ρ L a  ....................(2)

where a is accekeration

using (1) and (2), we get, acceleration a = g/L = 10/10 = 1 m/s2

time taken to get speed v = 20 m/s staring from rest and with  acceleration 1 m/s2 is ,  t = v/a = 20/1 = 20 s
Answered by Expert 12th October 2019, 12:59 PM
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