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Asked by Prashant DIGHE | 09 Oct, 2019, 10:05: PM
Expert Answer
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
PA = 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 Thiyagarajan K | 12 Oct, 2019, 12:59: PM
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