Question

A ball is immersed in water kept in container and released. At the same time container is accelerated
in horizontal direction with acceleration 144 m/s2. Acceleration of ball w.l.t. container (in m/s2) is
(specific gravity of ball =12/7, g= 10 m/s2)​

Answer 1

Answer:

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Answer 2

The efflux will be 5.48m/s

Explanation:

• Due to acceleration liquid profile gets tilted
• $tan\theta = a/g =1 \implies \theta =45^o$
• Therefore x = 0.5 m
• Now, height of the liquid in left most section =(1+ 0.5) = 1.5m
• So, FORMULA V= $\sqrt{2gh} = \sqrt {2*10*1.5} = \sqrt{30}$
• therefore V = 5.48 m/s

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