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A fluid container is containing a liquid of density r is is accelerating upward with acceleration a along the inclined place of inclination a as shwon. Then the angle of inclination q of free surface is :
                     
  • a)
    tan_1
  • b)
  • c)
     
  • d)
     
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A fluid container is containing a liquid of density r is is accelerati...
First resolve all components in the along and perpendicular to incline. Pressure difference is created in a vertical column full of liquid. This is because of gravity acting in downward direction. Similarly, pressure difference will be created too along the incline. So, p = h * d * g * cosa (in perpendicular direction) and
p = hd (a + g sina) (along incline).
So, tan(theta) = (a + gsina)/(gcosa)
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Most Upvoted Answer
A fluid container is containing a liquid of density r is is accelerati...
First resolve all components in the along and perpendicular to incline. Pressure difference is created in a vertical column full of liquid. This is because of gravity acting in downward direction. Similarly, pressure difference will be created too along the incline. So, p = h * d * g * cosa (in perpendicular direction) and
p = hd (a + g sina) (along incline).
So, tan(theta) = (a + gsina)/(gcosa)
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Community Answer
A fluid container is containing a liquid of density r is is accelerati...
First resolve all components in the along and perpendicular to incline. think about it, pressure difference is created in a vertical column full of liquid. why? because of gravity acting in downward direction. similarly pressure difference will be created too along the incline. so, p=hdgcosa ( in perpendicular direction) and p= xd (a +gsina) (along incline).
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