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A cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range, h is equal to
  • a)
    H/4
  • b)
    H/2
  • c)
    3 H/4
  • d)
    H
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A cylindrical vessel is filled with water up to height H. A hole is bo...
To find the maximum range of water from the hole, we need to determine the point where the water will exit the vessel with the highest velocity. This point is where the pressure at the hole is at a minimum, which occurs when the hole is at a depth h from the free surface of water.

Explanation:

- Bernoulli's principle states that as the velocity of a fluid increases, its pressure decreases. This principle is applicable to the flow of water through the hole in the vessel.
- As the water flows out of the hole, its velocity increases due to the gravitational force acting on it. This results in a decrease in pressure at the hole.
- The pressure at the hole is directly proportional to the depth of the hole from the free surface of water. This means that the deeper the hole, the higher the pressure at the hole.
- Therefore, the point where the pressure at the hole is at a minimum is where the depth of the hole is at a maximum. This occurs when the hole is at a depth h from the free surface of water, which is equal to half the height of the cylinder (H/2).
- At this point, the water will exit the vessel with the highest velocity, resulting in the maximum range.

Conclusion:

- For maximum range, the depth of the hole from the free surface of water should be equal to half the height of the cylinder (H/2).
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Community Answer
A cylindrical vessel is filled with water up to height H. A hole is bo...
The cylinder is filled upto a height of H and a hole is drilled at a depth h from the surface of water

Velocity of the water from v^2-u^2=2g(H-h) (Since,height(h)=(H-h))
v^2 =2g (H-h) (Since,u=0(or)the object is at rest)
v =√2g(H-h)
Time from s=ut +1/2at^2 (Since,u=0(or)the object is at rest)
h=0(t)+1/2gt^2
h=1/2gt^2
t=√2h/g
Range of water=flow Speed*time
=√2g(H-h)*√2h/g
=√2h(H-h)
Differentiate with respect to h
Differentiation of maximum will be zero
So,dx/dh=0
On solving we get height needed for maximum range of water will be H/2.
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A cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range, h is equal toa)H/4b)H/2c)3 H/4d)HCorrect answer is option 'B'. Can you explain this answer?
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