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A cylindrical vessel filled with water upto the height H becomes empty in time t0 due to a small hole at the bottom of the vessel. If water is filled to a height 4H it will flow out in time
- a)t0
- b)4t0
- c)8t0
- d)2t0
Correct answer is option 'D'. Can you explain this answer?
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A cylindrical vessel filled with water upto the height H becomes empty...
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A cylindrical vessel filled with water upto the height H becomes empty...
Explanation:
The time taken for a cylindrical vessel filled with water up to the height H becomes empty in time t0 due to a small hole at the bottom of the vessel is given by the Torricelli's law as:
t0 = (2πr²H)/A
where,
r = radius of the vessel
A = area of the hole
Now, if the water is filled to a height 4H, the time taken for it to flow out can be calculated as follows:
Let the height of the water column at any time t be h.
Therefore, the volume of water flowing out per unit time is given by the Torricelli's law as:
V/t = A√(2gh)
where,
V = volume of water in the vessel
t = time taken for the water to flow out
g = acceleration due to gravity
At the initial time t=0, h=4H and the volume of water in the vessel is V = πr²(4H) = 4πr²H
Substituting the values, we get:
V/t = A√(8gH)
Now, at any other time t, the height of the water column will be given by:
h = 4H - (gt²)/2
Substituting this value in the above equation, we get:
V/t = A√(8g(4H - (gt²)/2))
Squaring both sides and simplifying, we get:
t² = (16H)/g
or, t = 4√(H/g)
Therefore, the time taken for the water to flow out when the vessel is filled to a height 4H is 4 times the time taken for it to flow out when the vessel is filled to a height H. Hence, the correct option is D) 2t0.
The time taken for a cylindrical vessel filled with water up to the height H becomes empty in time t0 due to a small hole at the bottom of the vessel is given by the Torricelli's law as:
t0 = (2πr²H)/A
where,
r = radius of the vessel
A = area of the hole
Now, if the water is filled to a height 4H, the time taken for it to flow out can be calculated as follows:
Let the height of the water column at any time t be h.
Therefore, the volume of water flowing out per unit time is given by the Torricelli's law as:
V/t = A√(2gh)
where,
V = volume of water in the vessel
t = time taken for the water to flow out
g = acceleration due to gravity
At the initial time t=0, h=4H and the volume of water in the vessel is V = πr²(4H) = 4πr²H
Substituting the values, we get:
V/t = A√(8gH)
Now, at any other time t, the height of the water column will be given by:
h = 4H - (gt²)/2
Substituting this value in the above equation, we get:
V/t = A√(8g(4H - (gt²)/2))
Squaring both sides and simplifying, we get:
t² = (16H)/g
or, t = 4√(H/g)
Therefore, the time taken for the water to flow out when the vessel is filled to a height 4H is 4 times the time taken for it to flow out when the vessel is filled to a height H. Hence, the correct option is D) 2t0.
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