
There is a hole in area A at the bottom of the cylindrical vessel. Water is filled up to height h and water flows out in t second. If water is filled to a height of 4h, find the time at which the water will flow out.
A. t
B. 4t
C. 2t
D. \[\dfrac{t}{4}\]
Answer
160.8k+ views
Hint:Before going to solve this question let us discuss what data they have given. They have considered a cylindrical vessel where the water is filled to a height h and a hole is made at the bottom and water flows out in t seconds. Suppose if the water is filled to a height of 4h, we need to find how much time it takes to flow.
Formula Used:
To find the time required to empty the tank, the formula is,
\[t = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2H} }}{g}\]
Where,
A is area of the hole in a cylindrical vessel
\[{{\rm{A}}_{\rm{0}}}\]is area of a cylindrical vessel
H is height of a cylindrical vessel
g is acceleration due to gravity
Complete step by step solution:
Consider a cylindrical vessel of the area \[{{\rm{A}}_{\rm{0}}}\]and there is a hole at the bottom of the cylindrical vessel of area A. the water is filled up to height h and water flows out in the second t. suppose if water is filled to a height of 4h, then we need to find the time at which the water will flow out.
The time required to empty the tank can be written as,
\[t = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2H} }}{g}\]
Here,
\[{t_1} = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2{H_1}} }}{g}\]and
\[\Rightarrow {t_2} = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2{H_2}} }}{g}\]
Taking the ratios of time, we obtain
\[\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{\sqrt {{H_2}} }}{{\sqrt {{H_1}} }}\]
Here, \[{H_1} = h\] and \[{H_1} = 4h\]
Substitute the value in above equation we get,
\[\dfrac{{{t_2}}}{{{t_1}}} = \sqrt {\dfrac{{4h}}{h}} \]
\[\Rightarrow \dfrac{{{t_2}}}{{{t_1}}} = 2\]
\[\Rightarrow {t_2} = 2{t_1}\]
\[\therefore {t_2} = 2t\]
Therefore, the time at which the water will flow out is 2t.
Hence, option C is the correct answer.
Note:Here, it is important to remember Bernoulli's principle. It states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. This principle relates the pressure of a fluid to its elevation and its speed. This equation is used to approximate these parameters in water, air, or any fluid that has a very low viscosity.
Formula Used:
To find the time required to empty the tank, the formula is,
\[t = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2H} }}{g}\]
Where,
A is area of the hole in a cylindrical vessel
\[{{\rm{A}}_{\rm{0}}}\]is area of a cylindrical vessel
H is height of a cylindrical vessel
g is acceleration due to gravity
Complete step by step solution:
Consider a cylindrical vessel of the area \[{{\rm{A}}_{\rm{0}}}\]and there is a hole at the bottom of the cylindrical vessel of area A. the water is filled up to height h and water flows out in the second t. suppose if water is filled to a height of 4h, then we need to find the time at which the water will flow out.
The time required to empty the tank can be written as,
\[t = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2H} }}{g}\]
Here,
\[{t_1} = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2{H_1}} }}{g}\]and
\[\Rightarrow {t_2} = \dfrac{A}{{{A_0}}}\dfrac{{\sqrt {2{H_2}} }}{g}\]
Taking the ratios of time, we obtain
\[\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{\sqrt {{H_2}} }}{{\sqrt {{H_1}} }}\]
Here, \[{H_1} = h\] and \[{H_1} = 4h\]
Substitute the value in above equation we get,
\[\dfrac{{{t_2}}}{{{t_1}}} = \sqrt {\dfrac{{4h}}{h}} \]
\[\Rightarrow \dfrac{{{t_2}}}{{{t_1}}} = 2\]
\[\Rightarrow {t_2} = 2{t_1}\]
\[\therefore {t_2} = 2t\]
Therefore, the time at which the water will flow out is 2t.
Hence, option C is the correct answer.
Note:Here, it is important to remember Bernoulli's principle. It states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. This principle relates the pressure of a fluid to its elevation and its speed. This equation is used to approximate these parameters in water, air, or any fluid that has a very low viscosity.
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