Question

a large tank filled with water to a height h is to be emptied through a small hole at the bottom the ratio of time taken for the level of water to fall from h to (h/2) and from (h/2) to zero is

Answer 1

Hope it will help u...

Answer 2

Answer:

The ratio of time taken is √2-1.

Explanation:

We will use the following formula to find the time,

$T=\sqrt{\frac{2h}{g} }$       (1)

Where,

T=time taken to empty the tank

h=height from which water is falling

g=acceleration due to gravity

Case I:

Initial height(h₁)=h

Final height(h₂)=$\frac{h}{2}$

By using equation (1) we have,

$T_1=\sqrt{\frac{2}{g} }(h_1-h_2)$

$T_1=\sqrt{\frac{2}{g} }(\sqrt{h} -\sqrt{\frac{h}{2} } )$           (2)

Case II:

Initial height(h'₁)=$\frac{h}{2}$

Final height(h'₂)=0

By using equation (1) we have,

$T_2=\sqrt{\frac{2}{g} }(h'_1-h'_2)$

$T_2=\sqrt{\frac{2}{g} }(\sqrt{\frac{h}{2} } -\sqrt{0} )$

$T_2=\sqrt{\frac{2}{g} }(\sqrt{\frac{h}{2} })$       (3)

By taking the ratio of equations (2) and (3) we get;

$\frac{T_1}{T_2} =\frac{\sqrt{\frac{2}{g} }(\sqrt{h} -\sqrt{\frac{h}{2} } )}{\sqrt{\frac{2}{g} }(\sqrt{\frac{h}{2} })}$

$\frac{T_1}{T_2} =\frac{\frac{\sqrt{2}-1 }{\sqrt{2} } }{\frac{1}{\sqrt{2} } }$

$\frac{T_1}{T_2}=\sqrt{2}-1$        (4)

Hence,  the ratio of time taken is √2-1.

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