Water Bucket out of the world !

Bullshit

At any intermediate $x$ .

Firstly radius of the junction of water and air can be given as :

$r = a + \dfrac{(b-a)x}{H}$

Thanks to your values it comes out a neat :

$r = 1+x$

Secondly height of air column = $1-x$

Hence volume of air column(by using the formula for volume of frustum) as :

${V}_{air} = \dfrac{ \pi (1-x) }{3} ( {(1+x)}^{2} + {2}^{2} + 2(1+x) )$

$\Rightarrow {V}_{air} = \dfrac{ \pi (1-x) }{3}(x^{2}+4x+7)$

Also volume of air initially (i.e $x=\frac{1}{2}$ )

${V}_{i} = \frac{37\pi }{24}$

${P}_{i} = {P}_{0}$

Now since it is a polytropic process $\dfrac{P}{V}= K$

$\dfrac{{P}_{i}}{{V}_{i}} = \dfrac{ {P}_{air} }{ {V}_{air} }$

Hence we have :

${P}_{air} = \dfrac{8}{37}(1-x)(x^{2}+4x+7){P}_{0}$

Writing bernoulli equation for the given two points (one is a point at the junction and

another point is just outside the hole) we have :

${P}_{air} + \rho g x +\dfrac{1}{2}\rho {v}_{1}^{2} = {P}_{0} + \dfrac{1}{2}\rho {v}_{2}^{2}$

${v}_{1}$ = velocity of water level

${v}_{2}$ = velocity of water coming out of hole :

Also using equation of continuity we have :

${A}_{1} {v}_{1} = {A}_{2}{v}_{2}$

${A}_{1}$ = Area of junction( of air and water ) = $\pi {(1+x)}^{2}$

${A}_{2}$ = Area of hole

Solving we get the expression of ${v}_{1}$ as :

$\displaystyle ({ (\frac { { A }_{ 1 } }{ { A }_{ 2 } } })^{ 2 }-1){ v }_{ 1 }^{ 2 } = \frac { 2({ P }_{ air }-{ P }_{ 0 }) }{ \rho } +2gx$

Putting the values we have :

$\displaystyle ({ 10 }^{ 4 }{ \pi }^{ 2 }{ (1+x) }^{ 2 }-1){ v }_{ 1 }^{ 2 } = 200(\frac { 8 }{ 37 } (1-x)({ x }^{ 2 }+4x+7)-1)+20x$

Note that $-\dfrac{dx}{dt} = {v}_{1}$

Finally I get the expression of $T$ as :

$\displaystyle \int _{ 1/4 }^{ 1/2 }{ \sqrt { \frac { { 10 }^{ 4 }{ \pi }^{ 2 }{ (1+x) }^{4 }-1 }{ 20x+\frac { 200 }{ 37 } (8(1-x)({ x }^{ 2 }+4x+7)-37) } } dx } = 26.2564$