Leaky tank

Pressure increases linearly with depth. To balance the opposing force, the hole we create must have twice as much area, because it's at a depth with half as much pressure.

This leads to the answer (where $r'$ is the radius of the new hole we create, and $r$ is the radius of the existing hole: $r'^2 = 2r^2\implies r' = r\sqrt2 = \frac{\sqrt2}{2} = \boxed{0.707}$

If the tank is not to move, the momentum of water coming out of first hole must be equal and opposite of the water coming out of second hole.
Now as we know that velocity of water coming out of a hole at depth $h$ is
$v =\sqrt{2gh}$
Also mass coming out of the hole per second can be written as
$m = \rho \times$ Volume coming out per second.
As Volume coming out per second $=$ Area of hole $(A) \times$ velocity of water.
Therefore $m = \rho Av$ . And momentum of this water is
$p = mv = \rho A {v}^{2}$ .
As area of hole is $A = \pi {r}^{2}$ Therefore
$p = \rho \pi {r}^{2} {v}^{2}$ or $p \propto {r}^{2} {v}^{2}$ as $\rho, \pi$ are constants here.
As momentum of water coming out of first hole = momentum of water coming out of second hole in opposite direction, Therefore
$\frac{p_1}{p_2} = 1$ . But as from above derivation we can write
$\frac{p_1}{p_2} = \frac{{r_1}^{2} {v_1}^{2}}{{r_2}^{2} {v_2}^{2}} = 1$
Hence ${r_1}^{2}{v_1}^{2} = {r_2}^{2}{v_2}^{2}$ .
Also as $v =\sqrt{2gh}$ Therefore
$2 g {h_1} {{r_1}^{2}} = 2 g {h_2} {{r_2}^{2}}$ . or ${h_1} {{r_1}^{2}} = {h_2} {{r_2}^{2}}$ .
Plugging in the given values of ${h_1}$ , ${r_1}$ , ${h_2}$ ,
we get ${r_2} = \frac{1}{\sqrt{2}} = \fbox{0.707 cm}$

Clearly, the tank will be stationary only if no net force acts on it. So the forces on both the holes are equal(and opposite in direction). If the force applied by water on the first hole is $F_1$ , the hydrostatic pressure is $P_1$ , then we have

$F_1=A_1P_1$

$\Rightarrow F_1=\pi r_1^{2}P_1$

Now, the hydrostatic pressure is given by

$P=h\rho g$

So $F_1=h_1\rho g\pi r_1^{2}$

For no motion of the tank,

$F_1=F_2\Rightarrow h_1\rho g\pi r_1^{2}=h_2\rho g\pi r_2^{2}$

Putting the values given in appropriate units in the above equation, we get

$r_2=\boxed{0.7071\hspace{2mm}cm}$

F1 = F2

P1 x S1 = P2 x S2

d x h1 x S1 = s x h2 x S2

(h1)/(h2) = (S2)/(S1)

1/(0.5) x S1 = S2

2 x pi x (r1)^2 = pi x (r2)^2

(r2)^2 = 1/2

r2 = 0.707

First we will work out the relative velocities of water coming out.

Bernoulli's principle for incompressible fluid flow says: $\rho v^2 - 2\rho gh + 2P = C$ where $C$ is some constant, and $P$ is the pressure driving the flow, which is the air pressure on top of the tank, and $h$ is a height below the top of the tank, and $\rho$ is the water presure. We are assuming the difference in atmospheric pressure between the top of the tank and the hole to be negligible.

Since the velocity would be 0 at the top of the tank $(h = 0)$ , we get $C = 2P$ . Therefore $v^2 = 2gh$ .

Since $h_1 = 2h_2$ , we get $2v_2^2 = v_1^2$ .

Next, we can use conservation of momentum. If the tank does not move then the same momentum of water must flow out each side.

In a short time $t$ the volume of water that gets out is $v A t$ , where $A$ is the hole cross-section. So the momentum of this water is $\rho v^2 A t$ .

So we have two conditions: $v_1^2 A_1 = v_2^2 A_2$ $2v_2^2 = v_1^2$

Combining these two gives $2A_1 = A_2$ so $2r_1^2 =r_2^2$ Since $r_1 = 0.5$ , we get $r_2 = \sqrt{0.5} = \boxed{0.707...}$

Need to apply the idea of conservation of momentum to the tank. The momentum due to the flow at one end must equal the momentum due to flow at the other, i.e :

(1) $m_{1}v_{1}=m_{2}v_{2}$ where $v_{i}$ is the velocity of water out of the $i^{th}$ hole and so on...

(2) The mass of the flow can be written in terms of the radius of the holes and the velocity. $m_{i} = \pi r_{i}^{2} \rho v_{i}$ (can check by dimensional analysis)

3) The velocity of water is calculated using Toricelli's Law . Since the tank is so much larger than the radius of the holes, the change in height due to loss of water can be assumed negligible and the velocity of water at the top of the tank to be zero.

$v_{i} = \sqrt 2g \Delta h_{i}$

4) Then - $r_{1}^{2}v_{1}^{2}=r_{2}^{2}v_{2}^{2}$

$r_{2} = \sqrt \frac{\Delta h_{1}}{\Delta h_{2}} r_{1}$

5) $r_{2} =\frac{\sqrt 2}{2} \approx 0.707$

$a_{1}v_{1}^2 \rho = a_{2}v_{2}^2 \rho$ $\Rightarrow a_{2} = 2a_{1}$ or $r_{2} = \sqrt{2}r_{1} = 0.707 cm$

The pressure \­[P\­] is equal to divide the force \­[F\­] by the surface \­[ S\­]. Then: Then \­[P=\frac{F}{S}\­]­

In a fluid, pressure is the same in all directions. We also know the formula \­[P=\rho \times \g \times \h\­], Where \­[\rho\­] is the density of the fluid, \­[g\­] is the acceleration of gravity and \­[h\­] is the depth.

So he have: \­[\rho \times \g \times \h=\frac{F}{S}\­]­

The force defines the acceleration, so it has to be the same in hole 1 and in hole 2. \­[F = \rho \times \g \times \h \times S\­]­

\­[\rho\­]­ and \­[g\­]­ are constant so we have:

\­[\rho \times \g \times \h{1} \times S{1}\­= \rho \times \g \times \h{2} \times S{2}\­]­

Then: \­[h {1} \times S {1}­= \h{2} \times S{2}\­]­

\­[S=\pi \times r^{2} \­]­

So \­[h {1} \times \pi \times r {1}^{2}­ = h {2} \times \pi \times r {2}^{2}­]­

We know the two heights and one radius.

And solving thin equation we have [r_{1}^{2}=\sqrt{\frac{1}{2}} = 0.707\­]

My solution shows that the answer is only dependent on $r_{1}$ and independent of all other values given. Let height of the container be h, acceleration due to gravity be g and density of water be d. Now, force on both sides should be equal and force= $pressure \times area$ . So,

(hdg) $\pi$ $r^{2}$ = ( $\frac{h}{2}$ dg) $\pi$ $R^{2}$ where r= $r_{1}$ and R= radius of the new hole. This equation yields R= $\sqrt{2}$ r= $\sqrt{2}$ X 0.5cm=0.707cm

When a jet of water leaves the hole, it exerts a reaction force on the rest of the tank. If we want to stop the tank from moving, we need an equal force in the opposite direction on the tank. So, the water leaving the second hole must exert an equal force on the tank $Force = Pressure*Area$ Since force is same for both holes,

$P_{1}*A_{1}=P_{2}*A_{2}$

Using $P=h*d*g$ (d=density of water) and $A=pi*R^{2}$

$h_{1}*(r_{1})^{2}=h_{2}*(r_{2})^{2}$ Plugging h1=1 m, r1=0.5 cm, h2=0.5 m , we find **r2=0.707

Since there is no movement, there must be no net momentum. Since there is momentum at both holes, and the momentum vectors point in opposite directions, the magnitude of the momentum on either side must be the same.

Momentum=P, $P=m \times v$ .

By Toricelli's theorem, (a special case of Bernoulli's theorem), $v=\sqrt{2gy}$ , where y is the distance from the hole to the top of the tank.

Mass=m, area of hole=A, $m=V \times \rho =Av \rho$ .

$|P_1|=m_1 v_1= (A_1v_1 \rho)(v_1) =A_1v_1^2 \rho =(\pi 0.5^2)(2g \times 1) \rho=0.25 \times 2g \pi \rho$ .

$|P_2|=m_2 v_2= (A_2v_2 \rho)(v_2) =A_2v_2^2 \rho =(\pi r^2 )(2g \times 0.5)\rho=r^2 \times g \pi \rho$ .

$|P_1|=|P_2|$ , $0.25 \times 2g \pi \rho = r^2 \times g \pi \rho$ , $r^2=0.25 \times 2= 0.5$ , $r=\boxed{0.707 cm}$