Hole in the Wall!

To Double the Range,we will have to double the velocity of efflux

Let the initial velocity of efflux of water be $v$

then final velocity of efflux is $2v$

Let $P_{1}$ be the EXTRA PRESSURE required and $P_o$ be the atmospheric pressure

Applying Bernoulli's Equation(taking potential energy reference at height 10 metres below water level)

also velocity at top of the tank is negligible since $Av_{top}=av \Rightarrow v_{top}=\frac {a}{A}v... and A>>a$

making the equations for the two cases..we get

$P_{o}+\rho g(10)=P_{o}+\frac {1}{2}\rho (v)^2$

$(P_{1}+P_{o})+\rho g(10)=P_{o}+\frac {1}{2}\rho(2v)^2$

On Solving...we get

$P_{1}=3\times10\rho g$

$\Rightarrow \boxed{P_{1}=3 atm}(approx.)$

Firstly, we need to calculate the velocity of the water that gushes out of the small hole at $10m$ from the water surface to reach the range of $R$ . This can be done so, using Torricelli equation which is...

In order to travel twice the range, which is $2R$ , the velocity must also be doubled. Hence, using the previously calculated velocity, double it and equate that value with Torricelli equation again to find the new depth of the water from the surface.

Now, be careful , the question asks for the extra pressure, which implies the extra height of the water from its surface. Hence, you would need to substract the old depth from the new depth to find the change in height and it should yield $30m$ . Finally, use...

to find the required pressure in $atm$ and approximate that value to $\boxed{3}$ , which is the answer!

Let the speed with which the water rushes out be $v$ and the height of the tank be $\ell$ .
Then, applying fundamental Kinematics equations,
Clearly,

$\left(\ell-10\right) = \frac{1}{2} gt^2$ and $R = vt$

Eliminating $t$ , we get,

$R = v\sqrt{\frac{2\left(\ell-10\right)}{g}}$

It is obvious that, to double the Range, we have to double the speed with which the water ejects.

Now,
Taking the origin at the bottom of the tank,
Assume $P$ to be the Extra Pressure that we apply and $P_o$ to be the atmospheric pressure.
Also, as the tank is large , we can consider the velocity of water at the surface to be zero .

Applying Bernoulli's Equation , taking the streamline to start at the surface and ending just outside the hole, for the initial case, as well as the final case, we get,

$P_o + \rho g\ell = P_o + \rho g\left(\ell-10\right) + \frac{1}{2}\rho v^2$ , and

$\left(P_o+P\right) + \rho g\ell = P_o + \rho g\left(\ell-10\right) + \frac{1}{2}\rho\left(2v\right)^2$

Simplifying the two equations, we get,

$\rho g\left(10\right) = \frac{1}{2}\rho v^2$ , and

$P + \rho g\left(10\right) = 2\rho v^2$

Solving, we get,

$P = 3\rho g\left(10\right)$

$\Rightarrow P = 300000 Pa$

$\Rightarrow P \approx \boxed{3}$ atm

FOR DOUBLE RANGE EFFLUX VELOCITY SHOULD BE 2 TIMES USE BERNOULLI'S EQUATION