A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 kgm⁻³ . The velocity with which gasoline begins to shoot out of the hole is

The correct answer is 41.0 ms⁻¹.

To find the velocity with which gasoline begins to shoot out of the hole, we apply Bernoulli's principle. Let point 1 be at the top surface of the gasoline inside the sealed tank, and point 2 be at the hole where the gasoline shoots out into the atmosphere.

According to Bernoulli's equation:
$P_1+\frac{1}{2}\rho v_1^2+\mathrm{��}g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
where:
- $P_1$ is the pressure inside the sealed tank: $P_1=3.10\text{atm}$.
- $P_2$ is the atmospheric pressure outside the hole: $P_2=1.00\text{atm}$.
- $\rho$ is the density of the gasoline: $\rho =660{\text{kg m}}^{-3}$.
- $v_1$ is the velocity of the gasoline level at the top of the tank. Since the cross-sectional area of the tank is much larger than the area of the hole, we can assume $v_1\approx 0$.
- $v_2=v$ is the velocity of efflux (the shooting velocity).
- $h_1-h_2=h$ is the depth of the hole below the surface: $h=53.0\text{m}$.
- $g$ is the acceleration due to gravity: $g=9.8{\text{m s}}^{-2}$.

Rearranging the equation to solve for $v$:
$P_1+\rho gh=P_2+\frac{1}{2}\rho v^2$
$\frac{1}{2}\rho v^2=(P_1-P_2)+\rho gh$
$v^2=\frac{2(P_1-P_2)}{\rho }+2gh$

First, convert the pressure difference into Pascals (Pa). Note that $1\text{atm}=1.013\times {10}^5\text{Pa}$:
$P_1-P_2=3.10\text{atm}-1.00\text{atm}=2.10\text{atm}$
$P_1-P_2=2.10\times 1.013\times {10}^5\text{Pa}=2.1273\times {10}^5\text{Pa}$

Now, calculate each term inside the equation for $v^2$:
- Pressure term:
$\frac{2(P_1-P_2)}{\rho }=\frac{2\times 2.1273\times {10}^5}{660}\approx 644.6{\text{m}}^2{\text{s}}^{-2}$
- Gravity term:
$2gh=2\times 9.8\times 53.0=1038.8{\text{m}}^2{\text{s}}^{-2}$

Summing these terms:
$v^2=644.6+1038.8=1683.4{\text{m}}^2{\text{s}}^{-2}$
$v=\sqrt{1683.4}\approx 41.0{\text{m s}}^{-1}$

Thus, the gasoline shoots out of the hole with a velocity of approximately 41.0 ms⁻¹.