Question

A sniper fires a rifle bullet into a gasoline tank making a hole $53.0m$ below the surface of gasoline. The tank was sealed at $3.10atm$. The stored gasoline has a density of $660kg{m^{ - 3}}$. The velocity with which gasoline begins to shoot out of the hole is
(A) $27.8m{s^{ - 1}}$
(B) $41.0m{s^{ - 1}}$
(C) $9.6m{s^{ - 1}}$
(D) $19.7m{s^{ - 1}}$

Answer 1

Bernoulli’s equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container. Bernoulli’s principle can be derived from the principle of conservation of energy.
Formula Used: The formulae used in the solution are given here.
Bernoulli’s theorem:
${P_0} + h\rho g = \dfrac{1}{2}\rho {v^2}$ where ${P_0}$ is the sealing pressure of the tank, $h$ is the depth of hole below the surface of gasoline, $\rho$ is the density of the stored gasoline, $v$ is the velocity with which gasoline begins to shoot out of the hole.

Complete Step by Step Solution: Bernoulli’s principle states that, the total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant.
It has been given that a sniper fires a rifle bullet into a gasoline tank making a hole $53.0m$ below the surface of gasoline. The tank was sealed at $3.10atm$. The stored gasoline has a density of $660kg{m^{ - 3}}$. By Bernoulli’s theorem, we can write,
${P_0} + h\rho g = \dfrac{1}{2}\rho {v^2}$ where ${P_0}$ is the sealing pressure of the tank, $h$ is the depth of hole below the surface of gasoline, $\rho$ is the density of the stored gasoline, $v$ is the velocity with which gasoline begins to shoot out of the hole.
We assign values given in the question to the variables in the equation. Therefore,
${P_0} = 3.10 \times {10^5}$, $h = 53$, $\rho = 660kg{m^{ - 3}}$.
We assume that the acceleration due to gravity $g = 10m/{s^2}$.
Substituting the values, we get, $3.10 \times {10^5} + 53 \times 660 \times 10 = \dfrac{1}{2} \times 660 \times {v^2}$.
Simplifying the equation, we get, ${v^2} = \dfrac{{3.10 \times {{10}^5} + 53 \times 660 \times 10}}{{330}}$
$v = \sqrt {3.10 \times {{10}^5} + 530 \times 2} = 41m{s^{ - 1}}$

Hence, the correct answer is Option B.

Note: Bernoulli’s principle Although Bernoulli deduced the law, it was Leonhard Euler who derived Bernoulli’s equation in its usual form in the year 1752. Formulated by Daniel Bernoulli states that as the speed of a moving fluid increases (liquid or gas), the pressure within the fluid decreases.