A right circular cylindrical storage tank with height 10 meters and radius 8 meters is full of gasoline. Assume the density of gasoline is 720 kg/cu m. The work required to pump all the gasoline over the top of the tank, in kg-meters, is

A) $22,897,200 \pi$
B) $22,796,200 \pi$
C) $22,579,200 \pi$
D) $22,486,200 \pi$
E) $22,388,200 \pi$

Question

A right circular cylindrical storage tank with height 10 meters and radius 8 meters is full of gasoline. Assume the density of gasoline is 720 kg/cu m. The work required to pump all the gasoline over the top of the tank, in kg-meters, is

A) $22,897,200 \pi$
B) $22,796,200 \pi$
C) $22,579,200 \pi$
D) $22,486,200 \pi$
E) $22,388,200 \pi$

Quizplus · Accepted Answer

The work required to pump the gasoline over the top of the tank can be calculated using the formula for work, $W = \rho g h A d$, where $\rho$ is the density of the fluid (720 kg/m³ for gasoline), $g$ is the acceleration due to gravity (9.8 m/s², but it cancels out since we're asked for work in kg-meters), $h$ is the height at which the fluid is being lifted (which varies from 0 at the bottom to 10 meters at the top), $A$ is the cross-sectional area of the cylinder ($\pi r^2 = \pi \times 8^2 = 64\pi$ m²), and $d$ is the depth of the fluid element being considered.The work done to lift a thin slice of gasoline at a height $y$ to the top of the tank is $dW = \rho A g (10 - y) dy$, integrating this from $y = 0$ to $y = 10$ gives the total work:$$W = \int_0^{10} 720 \times 64\pi \times (10 - y) dy = 720 \times 64\pi \times \left[10y - \frac{y^2}{2}\right]_0^{10}$$$$W = 720 \times 64\pi \times \left[100 - \frac{100}{2}\right] = 720 \times 64\pi \times 50$$$$W = 720 \times 3200\pi = 2,304,000\pi$$However, this calculation does not match any of the provided options, indicating a mistake in the calculation process. The correct approach involves calculating the volume of gasoline and then multiplying by the average height it needs to be lifted (which is 5 meters, since the work required is equivalent to lifting the entire volume to the midpoint of the tank, then over the edge):Volume of the cylinder $V = \pi r^2 h = \pi \times 8^2 \times 10 = 640\pi$ m³.The mass of the gasoline is $m = \rho V = 720 \times 640\pi = 460,800\pi$ kg.The work required to lift this mass over the top of the tank (5 meters on average) is $W = mgh = 460,800\pi \times 5 = 2,304,000\pi$ kg-meters.This calculation still does not directly lead to the correct answer, indicating a misunderstanding in the explanation process. The correct method to find the work involves integrating the force needed to lift each slice of gasoline over the height of the tank, considering the density and the gravitational force. The discrepancy in the explanation suggests a need to correctly apply the principles of calculus and physics to find the total work done, which involves integrating the weight of each infinitesimal volume of gasoline over its lifting distance. Given the options and the typical approach to such a problem, the correct answer involves calculating the total energy required to move the entire volume of gasoline, taking into account the density and the height of the tank. The mistake in the initial explanation points to a miscalculation or misunderstanding of the problem's requirements. The correct calculation should directly relate to the volume of the tank, the density of gasoline, and the height over which the gasoline must be lifted, leading to one of the provided choices.

A Right Circular Cylindrical Storage Tank with Height 10 Meters 22,897,200π22,897,200 \pi22,897,200π

A Right Circular Cylindrical Storage Tank with Height 10 Meters $22,897,200 \pi$