Solve the problem.
-Find the force on one side of a cubical container $5 \mathrm {~cm}$ on an edge if the container is filled with mercury. The density of mercury is $133 \mathrm { kN } / \mathrm { m } ^ { 3 }$.
A) 1.7 N
B) 330 N
C) 8300 N
D) 8.3 N

Question

Quizplus · Accepted Answer

The force on one side of the container can be calculated using the formula for pressure due to a fluid, $P = \rho gh$, where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the fluid column. Here, $\rho = 133 \mathrm { kN } / \mathrm { m } ^ { 3 }$ (or $133000 \mathrm { N } / \mathrm { m } ^ { 3 }$), $g = 9.8 \mathrm { m/s } ^ { 2 }$, and $h = 0.05 \mathrm { m }$ (since $5 \mathrm { cm } = 0.05 \mathrm { m }$). The pressure exerted by the mercury is $P = 133000 \times 9.8 \times 0.05 = 65290 \mathrm { N/m } ^ { 2 }$. The force on one side of the container is the pressure multiplied by the area of that side, $F = PA$. The area of one side of the cube is $A = 0.05^2 = 0.0025 \mathrm { m } ^ { 2 }$. Therefore, the force is $F = 65290 \times 0.0025 = 163.225 \mathrm { N }$, which rounds to $163 \mathrm { N }$, but since none of the provided options match this calculation, it seems there was a mistake in the calculation process. Correctly, the force should be calculated considering the density directly as pressure due to the depth of mercury, and the correct calculation is as follows:Given the density of mercury is $133 \mathrm { kN } / \mathrm { m } ^ { 3 }$, and considering the depth of mercury is $5 \mathrm { cm } = 0.05 \mathrm { m }$, the force exerted by the mercury on one side of the container can be found by first calculating the weight of the mercury and then applying it to the area of one side.The volume of mercury is $V = 0.05^3 = 0.000125 \mathrm { m }^3$. The weight of the mercury (which is the force due to gravity) is $F = \rho V g$, but since we're dealing with pressure exerted by the static fluid, we simplify the calculation by considering the pressure exerted by the column of mercury directly, which is equivalent to the density of mercury times the height of the column (ignoring $g$ as it's already factored into the density unit $\mathrm { kN } / \mathrm { m } ^ { 3 }$).Thus, the force on one side of the container due to the pressure exerted by the mercury is directly related to the density and the height of the mercury column, not requiring the explicit calculation through the gravitational force formula. The mistake in the initial explanation was in the misapplication of formulas and units, leading to an incorrect force calculation.The correct approach is to calculate the pressure exerted at the bottom of the container due to the mercury and then use this to find the force on one side. However, the direct calculation involving the given options and the correct formula was not accurately represented in the initial explanation. The correct answer, based on the options provided and the typical understanding of such problems, would involve calculating the hydrostatic pressure and then applying it to find the force exerted on one side of the cube, considering the area of that side and the depth of the mercury. Given the misunderstanding in the calculation process, the correct answer should reflect the principles of hydrostatic pressure and force calculation based on the density of the fluid and the area of the container side.

Solve the Problem 5 cm5 \mathrm {~cm}5 cm On an Edge If the Container Is Filled with Mercury

Solve the Problem $5 \mathrm {~cm}$ On an Edge If the Container Is Filled with Mercury