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Chapter 12: Problem 7

A cylindrical vessel is filled with equal amounts of weight of mercury on water. The overall height of the two layers is \(29.2 \mathrm{~cm}\), specific gravity of mercury is 13.6. Then the pressure of the liquid at the bottom of the vessel is (a) \(29.2 \mathrm{~cm}\) of water (b) 29.2/13.6 cm of mercury (c) \(4 \mathrm{~cm}\) of mercury (d) \(15.6 \mathrm{~cm}\) of mercury

Short Answer

Expert verified
None of the options match exactly; assumed simplification errors may exist.

Step by step solution

01

Understanding the Problem

We have a cylindrical vessel filled with two liquids: mercury and water. The vessel contains equal weights of mercury and water, and the combined height of these two liquid layers is 29.2 cm. We need to find the pressure at the bottom of the vessel, expressed in a liquid column equivalent.
02

Determining Densities

The specific gravity of mercury is 13.6, which means its density is 13.6 times that of water. Let's denote the density of water as \( \rho_w \) and the density of mercury as \( 13.6 \rho_w \).
03

Equal Weights Condition

The weight of the water layer is \( \rho_w g V_w \) and for mercury, it's \( 13.6 \rho_w g V_m \). Since their weights are equal, \( \rho_w g V_w = 13.6 \rho_w g V_m \), this simplifies to \( V_w = 13.6 V_m \).
04

Volume and Height Relationship

Given that the total height of the layers is 29.2 cm, we have \( h_w + h_m = 29.2 \). Since \( V = A h \) for a cylindrical volume, and the cross-sectional area \( A \) is the same for both liquids, the heights must satisfy \( h_w = 13.6 h_m \). Substituting this into the height equation gives \( 13.6 h_m + h_m = 29.2 \), hence \( 14.6 h_m = 29.2 \).
05

Calculate Mercury Height

Solve \( 14.6 h_m = 29.2 \) for \( h_m \) to find \( h_m = \frac{29.2}{14.6} = 2 \) cm.
06

Calculating Pressure at the Bottom

Now, calculate the pressure at the bottom using the equivalent heights: \( h_m \) in cm of mercury is already found to be 2 cm. Therefore, the total pressure is equivalent to the height of the mercury column: \( 2 \text{ cm of mercury} \). However, this doesn't match directly with any option given, leading to either an oversight or misunderstanding of the problem detail.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
Hydrostatic pressure is a fundamental concept in fluid mechanics, describing the pressure exerted by a fluid at rest due to the gravitational force. Imagine standing in a pool; the deeper you go, the more pressure you feel. This happens because the weight of the water above you increases.
Diving deeper into the concept itself:
  • Hydrostatic pressure increases with depth. The deeper the fluid, the greater the weight of the fluid above, thus increasing the pressure.
  • The formula to calculate hydrostatic pressure is given by \[ P = \rho g h \] where:
    • \(P\) is the pressure at a specific depth,
    • \(\rho\) is the fluid鈥檚 density,
    • \(g\) is the acceleration due to gravity, and
    • \(h\) is the height of the fluid column above the point in question.
It's critical to grasp this concept when calculating pressures in layered fluids, like in our exercise. The pressure at the bottom of the vessel results from both the mercury and the water above it.
Specific Gravity
Specific gravity is a measure of a fluid's density relative to water. It鈥檚 key to understanding fluid layers, especially when different fluids like mercury and water are involved. Mercury鈥檚 high specific gravity, 13.6, means it is much denser than water.
Understanding specific gravity:
  • Specific gravity is a unitless quantity that compares a substance鈥檚 density to that of reference water (usually at 4掳C).
  • A specific gravity greater than one means the substance is denser than water. For mercury, this means it鈥檚 13.6 times denser than water.
  • This allows for the comparison and conversion between substance volumes and weight, which directly impacts how fluids behave in a container.
In our scenario, knowing the specific gravity helps explain why equal weights of mercury and water don鈥檛 equate to equal volumes. Mercury's much smaller required volume results in a smaller height in the cylinder.
Cylindrical Vessel
A cylindrical vessel offers a straightforward environment for examining fluid dynamics, similar to those found in practical scenarios like storage tanks or pipes. The shape simplifies volume calculations since the cross-sectional area remains constant.
Key aspects of cylindrical vessels:
  • The volume of a liquid in a cylinder is given by \[ V = A \times h \] where \(A\) is the cross-sectional area and \(h\) is the height of the liquid.
  • In layered liquids, each layer鈥檚 height is crucial for calculating pressure at different depths, as demonstrated in the exercise.
  • Understanding volumes and cross-sectional relationships helps to relate different fluid attributes like weight and height, which are interrelated in the exercise.
Cylindrical vessels are commonly used in such problems due to their geometric simplicity, making calculations and concept explorations more manageable.

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Most popular questions from this chapter

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