Problem 4
Question
A U-tube contains water and methylated spirit separated by mereury. The mercury columns in the two arms are in level with \(10.0 \mathrm{~cm}\) of water in one arm and \(12.5 \mathrm{~cm}\) of spirit in the other. The specific gravity of spirit would be. (a) \(0.70\) (b) \(0.80\) (c) \(0.90\) (d) \(0.60\)
Step-by-Step Solution
Verified Answer
The specific gravity of the spirit is 0.80 (option b).
1Step 1: Understanding the Problem
The U-tube is balanced, meaning the pressure due to the water on one arm is equal to the pressure due to the spirit on the other arm. We need to find the specific gravity of the spirit.
2Step 2: Calculate Water Pressure
The pressure exerted by the column of water can be calculated using the formula: \( P = h \cdot \rho \cdot g \), where \( h = 10.0 \mathrm{~cm} = 0.10 \mathrm{~m} \), \( \rho \) is the density of water (\(1000 \mathrm{~kg/m^3}\)), and \( g = 9.8 \mathrm{~m/s^2} \). Thus, \( P_{water} = 0.10 \cdot 1000 \cdot 9.8 = 980 \mathrm{~Pa} \).
3Step 3: Calculate Spirit Pressure
Use the same pressure formula for the spirit: \( P = h \cdot \rho_{spirit} \cdot g \). Given \( h = 12.5 \mathrm{~cm} = 0.125 \mathrm{~m} \), the equation becomes \( 0.125 \cdot \rho_{spirit} \cdot 9.8 = 980 \mathrm{~Pa} \).
4Step 4: Solve for Spirit Density
Rearrange the equation from the spirit pressure to solve for the density of spirit: \( \rho_{spirit} = \frac{980}{0.125 \cdot 9.8} = 800 \mathrm{~kg/m^3} \).
5Step 5: Calculate Specific Gravity of Spirit
Specific gravity is the ratio of the density of the spirit to the density of water. Therefore, \( \text{Specific Gravity} = \frac{\rho_{spirit}}{\rho_{water}} = \frac{800}{1000} = 0.80 \).
Key Concepts
Understanding Hydrostatic PressureUnderstanding Specific GravityBasics of Density Calculations
Understanding Hydrostatic Pressure
Hydrostatic pressure is an important concept in fluid dynamics. It refers to the pressure exerted by a fluid at equilibrium due to the force of gravity. When dealing with fluids in a U-tube, like water and methylated spirit separated by mercury, hydrostatic pressure is crucial for understanding fluid behavior.
The hydrostatic pressure can be calculated using the formula:
The hydrostatic pressure can be calculated using the formula:
- \( P = h \cdot \rho \cdot g \)
- \( P \) stands for pressure,
- \( h \) is the height of the fluid column,
- \( \rho \) represents the fluid's density, and
- \( g \) is the acceleration due to gravity, typically \( 9.8 \ \text{m/s}^2 \).
Understanding Specific Gravity
Specific gravity is an essential concept in fluid dynamics that allows for comparisons between the densities of different substances. It is a ratio, meaning it does not have any units, which makes it very useful.
Specific gravity is defined as:
Specific gravity is defined as:
- \( \text{Specific Gravity} = \frac{\rho_{\text{object}}}{\rho_{\text{reference}}} \)
- \( \rho_{\text{object}} \) is the density of the substance in question,
- \( \rho_{\text{reference}} \) is the density of the reference substance (usually water for liquids).
Basics of Density Calculations
Density plays a key role in understanding how different substances interact in fluid dynamics. Density is defined as mass per unit volume. It is often expressed in units of kg/m³.
To calculate density, use the formula:
To calculate density, use the formula:
- \( \rho = \frac{m}{V} \)
- \( \rho \) is the density,
- \( m \) is the mass, and
- \( V \) is the volume of the substance.
Other exercises in this chapter
Problem 2
A \(50 \mathrm{~kg}\) girl wearing high heel shoes balances on a single heel. If the heel is circular with a diameter \(1.0 \mathrm{~cm}\). What is the pressure
View solution Problem 3
The surface area of air bubble increases four times when it rises from bottom to top of a water tank where the temperature is uniform. If the atmospheric pressu
View solution 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
View solution Problem 7
A uniform rod of density \(\rho\) is placed in a wide tank containing a liquid o \((\sigma>\rho)\). The depth of liquid in the tank is half the length of the ro
View solution