/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 A hollow cylindrical container f... [FREE SOLUTION] | 91Ó°ÊÓ

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A hollow cylindrical container floats in water with half its length immersed. A liquid of specific gravity \(\sigma\) is slowly poured into the container until the levels of the liquio inside the container and of water outside are the same. It is observed that the container is submerged to two-thirds it length. Then the value of \(\sigma\) is (1) \(\frac{1}{2}\) (2) \(\frac{1}{3}\) (3) \(\frac{1}{4}\) (4) \(\frac{1}{5}\)

Short Answer

Expert verified
The specific gravity \( \sigma \) is \( \frac{1}{2} \) (option 1).

Step by step solution

01

Understand Initial Conditions

Initially, the cylinder floats in equilibrium with half of its length submerged in water. This implies that the buoyant force equals the weight of the container, which can be expressed as \( V_{submerged} \cdot \rho_{water} = W_{container} \). Since half its length is submerged, we can express the submerged volume as \( \frac{V}{2} \).
02

Set up Final State Conditions

After the liquid is added, both the inside liquid level and the water outside are equal, and the container is submerged two-thirds of its length. The buoyant force now includes the weight of the container and the weight of the liquid inside, so \( V'_{submerged} \cdot \rho_{water} = W_{container} + W_{liquid} \). The submerged volume is now \( \frac{2V}{3} \).
03

Define the Weight of Container and Liquid

Let \( V \) be the total volume of the container. Then initially, \( W_{container} = \frac{V}{2} \cdot \rho_{water} \). After adding the liquid, the weight of the liquid is \( \sigma \cdot \rho_{water} \cdot V_{half} = \sigma \cdot \rho_{water} \cdot \frac{V}{3} \) because two-thirds of the volume correlates with the weight of the fluid inside when container levels match external water.
04

Equate Final Buoyancy and Weights

The sum of the weight of the container and the liquid must equal the buoyant force due to the new submerged volume: \( \frac{2V}{3} \cdot \rho_{water} = \frac{V}{2} \cdot \rho_{water} + \sigma \cdot \rho_{water} \cdot \frac{V}{3} \). Cancel \( \rho_{water} \) and simplify: \( \frac{2V}{3} = \frac{V}{2} + \sigma \cdot \frac{V}{3} \).
05

Solve for \( \sigma \)

Rearrange the equation \( \frac{2}{3} = \frac{1}{2} + \sigma \cdot \frac{1}{3} \). Multiplying through by 3 to clear fractions gives \( 2 = \frac{3}{2} + \sigma \). Subtraction results in \( \sigma = 2 - \frac{3}{2} = \frac{1}{2} \).
06

Conclude the Correct Option

The calculated value of \( \sigma \) matches option (1) \( \frac{1}{2} \). Therefore, the specific gravity value required is \( \frac{1}{2} \).

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Key Concepts

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

Specific Gravity
Specific gravity is a crucial concept when discussing buoyancy and floating objects. It is defined as the ratio of the density of a substance to the density of a reference substance, typically, water. This ratio allows us to understand how heavy or light a substance is concerning water. Specific gravity is a dimensionless quantity because it is a ratio and therefore has no units. This property makes it quite handy in comparing the buoyancy of various materials without the need to focus on specific units.
  • If the specific gravity is less than 1, the substance is lighter than water and will float.
  • If it is exactly 1, the substance has the same density as water.
  • If greater than 1, the substance is denser and will sink.
In the problem, understanding specific gravity helps determine how the liquid added affects the buoyancy of the container. By calculating its value, you can assess the contribution of the liquid's weight to the immersed portion of the container.
Equilibrium in Fluids
Equilibrium in fluids occurs when the forces acting upon an object within a fluid are balanced. This typically involves the buoyant force and the gravitational force. An object floats when the buoyant force equals the gravitational force acting on it, often called "weight." In this state of equilibrium, the object neither sinks nor rises but stays at a consistent level within the fluid.
For a floating object like the cylindrical container in the problem, equilibrium is indicated by the level to which the container is submerged. Initially, the container floats stably with half of its length submerged, showing that the forces are balanced. The addition of liquid changes this balance, leading to a new state where the container is submerged to two-thirds of its length.
This adjustment showcases equilibrium, as the system reaches a new balance between the container's weight and the buoyant force. Observing how a system maintains or shifts equilibrium offers insight into fluid interactions and is a fundamental aspect of fluid dynamics.
Archimedes' Principle
Archimedes' Principle is the cornerstone of understanding buoyancy. It states that any object, wholly or partially submerged in a fluid, experiences a buoyant force equal to the weight of the fluid displaced by the object. This principle explains why objects sink, float, or remain neutrally buoyant in fluid environments.
  • When an object is fully submerged, the displaced fluid's weight equals the object's weight, causing it to float at a constant depth.
  • When partially submerged, the buoyant force corresponds to the volume of fluid displaced, which must balance the object's weight for it to float stably at the surface.
In the exercise, Archimedes' Principle helps us understand why the cylinder floats with half of its length submerged initially. As liquid is added, the system transitions to a new floating state where Archimedes' Principle still governs the relationship between displaced water weight and the cylinder's overall weight.
This principle provides a systematic method to address varying fluid mechanics problems involving floating or submerged objects, as seen when calculating the required specific gravity to achieve equilibrium after adding liquid to the container.

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Most popular questions from this chapter

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