/*! 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 41 When a body of density \(\rho\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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When a body of density \(\rho\) and volume \(V\) is lloating in a liquid of density \(\sigma\) (a) its true weight is \(V \rho g\) (b) loss in its weight is \(V \sigma g\) (c) its apparent weight is zero (d) its density \(\rho\) is lesser than that of liquid \(\sigma\)

Short Answer

Expert verified
All statements (a), (b), (c), and (d) are correct.

Step by step solution

01

Understanding the Problem

We need to determine the properties of a body floating in a liquid with given densities for the body and the liquid. This involves analyzing various physical conditions and verifying the given statements about the weight and density relationships between the body and the liquid.
02

Evaluate True Weight

The true weight of the body is the force due to gravity acting on its mass. The formula for weight is: \[ \text{True Weight} = V \rho g \]where \(V\) is the volume, \(\rho\) is the density of the body, and \(g\) is the acceleration due to gravity. Thus, statement (a) is correct.
03

Evaluate Loss in Weight

Archimedes' principle states that the buoyant force on a submerged or floating object is equal to the weight of the fluid displaced by the object. Therefore, the loss in weight due to buoyancy is given by:\[ \text{Loss in Weight} = V \sigma g \]where \(\sigma\) is the density of the liquid. Thus, statement (b) is also correct.
04

Evaluate Apparent Weight

When the body is floating, the buoyant force equals the weight of the object, making the net force zero. Therefore, the apparent weight of the floating body is zero. So, statement (c) is correct.
05

Evaluate Density Comparison

For a body to float, its average density must be less than the density of the liquid it is in. Hence, we must have \(\rho < \sigma\). This means statement (d) is correct.

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Key Concepts

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

Density
Density is a fundamental concept in physics that describes how compact a substance is. It's defined as mass per unit volume and is often represented by the Greek letter \( \rho \) (rho). You can calculate density using the formula:
  • \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
This means that density tells us how much matter exists in a given amount of space. In the context of floating bodies, understanding the density of both the floating object and the liquid it is in is crucial.

For a body to stay afloat, its density must be less than the density of the liquid. This principle is used daily in designing ships and understanding how various objects behave in water.
Buoyant Force
The buoyant force is an upward force exerted by a fluid, opposing the weight of an object that is either partially or fully immersed in it. This principle is central to understanding why some objects float and others do not. According to Archimedes' Principle,
  • The buoyant force is equal to the weight of the fluid displaced by the object.
Mathematically, it is represented as:
  • \( \text{Buoyant Force} = V \sigma g \)
where \( V \) is the volume of the fluid displaced, \( \sigma \) is the density of the fluid, and \( g \) is the acceleration due to gravity.

This force is what gives objects the ability to float. If the buoyant force is greater than or equal to the weight of the object, it will float. This is why heavier-than-water objects can still float if they displace enough water to equal their weight.
Apparent Weight
Apparent weight refers to the perceived weight of an object when it is submerged in a fluid, and it can change due to the buoyant force. When you're swimming or standing in water, you may notice that you feel lighter. This sensation is due to the buoyant force working against gravity.

For a floating object, the buoyant force is in perfect balance with the object's weight, resulting in an apparent weight of zero. Therefore, while a body is floating, it feels as if it doesn't weigh anything. This is why statement (c) in the exercise, which claims that the apparent weight is zero, is indeed correct.

Understanding apparent weight helps in designing things like ships and submarines, where balance and buoyancy are key considerations.
True Weight
True weight is the actual weight of an object regardless of the medium around it. It is the force exerted by gravity on that object. The true weight can be calculated using the formula:
  • \( \text{True Weight} = V \rho g \)
Here, \( V \) is the volume of the object, \( \rho \) is the object's density, and \( g \) is the acceleration due to gravity.

True weight is important for determining whether an object will float or sink when placed in a fluid. In the exercise, the true weight is part of the determining factor for the balance of forces responsible for the object's buoyancy. However, it's important to note that the true weight never changes while the object's apparent weight can vary based on its environment.

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