/*! 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 67 A body weigh \(50 \mathrm{~g}\) ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 67

A body weigh \(50 \mathrm{~g}\) in air and \(40 \mathrm{~g}\) in water. How much would it weight in a liquid of specific gravity \(1.5 ?\) |Karmataka CET 2008] [al \(65 \mathrm{~g}\) (b) \(45 \mathrm{~g}\) (c) \(30 \mathrm{~g}\) (d) \(35 \mathrm{~g}\)

Short Answer

Expert verified
The weight of the body in a liquid with specific gravity 1.5 is 35 g.

Step by step solution

01

Understand Buoyant Force

The weight of the body in air is 50 g, and in water it is 40 g. The difference, 10 g, is the buoyant force applied by the water. This force equals the weight of the water displaced by the body.
02

Calculate Volume of Displaced Water

Since the buoyant force in water is equal to the weight of the water displaced, we know the volume of water displaced also corresponds to 10 g of water, given water's specific gravity is 1 (since 1 g/cm鲁).
03

Find Buoyant Force in the Liquid

In a liquid with specific gravity 1.5, the weight of the liquid displaced would be 1.5 times the weight of the water displaced, meaning the buoyant force is 1.5 * 10 g = 15 g.
04

Calculate Weight in the Liquid

If the buoyant force in the liquid is 15 g, the weight of the body in the liquid is given by the weight in air minus this buoyant force: 50 g - 15 g = 35 g.

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Key Concepts

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

Buoyant Force
When an object is submerged in a fluid, it experiences an upward force called the buoyant force. This force is responsible for making objects feel lighter in water. Archimedes' Principle states that the buoyant force on an object is equal to the weight of the fluid that the object displaces.
In the given exercise, a body weighs 50 g in air but only 40 g in water. This 10 g difference represents the buoyant force exerted by the water, indicating that the object displaces a volume of water weighing 10 g. Understanding this concept is crucial because it helps determine how much less an object seems to weigh in a liquid compared to in air. The displaced fluid's weight plays a key role here, simplifying the calculations for weight changes in different fluids.
Specific Gravity
Specific gravity is a measure of how dense a substance is compared to water. It is a dimensionless number, which means it has no units, calculated as the ratio of the density of a substance to the density of water. For instance, water has a specific gravity of 1.
If a liquid has a specific gravity of 1.5, it indicates that the liquid is 1.5 times denser than water. This concept is important in our exercise because it directly affects the buoyant force experienced by the object. The higher the specific gravity, the greater the weight of the liquid displaced by a given volume of the object. As seen in the exercise, the object in a liquid with a specific gravity of 1.5 displaces more weight than it does in water, leading to a larger buoyant force.
Weight Measurement in Liquids
When measuring the weight of an object in different liquids, it is crucial to consider the buoyant force. This force reduces the object's apparent weight when submerged. To find the weight in a particular liquid, you subtract the buoyant force from the object's actual weight in air.
In our example, the object weighs 50 g in air. The buoyant force in another liquid with a specific gravity of 1.5 is calculated as follows:
  • Original water buoyant force = 10 g
  • New liquid buoyant force = 1.5 times water buoyant force
  • New buoyant force = 15 g
Thus, the object's weight in this denser liquid becomes 50 g - 15 g = 35 g. Understanding how to adjust for these forces allows for accurate weight measurement across different liquids, which is an important skill in many scientific and engineering applications.

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Most popular questions from this chapter

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