Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

An iron casting containing a number of cavities weighs 6000 N in air and 4000 N in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is $7.87 g / c m^{3}$ .

The total cavity volume in the casting is $0.126 m^{3}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The weight of iron casting in air, W=6000 N
The weight of iron casting in water, $W_{e f f} = 4000 N$
The density of solid iron, $ρ_{i r o n} = 7.87 \frac{g}{c m^{3}} o r 7.87 \times 10^{3} \frac{k g}{m^{3}}$

Step 2: Understanding the concept of Archimedes Principle

We can use the concept of Archimedes’ principle and expression of density. When an iron casting is fully submerged in water, then a buoyant force from the surrounding fluid acts on it. This force has a magnitude equal to the weight of the water displaced by the iron casting.

Formulae:

Force applied on body (or weight), $F_{b} = m_{f} g$ (i)

Density of a substance, $ρ = \frac{m}{V}$ (ii)

Step 3: Calculation of total volume of cavity

The volume of cavity is the difference between volume of casting and volume of iron. Hence,

$V_{c a v i t y} = V_{c a s t i n g} = V_{i r o n}$ (iii)

According to equation (ii), we get

The volume of iron as:

$V_{i r o n} = \frac{m_{i r o n}}{ρ_{i r o n}} = \frac{g m_{i r o n}}{g ρ_{i r o n}} = \frac{W}{g ρ_{i r o n}}$

When the iron casting is submerged in water, the forces acting on it are $F_{g} - F_{b}$ . Hence,the effective weight in water is given as:

$W_{e f f} = W - g ρ_{w} V_{c a s t i n g} V_{c a s t i n g} = \frac{W - W_{e f f}}{g ρ_{w}}$

Equation (iii) becomes and using the given values, we get

$V_{c a v i t y} = \frac{W - W_{e f f}}{g ρ_{w}} - \frac{W}{g ρ_{i r o n}} = (\frac{6000 N - 4000 N}{9.8 \frac{m}{s^{2}} \times 1000 \frac{k g}{m^{3}}}) - (\frac{6000 N}{9.8 \frac{m}{s^{2}} \times 7.87 \times 10^{3} \frac{k g}{m^{3}}}) = 0.126 m^{3}$

Hence, the volume of cavity is $0.126 m^{3}$

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Fundamentals Of Physics

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Short Answer

An iron casting containing a number of cavities weighs 6000 N in air and 4000 N in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is $7.87 g / c m^{3}$ .

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of Archimedes Principle

Step 3: Calculation of total volume of cavity

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Fundamentals Of Physics

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Short Answer

An iron casting containing a number of cavities weighs 6000 N in air and 4000 N in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is 7.87 g/cm3.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of Archimedes Principle

Step 3: Calculation of total volume of cavity

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An iron casting containing a number of cavities weighs 6000 N in air and 4000 N in water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is $7.87 g / c m^{3}$ .