StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q45P

Expert-verifiedFound in: Page 409

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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** ${\mathbf{7}}{\mathbf{.}}{\mathbf{87}}{\mathbf{}}{\mathit{g}}{\mathbf{/}}{\mathit{c}}{{\mathit{m}}}^{{\mathbf{3}}}$**.**

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

- The weight of iron casting in air, W=6000 N
- The weight of iron casting in water, ${W}_{eff}=4000N$
- The density of solid iron, ${\rho}_{iron}=7.87\frac{g}{c{m}^{3}}or7.87\times {10}^{3}\frac{kg}{{m}^{3}}$

**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, $\mathrm{\rho}=\frac{\mathrm{m}}{\mathrm{V}}$ (ii)

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

${V}_{cavity}={V}_{casting}={V}_{iron}$ (iii)

According to equation (ii), we get

The volume of iron as:

${V}_{iron}=\frac{{m}_{iron}}{{\rho}_{iron}}\phantom{\rule{0ex}{0ex}}=\frac{g{m}_{iron}}{g{\rho}_{iron}}\phantom{\rule{0ex}{0ex}}=\frac{W}{g{\rho}_{iron}}$

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:

${W}_{eff}=W-g{\rho}_{w}{V}_{casting}\phantom{\rule{0ex}{0ex}}{V}_{casting}=\frac{W-{W}_{eff}}{g{\rho}_{w}}$

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

${V}_{cavity}=\frac{W-{W}_{eff}}{g{\rho}_{w}}-\frac{W}{g{\rho}_{iron}}\phantom{\rule{0ex}{0ex}}=\left(\frac{6000N-4000N}{9.8\frac{m}{{s}^{2}}\times 1000\frac{kg}{{m}^{3}}}\right)-\left(\frac{6000N}{9.8\frac{m}{{s}^{2}}\times 7.87\times {10}^{3}\frac{kg}{{m}^{3}}}\right)\phantom{\rule{0ex}{0ex}}=0.126{m}^{3}$

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

94% of StudySmarter users get better grades.

Sign up for free