Chapter 8: Mechanical Properties of Solids (Properties of Bulk Matter)

Same load F and same extension ΔL, so $Y=\frac{FL}{A\Delta L}\propto \frac{L}{A}$:

$$\frac{Y_s}{Y_c}=\frac{L_s}{A_s}\cdot \frac{A_c}{L_c}=\frac{4.7}{3.0\times 10^{-5}}\times \frac{4.0\times 10^{-5}}{3.5}=\frac{18.8}{10.5}=1.79$$

About 1.8

(a) From the straight part: $Y=\frac{\text{stress}}{\text{strain}}=\frac{150\times 10^6}{0.002}=7.5\times 10^{10}{\text{N m}}^{-2}$.

(b) The graph stops being linear at about $300\times 10^6$ N m⁻², so the yield strength is about $3\times 10^8$ N m⁻².

(a) 7.5 × 10¹⁰ N m⁻² (b) about 3 × 10⁸ N m⁻²

(a) Young's modulus is the slope of the linear part. A's initial line is steeper, so A has the greater Young's modulus.

(b) Strength is the stress needed to break the material. A's curve reaches a higher stress before fracture, so A is stronger.

(a) A (b) A

(a) False. For the same stress rubber stretches far more than steel; Young's modulus of steel ($2\times 10^{11}$ Pa) is much larger.

(b) True. When a coil spring is stretched, its wire is twisted (sheared), not lengthened; so the stretching depends on the shear modulus.

(a) False (b) True

$A=\pi r^2=3.14\times (0.125\times 10^{-2}{)}^2=4.91\times 10^{-6}{\text{m}}^2$.

Steel carries both masses: $F_s=(4+6)\times 9.8=98$ N. Brass carries 6 kg: $F_b=58.8$ N.

$\Delta L_s=\frac{F_s{L}_s}{A{Y}_s}=\frac{98\times 1.5}{4.91\times 10^{-6}\times 2.0\times 10^{11}}=1.5\times 10^{-4}$ m.

$\Delta L_b=\frac{58.8\times 1.0}{4.91\times 10^{-6}\times 0.91\times 10^{11}}=1.3\times 10^{-4}$ m.

Steel 1.5 × 10⁻⁴ m; brass 1.3 × 10⁻⁴ m

Shearing force $F=100\times 9.8=980$ N on face area $A=(0.10{)}^2=0.01{\text{m}}^2$; $L=0.10$ m.

Shear strain $=\frac{F/A}{G}=\frac{9.8\times 10^4}{25\times 10^9}=3.92\times 10^{-6}$.

Deflection $\Delta x=\text{strain}\times L=3.92\times 10^{-6}\times 0.10=3.92\times 10^{-7}$ m.

3.92 × 10⁻⁷ m (the shear strain is 3.9 × 10⁻⁶, the value printed in the NCERT key)

Load on each column $=\frac{50000\times 9.8}{4}=1.225\times 10^5$ N.

Area $=\pi (R^2-r^2)=3.14\times (0.36-0.09)=0.848{\text{m}}^2$.

Strain $=\frac{F}{AY}=\frac{1.225\times 10^5}{0.848\times 2.0\times 10^{11}}=7.22\times 10^{-7}$.

7.22 × 10⁻⁷ per column (the NCERT key's 2.8 × 10⁻⁶ is the value if one column carried the whole load)

$A=15.2\times 10^{-3}\times 19.1\times 10^{-3}=2.90\times 10^{-4}{\text{m}}^2$; stress $=\frac{44500}{2.90\times 10^{-4}}=1.53\times 10^8{\text{N m}}^{-2}$.

Strain $=\frac{\text{stress}}{Y}=\frac{1.53\times 10^8}{1.1\times 10^{11}}=1.4\times 10^{-3}$.

About 1.4 × 10⁻³ (using Y = 110 GPa for copper)

$A=\pi (0.015{)}^2=7.07\times 10^{-4}{\text{m}}^2$; $F_{\max }=10^8\times 7.07\times 10^{-4}$.

7.07 × 10⁴ N

The bar is rigid and symmetric, so all three wires stretch by the same ΔL; they have the same length and the same tension F. From $Y=\frac{FL}{A\Delta L}$, $A\propto \frac{1}{Y}$, so $D^2\propto \frac{1}{Y}$:

$$\frac{D_{\text{Cu}}}{D_{\text{Fe}}}=\sqrt{\frac{Y_{\text{Fe}}}{Y_{\text{Cu}}}}=\sqrt{\frac{190}{110}}=1.31$$

D_copper : D_iron ≈ 1.31 (the NCERT key's 1.25 uses Y_Cu = 120 GPa)

At the lowest point the tension supports the weight and provides the centripetal force: $T=m(g+{\omega }^{2}r)$.

$\omega =2\text{rev/s}=4\pi {\text{rad s}}^{-1}$, ${\omega }^{2}r=157.9{\text{m s}}^{-2}$, so $T=14.5\times (9.8+157.9)=2432$ N.

$\Delta L=\frac{TL}{AY}=\frac{2432\times 1.0}{0.065\times 10^{-4}\times 2.0\times 10^{11}}=\frac{2432}{1.3\times 10^6}=1.87\times 10^{-3}$ m.

1.87 × 10⁻³ m (the NCERT key's 1.539 × 10⁻⁴ m comes from taking ω = 2 rad/s)

$\Delta p=100\times 1.013\times 10^5=1.013\times 10^7$ Pa; $\frac{|\Delta V|}{V}=\frac{0.5}{100}=5\times 10^{-3}$.

$B=\frac{1.013\times 10^7}{5\times 10^{-3}}$

2.026 × 10⁹ Pa

For air at constant temperature, $B=p=1.013\times 10^5$ Pa. So $\frac{B_{\text{water}}}{B_{\text{air}}}=\frac{2.026\times 10^9}{1.013\times 10^5}=2\times 10^4$.

In a liquid the molecules are closely packed and strongly bound, so it is very hard to compress. In a gas the molecules are far apart with weak forces between them, so it compresses easily.

About 2 × 10⁴; liquids are far less compressible than gases.

$\Delta p=80\times 1.013\times 10^5=8.10\times 10^6$ Pa; $\frac{\Delta V}{V}=\frac{\Delta p}{B}=\frac{8.10\times 10^6}{2.2\times 10^9}=3.68\times 10^{-3}$.

Mass is unchanged, so ${\rho }^{\prime }=\frac{\rho }{1-\Delta V/V}=\frac{1.03\times 10^3}{1-0.00368}$.

1.034 × 10³ kg m⁻³

$\frac{\Delta V}{V}=\frac{p}{B}=\frac{10\times 1.013\times 10^5}{37\times 10^9}=2.74\times 10^{-5}$.

2.7 × 10⁻⁵ (i.e. 0.0027 %)

$V=(0.10{)}^3=10^{-3}{\text{m}}^3$. $\Delta V=\frac{pV}{B}=\frac{7.0\times 10^6\times 10^{-3}}{140\times 10^9}=5.0\times 10^{-8}{\text{m}}^3$.

0.05 cm³ (the NCERT key's 0.058 cm³ uses B = 120 GPa)

$\frac{\Delta V}{V}=0.001$, so $\Delta p=B\times \frac{\Delta V}{V}=2.2\times 10^9\times 0.001$.

2.2 × 10⁶ Pa