Mechanical Properties of Solids

Given:
- Length of steel wire: $L_s = 4.7$ m
- Cross-sectional area of steel wire: $A_s = 3.0 \times 10^{-5}$ m²
- Length of copper wire: $L_c = 3.5$ m
- Cross-sectional area of copper wire: $A_c = 4.0 \times 10^{-5}$ m²
- Both wires stretch by the same amount ($\Delta L$) under the same load ($F$).

Concept/Formula:
$$Y = \frac{F \cdot L}{A \cdot \Delta L}$$

For steel:
$$Y_s = \frac{F \cdot L_s}{A_s \cdot \Delta L}$$

For copper:
$$Y_c = \frac{F \cdot L_c}{A_c \cdot \Delta L}$$

Working:
Dividing the two expressions:
$$\frac{Y_s}{Y_c} = \frac{F \cdot L_s}{A_s \cdot \Delta L} \times \frac{A_c \cdot \Delta L}{F \cdot L_c}$$

$$\frac{Y_s}{Y_c} = \frac{L_s}{L_c} \times \frac{A_c}{A_s}$$

$$\frac{Y_s}{Y_c} = \frac{4.7}{3.5} \times \frac{4.0 \times 10^{-5}}{3.0 \times 10^{-5}}$$

$$\frac{Y_s}{Y_c} = \frac{4.7}{3.5} \times \frac{4.0}{3.0} = 1.343 \times 1.333 \approx 1.79$$

Answer: The ratio of Young's modulus of steel to that of copper is approximately $\boxed{1.79}$.

Note: From the standard Fig. 8.9 in NCERT, the stress-strain graph shows a linear region up to a stress of about $150 \times 10^6$ N m⁻² at a strain of $0.002$ (i.e., $2 \times 10^{-3}$), and the curve departs from linearity (yield point) at approximately $300 \times 10^6$ N m⁻².

(a) Young's Modulus:

Concept: Young's modulus is the slope of the linear (elastic) portion of the stress-strain curve.

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{150 \times 10^6 \text{ N m}^{-2}}{0.002}$$

$$\boxed{Y = 7.5 \times 10^{10} \text{ N m}^{-2}}$$

(b) Approximate Yield Strength:

The yield strength is the stress at which the material begins to deform plastically (where the curve departs from linearity).

From the graph:
$$\boxed{\text{Yield Strength} \approx 300 \times 10^6 \text{ N m}^{-2} = 3 \times 10^8 \text{ N m}^{-2}}$$

Note: From the standard NCERT Fig. 8.10, material A has a steeper slope in the linear region, and material B has a higher stress at the fracture/breaking point.

(a) Greater Young's Modulus:

Concept: Young's modulus = slope of the linear portion of the stress-strain curve. A steeper slope means a greater Young's modulus.

Since material A has a greater slope (steeper linear region) than material B, material A has the greater Young's modulus.

(b) Stronger Material:

Concept: The strength of a material is determined by the stress it can withstand before fracture (breaking stress). The material that fractures at a higher stress value is stronger.

From the graph, material B fractures at a higher value of stress than material A. Therefore, material B is the stronger material.

(a) The Young's modulus of rubber is greater than that of steel.

Answer: FALSE

Reason: Young's modulus measures the resistance of a material to elastic deformation under longitudinal stress. Steel requires a very large force to produce a small elongation, while rubber stretches easily under a small force. Therefore, steel has a much larger Young's modulus ($Y_{\text{steel}} \approx 2 \times 10^{11}$ N m⁻²) compared to rubber ($Y_{\text{rubber}} \approx 10^6$ – $10^7$ N m⁻²). A material that stretches less for a given load is considered more elastic (has higher Young's modulus).

(b) The stretching of a coil is determined by its shear modulus.

Answer: TRUE

Reason: When a coil (helical spring) is stretched or compressed, the individual elements of the wire making up the coil are not under simple tension or compression — they undergo twisting, which is a shear deformation. Therefore, the stretching (extension) of a coil spring is governed by the shear modulus (modulus of rigidity) of the material, not by the Young's modulus.

Given:
- Diameter of each wire: $d = 0.25$ cm $= 0.25 \times 10^{-2}$ m
- Radius: $r = 0.125 \times 10^{-2}$ m
- Cross-sectional area: $A = \pi r^2 = \pi \times (0.125 \times 10^{-2})^2 = \pi \times 1.5625 \times 10^{-6}$
$$A = 4.91 \times 10^{-6} \text{ m}^2$$
- Length of steel wire: $L_s = 1.5$ m
- Length of brass wire: $L_b = 1.0$ m
- $Y_{\text{steel}} = 2.0 \times 10^{11}$ N m⁻²
- $Y_{\text{brass}} = 0.91 \times 10^{11}$ N m⁻²

Note: From Fig. 8.11 (standard NCERT figure), the steel wire supports a load of 6 kg (from the 6 kg mass) plus the 4 kg mass below it, so total load on steel = 10 kg. The brass wire supports only the 6 kg mass.

Actually, from the standard figure: Steel wire is at the top, carrying both masses (6 kg + 4 kg = 10 kg). Brass wire is below, carrying only 6 kg.

Load on steel wire: $F_s = (4 + 6) \times 9.8 = 10 \times 9.8 = 98$ N

Load on brass wire: $F_b = 6 \times 9.8 = 58.8$ N

Formula: $\Delta L = \frac{F L}{A Y}$

Elongation of steel wire:
$$\Delta L_s = \frac{F_s \cdot L_s}{A \cdot Y_s} = \frac{98 \times 1.5}{4.91 \times 10^{-6} \times 2.0 \times 10^{11}}$$

$$\Delta L_s = \frac{147}{9.82 \times 10^{5}} = 1.497 \times 10^{-4} \text{ m}$$

$$\boxed{\Delta L_s \approx 1.5 \times 10^{-4} \text{ m}}$$

Elongation of brass wire:
$$\Delta L_b = \frac{F_b \cdot L_b}{A \cdot Y_b} = \frac{58.8 \times 1.0}{4.91 \times 10^{-6} \times 0.91 \times 10^{11}}$$

$$\Delta L_b = \frac{58.8}{4.468 \times 10^{5}} = 1.316 \times 10^{-4} \text{ m}$$

$$\boxed{\Delta L_b \approx 1.3 \times 10^{-4} \text{ m}}$$

Given:
- Edge of cube: $L = 10$ cm $= 0.1$ m
- Mass attached: $m = 100$ kg
- Shear modulus of aluminium: $G = 25$ GPa $= 25 \times 10^9$ N m⁻²
- $g = 9.8$ m s⁻²

Concept/Formula:
Shear stress $= \frac{F}{A}$, Shear strain $= \frac{\Delta L}{L}$

$$G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\Delta L / L}$$

$$\Delta L = \frac{F \cdot L}{A \cdot G}$$

Working:
- Force applied (weight): $F = mg = 100 \times 9.8 = 980$ N
- Area of the face: $A = L^2 = (0.1)^2 = 0.01$ m²
- Height of cube (= $L$): $L = 0.1$ m

$$\Delta L = \frac{980 \times 0.1}{0.01 \times 25 \times 10^9}$$

$$\Delta L = \frac{98}{2.5 \times 10^8} = 3.92 \times 10^{-7} \text{ m}$$

Answer: The vertical deflection of the face is $\boxed{\Delta L \approx 3.92 \times 10^{-7} \text{ m}}$.

Given:
- Total mass of structure: $M = 50{,}000$ kg
- Number of columns: $n = 4$
- Inner radius: $r = 30$ cm $= 0.30$ m
- Outer radius: $R = 60$ cm $= 0.60$ m
- Young's modulus of mild steel: $Y = 2 \times 10^{11}$ N m⁻²
- $g = 9.8$ m s⁻²

Working:

Load on each column:
$$F = \frac{Mg}{4} = \frac{50000 \times 9.8}{4} = \frac{490000}{4} = 1.225 \times 10^5 \text{ N}$$

Cross-sectional area of each hollow column:
$$A = \pi(R^2 - r^2) = \pi\left[(0.60)^2 - (0.30)^2\right]$$
$$A = \pi[0.36 - 0.09] = \pi \times 0.27 = 0.8482 \text{ m}^2$$

Compressional stress on each column:
$$\text{Stress} = \frac{F}{A} = \frac{1.225 \times 10^5}{0.8482} = 1.445 \times 10^5 \text{ N m}^{-2}$$

Compressional strain:
$$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{1.445 \times 10^5}{2 \times 10^{11}}$$

$$\boxed{\text{Strain} = 7.22 \times 10^{-7}}$$

Given:
- Cross-section dimensions: $15.2$ mm $\times$ $19.1$ mm
- Tensile force: $F = 44{,}500$ N
- Young's modulus of copper: $Y = 1.2 \times 10^{11}$ N m⁻²

Working:

Cross-sectional area:
$$A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 290.32 \times 10^{-6} \text{ m}^2$$
$$A = 2.9032 \times 10^{-4} \text{ m}^2$$

Tensile stress:
$$\text{Stress} = \frac{F}{A} = \frac{44500}{2.9032 \times 10^{-4}} = 1.532 \times 10^{8} \text{ N m}^{-2}$$

Resulting strain:
$$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{1.532 \times 10^8}{1.2 \times 10^{11}}$$

$$\boxed{\text{Strain} = 1.277 \times 10^{-3} \approx 1.28 \times 10^{-3}}$$

Given:
- Radius of steel cable: $r = 1.5$ cm $= 1.5 \times 10^{-2}$ m
- Maximum allowable stress: $\sigma_{\max} = 10^8$ N m⁻²

Working:

Cross-sectional area of cable:
$$A = \pi r^2 = \pi \times (1.5 \times 10^{-2})^2 = \pi \times 2.25 \times 10^{-4}$$
$$A = 7.069 \times 10^{-4} \text{ m}^2$$

Maximum load:
$$F_{\max} = \sigma_{\max} \times A = 10^8 \times 7.069 \times 10^{-4}$$

$$\boxed{F_{\max} = 7.07 \times 10^4 \text{ N}}$$

This is the maximum load (force) the cable can support.

Given:
- Mass of bar: $M = 15$ kg (supported symmetrically, so each wire has the same tension $T$)
- Length of each wire: $L = 2.0$ m
- End wires: copper; Middle wire: iron
- Each wire has the same tension.
- $Y_{\text{iron}} = 1.9 \times 10^{11}$ N m⁻²
- $Y_{\text{copper}} = 1.2 \times 10^{11}$ N m⁻²

Concept:
Since the bar is rigid and supported symmetrically, all three wires have the same length and must undergo the same elongation $\Delta L$.

For each wire: $\Delta L = \frac{T L}{A Y}$

Since $T$, $L$, and $\Delta L$ are the same for all wires:
$$\frac{T L}{A_{\text{iron}} Y_{\text{iron}}} = \frac{T L}{A_{\text{copper}} Y_{\text{copper}}}$$

$$A_{\text{iron}} \cdot Y_{\text{iron}} = A_{\text{copper}} \cdot Y_{\text{copper}}$$

$$\frac{A_{\text{iron}}}{A_{\text{copper}}} = \frac{Y_{\text{copper}}}{Y_{\text{iron}}}$$

Since $A = \frac{\pi d^2}{4}$:
$$\frac{d_{\text{iron}}^2}{d_{\text{copper}}^2} = \frac{Y_{\text{copper}}}{Y_{\text{iron}}} = \frac{1.2 \times 10^{11}}{1.9 \times 10^{11}} = \frac{1.2}{1.9}$$

$$\frac{d_{\text{iron}}}{d_{\text{copper}}} = \sqrt{\frac{1.2}{1.9}} = \sqrt{0.6316} = 0.794$$

$$\boxed{\frac{d_{\text{iron}}}{d_{\text{copper}}} \approx 1 : 1.26}$$

Or equivalently, $d_{\text{copper}} : d_{\text{iron}} \approx 1.26 : 1$.

Given:
- Mass: $m = 14.5$ kg
- Unstretched length of steel wire: $L = 1.0$ m
- Angular velocity: $\omega = 2$ rev/s $= 2 \times 2\pi$ rad/s $= 4\pi$ rad/s
- Cross-sectional area: $A = 0.065$ cm² $= 0.065 \times 10^{-4}$ m² $= 6.5 \times 10^{-6}$ m²
- Young's modulus of steel: $Y = 2 \times 10^{11}$ N m⁻²
- $g = 9.8$ m s⁻²

Concept:
At the lowest point of the vertical circle, the net upward force provides centripetal acceleration. The tension $T$ in the wire must support the weight and provide centripetal force:

$$T - mg = m\omega^2 L$$

$$T = m(g + \omega^2 L)$$

Working:
$$\omega^2 = (4\pi)^2 = 16\pi^2 \approx 157.9 \text{ rad}^2\text{s}^{-2}$$

$$T = 14.5 \times (9.8 + 157.9 \times 1.0)$$
$$T = 14.5 \times (9.8 + 157.9)$$
$$T = 14.5 \times 167.7 = 2431.65 \text{ N}$$

Elongation:
$$\Delta L = \frac{T \cdot L}{A \cdot Y} = \frac{2431.65 \times 1.0}{6.5 \times 10^{-6} \times 2 \times 10^{11}}$$

$$\Delta L = \frac{2431.65}{1.3 \times 10^{6}} = 1.87 \times 10^{-3} \text{ m}$$

$$\boxed{\Delta L \approx 1.87 \times 10^{-3} \text{ m} \approx 1.87 \text{ mm}}$$

Given:
- Initial volume: $V = 100.0$ litre $= 100.0 \times 10^{-3}$ m³
- Final volume: $V' = 100.5$ litre $= 100.5 \times 10^{-3}$ m³
- Pressure increase: $\Delta p = 100.0$ atm $= 100 \times 1.013 \times 10^5 = 1.013 \times 10^7$ Pa

Working:

Change in volume:
$$\Delta V = V' - V = 100.5 - 100.0 = 0.5 \text{ litre} = 0.5 \times 10^{-3} \text{ m}^3$$

Volume strain:
$$\frac{\Delta V}{V} = \frac{0.5 \times 10^{-3}}{100.0 \times 10^{-3}} = 5 \times 10^{-3}$$

Bulk modulus of water:
$$B = \frac{\Delta p}{\Delta V / V} = \frac{1.013 \times 10^7}{5 \times 10^{-3}}$$

$$\boxed{B_{\text{water}} = 2.026 \times 10^9 \text{ N m}^{-2} \approx 2.026 \text{ GPa}}$$

Bulk modulus of air (at constant temperature, isothermal): $B_{\text{air}} = P_{\text{atm}} = 1.013 \times 10^5$ N m⁻²

Ratio:
$$\frac{B_{\text{water}}}{B_{\text{air}}} = \frac{2.026 \times 10^9}{1.013 \times 10^5} \approx 2 \times 10^4$$

Explanation: The ratio is very large because air is highly compressible — its molecules are far apart and can be pushed closer together easily. Water molecules, on the other hand, are already closely packed (intermolecular distances are small), and the strong intermolecular repulsive forces resist compression strongly. Hence, water requires a much larger pressure to achieve the same fractional change in volume as air.

Given:
- Pressure at depth: $p = 80.0$ atm $= 80.0 \times 1.013 \times 10^5 = 8.104 \times 10^6$ Pa
- Surface density: $\rho_0 = 1.03 \times 10^3$ kg m⁻³
- Bulk modulus of water: $B = 2.2 \times 10^9$ N m⁻² (standard value)

Concept:
$$B = \frac{\Delta p}{\Delta V / V}$$

$$\frac{\Delta V}{V} = \frac{\Delta p}{B} = \frac{8.104 \times 10^6}{2.2 \times 10^9} = 3.684 \times 10^{-3}$$

Working:
Let $V_0$ be the initial volume and $V$ be the volume at depth.
$$V = V_0 - \Delta V = V_0\left(1 - \frac{\Delta V}{V_0}\right) = V_0(1 - 3.684 \times 10^{-3})$$

Since mass is constant: $\rho V = \rho_0 V_0$

$$\rho = \frac{\rho_0 V_0}{V} = \frac{\rho_0}{1 - \Delta V/V_0}$$

$$\rho = \frac{1.03 \times 10^3}{1 - 3.684 \times 10^{-3}} = \frac{1.03 \times 10^3}{0.99632}$$

$$\boxed{\rho \approx 1.034 \times 10^3 \text{ kg m}^{-3}}$$

Given:
- Hydraulic pressure: $p = 10$ atm $= 10 \times 1.013 \times 10^5 = 1.013 \times 10^6$ N m⁻²
- Bulk modulus of glass: $B = 37 \times 10^9$ N m⁻² (standard value)

Formula:
$$\frac{\Delta V}{V} = \frac{p}{B}$$

Working:
$$\frac{\Delta V}{V} = \frac{1.013 \times 10^6}{37 \times 10^9}$$

$$\frac{\Delta V}{V} = 2.74 \times 10^{-5}$$

$$\boxed{\frac{\Delta V}{V} \approx 2.74 \times 10^{-5}}$$

The fractional change in volume is $2.74 \times 10^{-5}$ (a decrease).

Given:
- Edge of copper cube: $a = 10$ cm $= 0.1$ m
- Hydraulic pressure: $p = 7.0 \times 10^6$ Pa
- Bulk modulus of copper: $B = 140 \times 10^9$ N m⁻² $= 1.4 \times 10^{11}$ N m⁻²

Working:

Initial volume:
$$V = a^3 = (0.1)^3 = 10^{-3} \text{ m}^3$$

Volume contraction:
$$\Delta V = \frac{p \cdot V}{B} = \frac{7.0 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}}$$

$$\Delta V = \frac{7.0 \times 10^3}{1.4 \times 10^{11}} = 5.0 \times 10^{-8} \text{ m}^3$$

$$\boxed{\Delta V = 5.0 \times 10^{-8} \text{ m}^3}$$

Given:
- Volume of water: $V = 1$ litre $= 10^{-3}$ m³
- Fractional compression: $\frac{\Delta V}{V} = 0.10\% = \frac{0.10}{100} = 1 \times 10^{-3}$
- Bulk modulus of water: $B = 2.2 \times 10^9$ N m⁻²

Formula:
$$\Delta p = B \times \frac{\Delta V}{V}$$

Working:
$$\Delta p = 2.2 \times 10^9 \times 1 \times 10^{-3}$$

$$\boxed{\Delta p = 2.2 \times 10^6 \text{ Pa} = 2.2 \text{ MPa}}$$

The pressure on the litre of water must be increased by $2.2 \times 10^6$ Pa to compress it by 0.10%.