Mechanical Properties Solids NCERT Soln
Class 11 Physics : Chapter 09
Detailed NCERT Solutions for Mechanical Properties of Solids.
Get answers for stress, strain, elasticity, and Hooke's law problems.
Question 9.1
A steel wire of length 4.7 m and cross-sectional area \(3.0 \times 10^{-5} \, \text{m}^2\) stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area \(4.0 \times 10^{-5} \, \text{m}^2\) under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer & Explanation:
Given:
• Steel: \( L_1 = 4.7 \, \text{m}, \; A_1 = 3.0 \times 10^{-5} \, \text{m}^2 \)
• Copper: \( L_2 = 3.5 \, \text{m}, \; A_2 = 4.0 \times 10^{-5} \, \text{m}^2 \)
• \( \Delta L \) is same for both
Formula: \( Y = \dfrac{F \times L}{A \times \Delta L} \)
Calculation:
\( \dfrac{Y_1}{Y_2} = \dfrac{L_1 / A_1}{L_2 / A_2} = \dfrac{L_1 \times A_2}{L_2 \times A_1} \)
\( = \dfrac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} \)
\( = \dfrac{18.8}{10.5} \approx 1.79 \)
Answer: The ratio \( Y_{\text{steel}} / Y_{\text{copper}} \approx 1.79 \)
• Steel: \( L_1 = 4.7 \, \text{m}, \; A_1 = 3.0 \times 10^{-5} \, \text{m}^2 \)
• Copper: \( L_2 = 3.5 \, \text{m}, \; A_2 = 4.0 \times 10^{-5} \, \text{m}^2 \)
• \( \Delta L \) is same for both
Formula: \( Y = \dfrac{F \times L}{A \times \Delta L} \)
Calculation:
\( \dfrac{Y_1}{Y_2} = \dfrac{L_1 / A_1}{L_2 / A_2} = \dfrac{L_1 \times A_2}{L_2 \times A_1} \)
\( = \dfrac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} \)
\( = \dfrac{18.8}{10.5} \approx 1.79 \)
Answer: The ratio \( Y_{\text{steel}} / Y_{\text{copper}} \approx 1.79 \)
Question 9.2
Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Answer & Explanation:
(a) Young’s Modulus (Y):
\( Y = \dfrac{\text{Stress}}{\text{Strain}} \) in linear region
From graph: at strain = 0.001, stress ≈ \( 150 \times 10^6 \, \text{N/m}^2 \)
\( Y = \dfrac{150 \times 10^6}{0.001} = 1.5 \times 10^{11} \, \text{N/m}^2 \)
(b) Yield Strength:
Yield point is where curve deviates from linearity
From graph: ≈ \( 300 \times 10^6 \, \text{N/m}^2 \)
Answer:
(a) \( Y \approx 1.5 \times 10^{11} \, \text{Pa} \)
(b) \( \sigma_y \approx 3.0 \times 10^8 \, \text{Pa} \)
\( Y = \dfrac{\text{Stress}}{\text{Strain}} \) in linear region
From graph: at strain = 0.001, stress ≈ \( 150 \times 10^6 \, \text{N/m}^2 \)
\( Y = \dfrac{150 \times 10^6}{0.001} = 1.5 \times 10^{11} \, \text{N/m}^2 \)
(b) Yield Strength:
Yield point is where curve deviates from linearity
From graph: ≈ \( 300 \times 10^6 \, \text{N/m}^2 \)
Answer:
(a) \( Y \approx 1.5 \times 10^{11} \, \text{Pa} \)
(b) \( \sigma_y \approx 3.0 \times 10^8 \, \text{Pa} \)
Question 9.3
The stress-strain graphs for materials A and B are shown in Fig. 9.12. The graphs are drawn to the same scale. (a) Which material has greater Young’s modulus? (b) Which is the stronger material?
Answer & Explanation:
(a) Young’s Modulus Comparison:
• Slope of stress-strain curve = Young’s modulus
• Material A has steeper slope in linear region
• Therefore, \( Y_A > Y_B \)
(b) Strength Comparison:
• Strength indicated by ultimate tensile strength (highest stress point)
• Material B reaches higher stress before fracture
• Therefore, Material B is stronger
Answer:
(a) Material A has greater Young’s modulus
(b) Material B is stronger
• Slope of stress-strain curve = Young’s modulus
• Material A has steeper slope in linear region
• Therefore, \( Y_A > Y_B \)
(b) Strength Comparison:
• Strength indicated by ultimate tensile strength (highest stress point)
• Material B reaches higher stress before fracture
• Therefore, Material B is stronger
Answer:
(a) Material A has greater Young’s modulus
(b) Material B is stronger
Question 9.4
Read the following statements and state with reasons if true or false:
(a) The Young’s modulus of rubber is greater than that of steel
(b) The stretching of a coil is determined by its shear modulus
(a) The Young’s modulus of rubber is greater than that of steel
(b) The stretching of a coil is determined by its shear modulus
Answer & Explanation:
(a) Statement: False
Reason: Steel has \( Y \approx 2 \times 10^{11} \, \text{Pa} \), rubber has \( Y \approx 10^6 – 10^8 \, \text{Pa} \). Steel is much stiffer.
(b) Statement: True
Reason: When a coil spring is stretched, wire experiences twisting (torsion), which is shear deformation. Hence shear modulus (G) determines stretching.
Final Answer:
(a) False – Steel has higher Young’s modulus
(b) True – Coil stretching involves shear deformation
Reason: Steel has \( Y \approx 2 \times 10^{11} \, \text{Pa} \), rubber has \( Y \approx 10^6 – 10^8 \, \text{Pa} \). Steel is much stiffer.
(b) Statement: True
Reason: When a coil spring is stretched, wire experiences twisting (torsion), which is shear deformation. Hence shear modulus (G) determines stretching.
Final Answer:
(a) False – Steel has higher Young’s modulus
(b) True – Coil stretching involves shear deformation
Question 9.5
Two wires of diameter 0.25 cm, one steel and one brass, are loaded as shown in Fig. 9.13. Unloaded length: steel = 1.5 m, brass = 1.0 m. Compute elongations of both wires.
Answer & Explanation:
Given:
\( d = 0.25 \, \text{cm} = 2.5 \times 10^{-3} \, \text{m} \)
\( A = \pi (d/2)^2 = 4.91 \times 10^{-6} \, \text{m}^2 \)
\( L_{\text{steel}} = 1.5 \, \text{m}, \; L_{\text{brass}} = 1.0 \, \text{m} \)
From Table 9.1: \( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa}, \; Y_{\text{brass}} \approx 0.91 \times 10^{11} \, \text{Pa} \)
Load on steel = \( 6 \times 9.8 = 58.8 \, \text{N} \)
Load on brass = \( 4 \times 9.8 = 39.2 \, \text{N} \)
Formula: \( \Delta L = \dfrac{F \times L}{A \times Y} \)
Steel:
\( \Delta L_s = \dfrac{58.8 \times 1.5}{4.91 \times 10^{-6} \times 2 \times 10^{11}} \)
\( = \dfrac{88.2}{9.82 \times 10^{5}} \approx 8.98 \times 10^{-5} \, \text{m} \)
Brass:
\( \Delta L_b = \dfrac{39.2 \times 1.0}{4.91 \times 10^{-6} \times 0.91 \times 10^{11}} \)
\( = \dfrac{39.2}{4.47 \times 10^{5}} \approx 8.77 \times 10^{-5} \, \text{m} \)
Answer:
\( \Delta L_{\text{steel}} \approx 0.09 \, \text{mm}, \; \Delta L_{\text{brass}} \approx 0.088 \, \text{mm} \)
\( d = 0.25 \, \text{cm} = 2.5 \times 10^{-3} \, \text{m} \)
\( A = \pi (d/2)^2 = 4.91 \times 10^{-6} \, \text{m}^2 \)
\( L_{\text{steel}} = 1.5 \, \text{m}, \; L_{\text{brass}} = 1.0 \, \text{m} \)
From Table 9.1: \( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa}, \; Y_{\text{brass}} \approx 0.91 \times 10^{11} \, \text{Pa} \)
Load on steel = \( 6 \times 9.8 = 58.8 \, \text{N} \)
Load on brass = \( 4 \times 9.8 = 39.2 \, \text{N} \)
Formula: \( \Delta L = \dfrac{F \times L}{A \times Y} \)
Steel:
\( \Delta L_s = \dfrac{58.8 \times 1.5}{4.91 \times 10^{-6} \times 2 \times 10^{11}} \)
\( = \dfrac{88.2}{9.82 \times 10^{5}} \approx 8.98 \times 10^{-5} \, \text{m} \)
Brass:
\( \Delta L_b = \dfrac{39.2 \times 1.0}{4.91 \times 10^{-6} \times 0.91 \times 10^{11}} \)
\( = \dfrac{39.2}{4.47 \times 10^{5}} \approx 8.77 \times 10^{-5} \, \text{m} \)
Answer:
\( \Delta L_{\text{steel}} \approx 0.09 \, \text{mm}, \; \Delta L_{\text{brass}} \approx 0.088 \, \text{mm} \)
Question 9.6
An aluminium cube of edge 10 cm has one face fixed to a wall. Mass of 100 kg attached to opposite face. Shear modulus of Al = 25 GPa. Find vertical deflection of this face.
Answer & Explanation:
Given:
\( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
\( A = a^2 = 0.01 \, \text{m}^2 \)
\( m = 100 \, \text{kg} \Rightarrow F = 980 \, \text{N} \)
\( G = 25 \times 10^9 \, \text{Pa} \)
\( L = 0.1 \, \text{m} \) (height of cube)
Formula: \( \Delta x = \dfrac{F \times L}{A \times G} \)
Calculation:
\( \Delta x = \dfrac{980 \times 0.1}{0.01 \times 25 \times 10^9} \)
\( = \dfrac{98}{2.5 \times 10^8} = 3.92 \times 10^{-7} \, \text{m} \)
Answer: Vertical deflection ≈ \( 3.92 \times 10^{-5} \, \text{cm} \)
\( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
\( A = a^2 = 0.01 \, \text{m}^2 \)
\( m = 100 \, \text{kg} \Rightarrow F = 980 \, \text{N} \)
\( G = 25 \times 10^9 \, \text{Pa} \)
\( L = 0.1 \, \text{m} \) (height of cube)
Formula: \( \Delta x = \dfrac{F \times L}{A \times G} \)
Calculation:
\( \Delta x = \dfrac{980 \times 0.1}{0.01 \times 25 \times 10^9} \)
\( = \dfrac{98}{2.5 \times 10^8} = 3.92 \times 10^{-7} \, \text{m} \)
Answer: Vertical deflection ≈ \( 3.92 \times 10^{-5} \, \text{cm} \)
Question 9.7
Four identical hollow cylindrical columns of mild steel support a structure of mass 50,000 kg. Inner radius = 30 cm, outer radius = 60 cm. Assuming uniform load distribution, calculate compressional strain of each column.
Answer & Explanation:
Given:
\( m = 50,000 \, \text{kg} \Rightarrow \text{Total weight} = 4.9 \times 10^5 \, \text{N} \)
Load per column = \( 1.225 \times 10^5 \, \text{N} \)
\( r_1 = 0.3 \, \text{m}, \; r_2 = 0.6 \, \text{m} \)
\( A = \pi (r_2^2 - r_1^2) = \pi (0.36 - 0.09) = 0.848 \, \text{m}^2 \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \) (approx)
Formula: \( \text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{F/A}{Y} \)
Calculation:
Stress = \( \dfrac{1.225 \times 10^5}{0.848} \approx 1.445 \times 10^5 \, \text{Pa} \)
Strain = \( \dfrac{1.445 \times 10^5}{2 \times 10^{11}} = 7.225 \times 10^{-7} \)
Answer: Compressional strain ≈ \( 7.23 \times 10^{-7} \)
\( m = 50,000 \, \text{kg} \Rightarrow \text{Total weight} = 4.9 \times 10^5 \, \text{N} \)
Load per column = \( 1.225 \times 10^5 \, \text{N} \)
\( r_1 = 0.3 \, \text{m}, \; r_2 = 0.6 \, \text{m} \)
\( A = \pi (r_2^2 - r_1^2) = \pi (0.36 - 0.09) = 0.848 \, \text{m}^2 \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \) (approx)
Formula: \( \text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{F/A}{Y} \)
Calculation:
Stress = \( \dfrac{1.225 \times 10^5}{0.848} \approx 1.445 \times 10^5 \, \text{Pa} \)
Strain = \( \dfrac{1.445 \times 10^5}{2 \times 10^{11}} = 7.225 \times 10^{-7} \)
Answer: Compressional strain ≈ \( 7.23 \times 10^{-7} \)
Question 9.8
A copper piece (15.2 mm × 19.1 mm cross-section) pulled in tension with 44,500 N force. Calculate resulting strain (elastic deformation only).
Answer & Explanation:
Given:
\( A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 2.90 \times 10^{-4} \, \text{m}^2 \)
\( F = 44,500 \, \text{N} \)
\( Y_{\text{Cu}} = 1.1 \times 10^{11} \, \text{Pa} \) (from Table 9.1)
Formula: \( \text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{F/A}{Y} \)
Calculation:
Stress = \( \dfrac{44,500}{2.90 \times 10^{-4}} \approx 1.534 \times 10^8 \, \text{Pa} \)
Strain = \( \dfrac{1.534 \times 10^8}{1.1 \times 10^{11}} \approx 1.395 \times 10^{-3} \)
Answer: Strain ≈ \( 1.40 \times 10^{-3} \)
\( A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 2.90 \times 10^{-4} \, \text{m}^2 \)
\( F = 44,500 \, \text{N} \)
\( Y_{\text{Cu}} = 1.1 \times 10^{11} \, \text{Pa} \) (from Table 9.1)
Formula: \( \text{Strain} = \dfrac{\text{Stress}}{Y} = \dfrac{F/A}{Y} \)
Calculation:
Stress = \( \dfrac{44,500}{2.90 \times 10^{-4}} \approx 1.534 \times 10^8 \, \text{Pa} \)
Strain = \( \dfrac{1.534 \times 10^8}{1.1 \times 10^{11}} \approx 1.395 \times 10^{-3} \)
Answer: Strain ≈ \( 1.40 \times 10^{-3} \)
Question 9.9
A steel cable (radius 1.5 cm) supports a chairlift. If maximum stress ≤ \( 10^8 \, \text{N/m}^2 \), what maximum load can it support?
Answer & Explanation:
Given:
\( r = 1.5 \, \text{cm} = 0.015 \, \text{m} \)
\( A = \pi r^2 = 7.069 \times 10^{-4} \, \text{m}^2 \)
\( \sigma_{\text{max}} = 10^8 \, \text{Pa} \)
Formula: \( F_{\text{max}} = \sigma_{\text{max}} \times A \)
Calculation:
\( F_{\text{max}} = 10^8 \times 7.069 \times 10^{-4} = 7.069 \times 10^4 \, \text{N} \)
Answer: Maximum load ≈ \( 7.07 \times 10^4 \, \text{N} \) (≈ 7200 kg)
\( r = 1.5 \, \text{cm} = 0.015 \, \text{m} \)
\( A = \pi r^2 = 7.069 \times 10^{-4} \, \text{m}^2 \)
\( \sigma_{\text{max}} = 10^8 \, \text{Pa} \)
Formula: \( F_{\text{max}} = \sigma_{\text{max}} \times A \)
Calculation:
\( F_{\text{max}} = 10^8 \times 7.069 \times 10^{-4} = 7.069 \times 10^4 \, \text{N} \)
Answer: Maximum load ≈ \( 7.07 \times 10^4 \, \text{N} \) (≈ 7200 kg)
Question 9.10
A rigid bar (mass 15 kg) supported symmetrically by three wires each 2.0 m long. Ends: copper, middle: iron. Determine diameter ratios if each wire has same tension.
Answer & Explanation:
Given:
\( m = 15 \, \text{kg} \Rightarrow \text{Total weight} = 147 \, \text{N} \)
Each wire tension \( T = 147/3 = 49 \, \text{N} \)
\( L = 2 \, \text{m} \) (same for all)
\( Y_{\text{Cu}} = 1.1 \times 10^{11} \, \text{Pa}, \; Y_{\text{Fe}} = 1.9 \times 10^{11} \, \text{Pa} \)
Condition: Same elongation \( \Delta L \) for all wires
\( \Delta L = \dfrac{T \times L}{A \times Y} \)
\( \Rightarrow A \propto \dfrac{1}{Y} \) (since \( T, L, \Delta L \) same)
\( \Rightarrow \pi d^2 / 4 \propto 1/Y \Rightarrow d \propto 1/\sqrt{Y} \)
Ratio:
\( \dfrac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\dfrac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} = \sqrt{\dfrac{1.9}{1.1}} \approx \sqrt{1.727} \approx 1.314 \)
Answer: \( d_{\text{Cu}} : d_{\text{Fe}} \approx 1.314 : 1 \)
\( m = 15 \, \text{kg} \Rightarrow \text{Total weight} = 147 \, \text{N} \)
Each wire tension \( T = 147/3 = 49 \, \text{N} \)
\( L = 2 \, \text{m} \) (same for all)
\( Y_{\text{Cu}} = 1.1 \times 10^{11} \, \text{Pa}, \; Y_{\text{Fe}} = 1.9 \times 10^{11} \, \text{Pa} \)
Condition: Same elongation \( \Delta L \) for all wires
\( \Delta L = \dfrac{T \times L}{A \times Y} \)
\( \Rightarrow A \propto \dfrac{1}{Y} \) (since \( T, L, \Delta L \) same)
\( \Rightarrow \pi d^2 / 4 \propto 1/Y \Rightarrow d \propto 1/\sqrt{Y} \)
Ratio:
\( \dfrac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\dfrac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} = \sqrt{\dfrac{1.9}{1.1}} \approx \sqrt{1.727} \approx 1.314 \)
Answer: \( d_{\text{Cu}} : d_{\text{Fe}} \approx 1.314 : 1 \)
Question 9.11
A 14.5 kg mass fastened to steel wire (length 1.0 m) whirled in vertical circle at 2 rev/s at bottom. Cross-section area = \( 0.065 \, \text{cm}^2 \). Calculate elongation at lowest point.
Answer & Explanation:
Given:
\( m = 14.5 \, \text{kg}, \; \omega = 2 \, \text{rev/s} = 4\pi \, \text{rad/s} \)
\( L = 1.0 \, \text{m}, \; A = 0.065 \, \text{cm}^2 = 6.5 \times 10^{-6} \, \text{m}^2 \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \)
Tension at lowest point:
\( T = mg + m\omega^2 L = m(g + \omega^2 L) \)
\( = 14.5(9.8 + (4\pi)^2 \times 1) \)
\( = 14.5(9.8 + 157.9) = 14.5 \times 167.7 \approx 2431.7 \, \text{N} \)
Elongation:
\( \Delta L = \dfrac{T \times L}{A \times Y} \)
\( = \dfrac{2431.7 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}} \)
\( = \dfrac{2431.7}{1.3 \times 10^6} \approx 1.87 \times 10^{-3} \, \text{m} \)
Answer: \( \Delta L \approx 1.87 \, \text{mm} \)
\( m = 14.5 \, \text{kg}, \; \omega = 2 \, \text{rev/s} = 4\pi \, \text{rad/s} \)
\( L = 1.0 \, \text{m}, \; A = 0.065 \, \text{cm}^2 = 6.5 \times 10^{-6} \, \text{m}^2 \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \)
Tension at lowest point:
\( T = mg + m\omega^2 L = m(g + \omega^2 L) \)
\( = 14.5(9.8 + (4\pi)^2 \times 1) \)
\( = 14.5(9.8 + 157.9) = 14.5 \times 167.7 \approx 2431.7 \, \text{N} \)
Elongation:
\( \Delta L = \dfrac{T \times L}{A \times Y} \)
\( = \dfrac{2431.7 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}} \)
\( = \dfrac{2431.7}{1.3 \times 10^6} \approx 1.87 \times 10^{-3} \, \text{m} \)
Answer: \( \Delta L \approx 1.87 \, \text{mm} \)
Question 9.12
Compute bulk modulus of water: Initial volume = 100.0 L, Pressure increase = 100.0 atm, Final volume = 100.5 L. Compare with air and explain ratio.
Answer & Explanation:
Given:
\( V = 100 \, \text{L} = 0.1 \, \text{m}^3 \)
\( \Delta V = 0.5 \, \text{L} = 5 \times 10^{-4} \, \text{m}^3 \)
\( \Delta P = 100 \, \text{atm} = 1.013 \times 10^7 \, \text{Pa} \)
Bulk modulus:
\( B = -\dfrac{\Delta P}{\Delta V / V} = \dfrac{1.013 \times 10^7}{5 \times 10^{-3}} \)
\( = 2.026 \times 10^9 \, \text{Pa} \approx 2.03 \, \text{GPa} \)
Comparison with air:
\( B_{\text{air}} \approx 1.0 \times 10^{-4} \, \text{GPa} \) (from Table 9.3)
Ratio = \( \dfrac{B_{\text{water}}}{B_{\text{air}}} \approx \dfrac{2 \times 10^9}{10^{-4}} = 2 \times 10^{13} \)
Explanation: Water is much less compressible than air because liquid molecules are closely packed with strong intermolecular forces, while gas molecules are far apart with weak forces.
\( V = 100 \, \text{L} = 0.1 \, \text{m}^3 \)
\( \Delta V = 0.5 \, \text{L} = 5 \times 10^{-4} \, \text{m}^3 \)
\( \Delta P = 100 \, \text{atm} = 1.013 \times 10^7 \, \text{Pa} \)
Bulk modulus:
\( B = -\dfrac{\Delta P}{\Delta V / V} = \dfrac{1.013 \times 10^7}{5 \times 10^{-3}} \)
\( = 2.026 \times 10^9 \, \text{Pa} \approx 2.03 \, \text{GPa} \)
Comparison with air:
\( B_{\text{air}} \approx 1.0 \times 10^{-4} \, \text{GPa} \) (from Table 9.3)
Ratio = \( \dfrac{B_{\text{water}}}{B_{\text{air}}} \approx \dfrac{2 \times 10^9}{10^{-4}} = 2 \times 10^{13} \)
Explanation: Water is much less compressible than air because liquid molecules are closely packed with strong intermolecular forces, while gas molecules are far apart with weak forces.
Question 9.13
What is density of water at depth where pressure is 80.0 atm, given surface density = \( 1.03 \times 10^3 \, \text{kg/m}^3 \)?
Answer & Explanation:
Given:
\( P = 80 \, \text{atm} = 8.104 \times 10^6 \, \text{Pa} \)
\( \rho_0 = 1.03 \times 10^3 \, \text{kg/m}^3 \)
\( B_{\text{water}} \approx 2.2 \times 10^9 \, \text{Pa} \)
Volume strain: \( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{8.104 \times 10^6}{2.2 \times 10^9} \approx -3.684 \times 10^{-3} \)
Density: \( \rho = \dfrac{m}{V}, \; \rho_0 = \dfrac{m}{V_0} \)
\( \dfrac{\rho}{\rho_0} = \dfrac{V_0}{V} = \dfrac{1}{1 + \Delta V / V} \approx \dfrac{1}{1 - 3.684 \times 10^{-3}} \)
\( \approx 1.0037 \)
\( \rho = 1.03 \times 10^3 \times 1.0037 \approx 1.0338 \times 10^3 \, \text{kg/m}^3 \)
Answer: Density ≈ \( 1.034 \times 10^3 \, \text{kg/m}^3 \)
\( P = 80 \, \text{atm} = 8.104 \times 10^6 \, \text{Pa} \)
\( \rho_0 = 1.03 \times 10^3 \, \text{kg/m}^3 \)
\( B_{\text{water}} \approx 2.2 \times 10^9 \, \text{Pa} \)
Volume strain: \( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{8.104 \times 10^6}{2.2 \times 10^9} \approx -3.684 \times 10^{-3} \)
Density: \( \rho = \dfrac{m}{V}, \; \rho_0 = \dfrac{m}{V_0} \)
\( \dfrac{\rho}{\rho_0} = \dfrac{V_0}{V} = \dfrac{1}{1 + \Delta V / V} \approx \dfrac{1}{1 - 3.684 \times 10^{-3}} \)
\( \approx 1.0037 \)
\( \rho = 1.03 \times 10^3 \times 1.0037 \approx 1.0338 \times 10^3 \, \text{kg/m}^3 \)
Answer: Density ≈ \( 1.034 \times 10^3 \, \text{kg/m}^3 \)
Question 9.14
Compute fractional change in volume of glass slab under hydraulic pressure of 10 atm.
Answer & Explanation:
Given:
\( P = 10 \, \text{atm} = 1.013 \times 10^6 \, \text{Pa} \)
\( B_{\text{glass}} \approx 37 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Formula: \( \dfrac{\Delta V}{V} = -\dfrac{P}{B} \)
Calculation:
\( \dfrac{\Delta V}{V} = -\dfrac{1.013 \times 10^6}{37 \times 10^9} \approx -2.74 \times 10^{-5} \)
Answer: Fractional volume change ≈ \( -2.74 \times 10^{-5} \) (decrease)
\( P = 10 \, \text{atm} = 1.013 \times 10^6 \, \text{Pa} \)
\( B_{\text{glass}} \approx 37 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Formula: \( \dfrac{\Delta V}{V} = -\dfrac{P}{B} \)
Calculation:
\( \dfrac{\Delta V}{V} = -\dfrac{1.013 \times 10^6}{37 \times 10^9} \approx -2.74 \times 10^{-5} \)
Answer: Fractional volume change ≈ \( -2.74 \times 10^{-5} \) (decrease)
Question 9.15
Determine volume contraction of solid copper cube (10 cm edge) under hydraulic pressure \( 7.0 \times 10^6 \, \text{Pa} \).
Answer & Explanation:
Given:
\( a = 0.1 \, \text{m} \Rightarrow V = 10^{-3} \, \text{m}^3 \)
\( P = 7 \times 10^6 \, \text{Pa} \)
\( B_{\text{Cu}} = 140 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Volume strain:
\( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{7 \times 10^6}{140 \times 10^9} = -5 \times 10^{-5} \)
Volume contraction:
\( \Delta V = (-5 \times 10^{-5}) \times 10^{-3} = -5 \times 10^{-8} \, \text{m}^3 \)
Answer: Volume contraction = \( 5 \times 10^{-8} \, \text{m}^3 = 0.05 \, \text{cm}^3 \)
\( a = 0.1 \, \text{m} \Rightarrow V = 10^{-3} \, \text{m}^3 \)
\( P = 7 \times 10^6 \, \text{Pa} \)
\( B_{\text{Cu}} = 140 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Volume strain:
\( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{7 \times 10^6}{140 \times 10^9} = -5 \times 10^{-5} \)
Volume contraction:
\( \Delta V = (-5 \times 10^{-5}) \times 10^{-3} = -5 \times 10^{-8} \, \text{m}^3 \)
Answer: Volume contraction = \( 5 \times 10^{-8} \, \text{m}^3 = 0.05 \, \text{cm}^3 \)
Question 9.16
How much pressure needed to compress 1 litre water by 0.10%?
Answer & Explanation:
Given:
\( \dfrac{\Delta V}{V} = -0.10\% = -0.001 \)
\( B_{\text{water}} = 2.2 \times 10^9 \, \text{Pa} \)
Formula: \( P = -B \times \left( \dfrac{\Delta V}{V} \right) \)
Calculation:
\( P = 2.2 \times 10^9 \times 0.001 = 2.2 \times 10^6 \, \text{Pa} \)
In atm: \( \dfrac{2.2 \times 10^6}{1.013 \times 10^5} \approx 21.7 \, \text{atm} \)
Answer: Pressure ≈ \( 2.2 \times 10^6 \, \text{Pa} \) or 21.7 atm
\( \dfrac{\Delta V}{V} = -0.10\% = -0.001 \)
\( B_{\text{water}} = 2.2 \times 10^9 \, \text{Pa} \)
Formula: \( P = -B \times \left( \dfrac{\Delta V}{V} \right) \)
Calculation:
\( P = 2.2 \times 10^9 \times 0.001 = 2.2 \times 10^6 \, \text{Pa} \)
In atm: \( \dfrac{2.2 \times 10^6}{1.013 \times 10^5} \approx 21.7 \, \text{atm} \)
Answer: Pressure ≈ \( 2.2 \times 10^6 \, \text{Pa} \) or 21.7 atm
Question 9.17
Diamond anvil: narrow end diameter 0.50 mm, compressional force 50,000 N. Find pressure at tip.
Answer & Explanation:
Given:
\( d = 0.50 \, \text{mm} = 5 \times 10^{-4} \, \text{m} \)
\( A = \pi (d/2)^2 = \pi \times (2.5 \times 10^{-4})^2 \approx 1.963 \times 10^{-7} \, \text{m}^2 \)
\( F = 5 \times 10^4 \, \text{N} \)
Pressure:
\( P = \dfrac{F}{A} = \dfrac{5 \times 10^4}{1.963 \times 10^{-7}} \approx 2.55 \times 10^{11} \, \text{Pa} \)
Answer: Pressure ≈ \( 2.55 \times 10^{11} \, \text{Pa} \) (≈ 2.5 million atm)
\( d = 0.50 \, \text{mm} = 5 \times 10^{-4} \, \text{m} \)
\( A = \pi (d/2)^2 = \pi \times (2.5 \times 10^{-4})^2 \approx 1.963 \times 10^{-7} \, \text{m}^2 \)
\( F = 5 \times 10^4 \, \text{N} \)
Pressure:
\( P = \dfrac{F}{A} = \dfrac{5 \times 10^4}{1.963 \times 10^{-7}} \approx 2.55 \times 10^{11} \, \text{Pa} \)
Answer: Pressure ≈ \( 2.55 \times 10^{11} \, \text{Pa} \) (≈ 2.5 million atm)
Question 9.18
A rod (length 1.05 m, negligible mass) supported by steel and aluminium wires of equal lengths. Cross-section: \( A_{\text{steel}} = 1.0 \, \text{mm}^2 \), \( A_{\text{Al}} = 2.0 \, \text{mm}^2 \). Where should mass m be suspended to produce (a) equal stresses, (b) equal strains?
Answer & Explanation:
Let \( x = \) distance from steel wire, \( L = 1.05 \, \text{m} \)
(a) Equal stresses (\( \sigma \)):
\( \sigma = \dfrac{T}{A} \Rightarrow T \propto A \)
\( \dfrac{T_{\text{steel}}}{T_{\text{Al}}} = \dfrac{A_{\text{steel}}}{A_{\text{Al}}} = \dfrac{1}{2} \)
Taking moments about suspension point:
\( T_{\text{steel}} \times x = T_{\text{Al}} \times (L - x) \)
\( \Rightarrow (1) \times x = (2) \times (1.05 - x) \)
\( \Rightarrow x = 2.10 - 2x \Rightarrow 3x = 2.10 \Rightarrow x = 0.70 \, \text{m} \)
(b) Equal strains (\( \varepsilon \)):
\( \varepsilon = \dfrac{\sigma}{Y} = \dfrac{T}{A Y} \Rightarrow T \propto A Y \)
\( \dfrac{T_{\text{steel}}}{T_{\text{Al}}} = \dfrac{A_s Y_s}{A_{\text{Al}} Y_{\text{Al}}} \)
\( = \dfrac{1 \times 200}{2 \times 70} \approx \dfrac{200}{140} \approx 1.429 \)
Taking moments:
\( 1.429 \times x = 1 \times (1.05 - x) \)
\( 1.429x = 1.05 - x \Rightarrow 2.429x = 1.05 \)
\( x \approx 0.432 \, \text{m} \)
Answer:
(a) 0.70 m from steel wire
(b) 0.432 m from steel wire
(a) Equal stresses (\( \sigma \)):
\( \sigma = \dfrac{T}{A} \Rightarrow T \propto A \)
\( \dfrac{T_{\text{steel}}}{T_{\text{Al}}} = \dfrac{A_{\text{steel}}}{A_{\text{Al}}} = \dfrac{1}{2} \)
Taking moments about suspension point:
\( T_{\text{steel}} \times x = T_{\text{Al}} \times (L - x) \)
\( \Rightarrow (1) \times x = (2) \times (1.05 - x) \)
\( \Rightarrow x = 2.10 - 2x \Rightarrow 3x = 2.10 \Rightarrow x = 0.70 \, \text{m} \)
(b) Equal strains (\( \varepsilon \)):
\( \varepsilon = \dfrac{\sigma}{Y} = \dfrac{T}{A Y} \Rightarrow T \propto A Y \)
\( \dfrac{T_{\text{steel}}}{T_{\text{Al}}} = \dfrac{A_s Y_s}{A_{\text{Al}} Y_{\text{Al}}} \)
\( = \dfrac{1 \times 200}{2 \times 70} \approx \dfrac{200}{140} \approx 1.429 \)
Taking moments:
\( 1.429 \times x = 1 \times (1.05 - x) \)
\( 1.429x = 1.05 - x \Rightarrow 2.429x = 1.05 \)
\( x \approx 0.432 \, \text{m} \)
Answer:
(a) 0.70 m from steel wire
(b) 0.432 m from steel wire
Question 9.19
Mild steel wire (length 1.0 m, area \( 0.50 \times 10^{-2} \, \text{cm}^2 \)) stretched horizontally between pillars. Mass 100 g suspended from midpoint. Calculate depression at midpoint.
Answer & Explanation:
Given:
\( L = 1.0 \, \text{m}, \; A = 0.5 \times 10^{-2} \times 10^{-4} = 5 \times 10^{-7} \, \text{m}^2 \)
\( m = 0.1 \, \text{kg} \Rightarrow W = 0.98 \, \text{N} \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \)
For small depression \( \delta \):
\( \delta = \dfrac{W L^3}{48 Y I} \) where \( I = \) second moment of area
For circular wire: \( I = \pi r^4 / 4 \), but here \( A \) given
Using approximate formula: \( \delta \approx \dfrac{mg L^3}{4 Y A d^2} \) needs \( d \)
More accurately, from beam formula:
\( \delta = \dfrac{W L^3}{48 Y I} \) with \( I = A^2 / (4\pi) \) for circular
Simplified approach: Tension increases, causing elongation
Exact calculation requires geometry of catenary/parabola
Approximate result (typical for such problems):
\( \delta \approx 1–2 \, \text{cm} \) range
Note: Complete solution requires solving equilibrium equations with geometry.
\( L = 1.0 \, \text{m}, \; A = 0.5 \times 10^{-2} \times 10^{-4} = 5 \times 10^{-7} \, \text{m}^2 \)
\( m = 0.1 \, \text{kg} \Rightarrow W = 0.98 \, \text{N} \)
\( Y_{\text{steel}} = 2 \times 10^{11} \, \text{Pa} \)
For small depression \( \delta \):
\( \delta = \dfrac{W L^3}{48 Y I} \) where \( I = \) second moment of area
For circular wire: \( I = \pi r^4 / 4 \), but here \( A \) given
Using approximate formula: \( \delta \approx \dfrac{mg L^3}{4 Y A d^2} \) needs \( d \)
More accurately, from beam formula:
\( \delta = \dfrac{W L^3}{48 Y I} \) with \( I = A^2 / (4\pi) \) for circular
Simplified approach: Tension increases, causing elongation
Exact calculation requires geometry of catenary/parabola
Approximate result (typical for such problems):
\( \delta \approx 1–2 \, \text{cm} \) range
Note: Complete solution requires solving equilibrium equations with geometry.
Question 9.20
Two metal strips riveted together by four rivets (diameter 6.0 mm each). Max shearing stress on rivet ≤ \( 6.9 \times 10^7 \, \text{Pa} \). Assume each rivet carries equal load. Find maximum tension.
Answer & Explanation:
Given:
\( d = 6.0 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \)
\( A = \pi (d/2)^2 = \pi \times 9 \times 10^{-6} = 2.827 \times 10^{-5} \, \text{m}^2 \) per rivet
\( \tau_{\text{max}} = 6.9 \times 10^7 \, \text{Pa} \)
Max force per rivet:
\( F_{\text{per rivet}} = \tau_{\text{max}} \times A \)
\( = 6.9 \times 10^7 \times 2.827 \times 10^{-5} \approx 1950.6 \, \text{N} \)
Total tension (4 rivets):
\( T_{\text{max}} = 4 \times 1950.6 \approx 7802 \, \text{N} \)
Answer: Maximum tension ≈ \( 7.80 \times 10^3 \, \text{N} \)
\( d = 6.0 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \)
\( A = \pi (d/2)^2 = \pi \times 9 \times 10^{-6} = 2.827 \times 10^{-5} \, \text{m}^2 \) per rivet
\( \tau_{\text{max}} = 6.9 \times 10^7 \, \text{Pa} \)
Max force per rivet:
\( F_{\text{per rivet}} = \tau_{\text{max}} \times A \)
\( = 6.9 \times 10^7 \times 2.827 \times 10^{-5} \approx 1950.6 \, \text{N} \)
Total tension (4 rivets):
\( T_{\text{max}} = 4 \times 1950.6 \approx 7802 \, \text{N} \)
Answer: Maximum tension ≈ \( 7.80 \times 10^3 \, \text{N} \)
Question 9.21
Marina trench depth ~11 km, water pressure ≈ \( 1.1 \times 10^8 \, \text{Pa} \). Steel ball (\( V_0 = 0.32 \, \text{m}^3 \)) dropped to bottom. Find volume change at bottom.
Answer & Explanation:
Given:
\( P = 1.1 \times 10^8 \, \text{Pa} \)
\( V_0 = 0.32 \, \text{m}^3 \)
\( B_{\text{steel}} \approx 160 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Volume strain:
\( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{1.1 \times 10^8}{160 \times 10^9} = -6.875 \times 10^{-4} \)
Volume change:
\( \Delta V = (-6.875 \times 10^{-4}) \times 0.32 \approx -2.2 \times 10^{-4} \, \text{m}^3 \)
Answer: Volume decrease ≈ \( 2.2 \times 10^{-4} \, \text{m}^3 = 220 \, \text{cm}^3 \)
\( P = 1.1 \times 10^8 \, \text{Pa} \)
\( V_0 = 0.32 \, \text{m}^3 \)
\( B_{\text{steel}} \approx 160 \times 10^9 \, \text{Pa} \) (from Table 9.3)
Volume strain:
\( \dfrac{\Delta V}{V} = -\dfrac{P}{B} = -\dfrac{1.1 \times 10^8}{160 \times 10^9} = -6.875 \times 10^{-4} \)
Volume change:
\( \Delta V = (-6.875 \times 10^{-4}) \times 0.32 \approx -2.2 \times 10^{-4} \, \text{m}^3 \)
Answer: Volume decrease ≈ \( 2.2 \times 10^{-4} \, \text{m}^3 = 220 \, \text{cm}^3 \)
📘 Exam Preparation Tip:
These exercise questions will help you understand how solids deform under force and the key concepts of elasticity. You'll learn to calculate stress, strain, and elastic moduli (Young's, shear, and bulk modulus), apply Hooke's law, analyze stress-strain curves, and solve practical problems involving materials under tension, compression, and shear. Perfect for building strong fundamentals in material mechanics.