Question 1

Q1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a giv

Accepted Answer

The ratio of the Young’s modulus of steel to that of copper is approximately 1.79.

Question 2

Q2. Figure 9.11 shows the stress-strain curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

Accepted Answer

(a) Young's modulus for the material is approximately 3.0 × 10¹⁰ N/m².
(b) The approximate yield strength for this material is 9.0 × 10⁷ N/m².

Question 3

Q3. The stress-strain graphs for two materials A and B are shown in Fig. 9.12.
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b

Accepted Answer

Based on the provided stress-strain graphs for materials A and B:

**(a) Which of the materials has the greater Young’s modulus?**
Material A has a greater Young’s modulus.

**Reason:** Young's modulus is represented by the slope of the linear elastic region of the stress-strain curve (Stress/Strain). A steeper slope indicates a higher Young's modulus. From Figure 9.12, the initial linear portion of curve A is steeper than that of curve B, indicating that material A requires more stress to produce a given amount of strain in its elastic region, hence it is more elastic or stiffer.

**(b) Which of the two materials is stronger?**
Material A is stronger.

**Reason:** The strength of a material is determined by its ultimate tensile strength (UTS), which is the maximum stress the material can withstand before it begins to neck or fracture. From Figure 9.12, the peak stress (ultimate tensile strength) for material A is higher than that for material B. Therefore, material A can withstand a greater maximum load before failure, making it stronger.

**(c) Which of the two materials is more ductile?**
Material B is more ductile.

**Reason:** Ductility is the ability of a material to undergo significant plastic deformation before fracturing. On a stress-strain curve, a ductile material shows a large amount of strain (elongation) after the elastic limit and before the fracture point. Material B exhibits a much larger plastic deformation (larger strain at fracture) compared to material A, indicating it can be drawn into wires or stretched considerably without breaking. Material A fractures at a relatively smaller strain after its elastic limit.

**(d) Which of the two materials is more brittle?**
Material A is more brittle.

**Reason:** Brittleness is the property of a material to fracture with little or no plastic deformation. A brittle material breaks suddenly once the elastic limit is exceeded, showing a small strain at fracture. Material A shows very little plastic deformation (small strain range between the elastic limit and fracture point) compared to material B, making it more brittle. It fractures relatively soon after reaching its ultimate strength.

Question 4

Q4. Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretc

Accepted Answer

**(a) The Young’s modulus of rubber is greater than that of steel;**
**False.**

**Reason:** Young's modulus (Y = Stress/Strain) is a measure of a material's stiffness or resistance to elastic deformation under tensile or compressive stress. A material with a higher Young's modulus is stiffer and requires more stress to produce a given strain. Steel is much stiffer than rubber. For the same amount of stress, steel will undergo a much smaller strain (elongation) compared to rubber. Conversely, to produce the same strain, rubber requires significantly less stress than steel. Therefore, the Young's modulus of steel is much greater than that of rubber (typically, Y_steel ≈ 200 × 10⁹ N/m², while Y_rubber ≈ 0.01-0.1 × 10⁹ N/m²).

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

**Reason:** When a helical spring (coil) is stretched, the primary deformation in the wire of the coil is not simple tension or compression along its length. Instead, each turn of the coil experiences a twisting or torsional deformation. This twisting action involves a change in the shape of the material rather than a change in its volume, which is characteristic of shear deformation. Therefore, the stretching of a coil is primarily governed by the shear modulus (G) of the material of its wire, not its Young's modulus. The shear modulus relates shear stress to shear strain.

Question 5

Q5. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their dia

Accepted Answer

The ratio of the diameter of the copper wire to the diameter of the iron wire (d_c / d_i) is approximately 1.31.

Question 6

Q6. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a giv

Accepted Answer

The ratio of the Young's modulus of steel to that of copper is 1.79.

Question 7

Q7. In the previous problem, (9.6), if the stress in the steel wire is 1.5 × 10⁸ N m⁻², calculate the stress in the copper wire.

Accepted Answer

The stress in the copper wire is 1.5 × 10⁸ N m⁻².

Question 8

Q8. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their dia

Accepted Answer

The ratio of the diameter of the copper wire to the diameter of the iron wire (d_c / d_i) is approximately 1.31.

Question 9

Q9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress to be tolerated by the cable is 10^8 N m^-2, what is the maximum load the cable can

Accepted Answer

The maximum load the steel cable can support is 7.065 × 10^4 N.

Question 10

Q10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their dia

Accepted Answer

The ratio of the diameter of the copper wire to the diameter of the iron wire (d_copper / d_iron) is approximately 1.31.

Question 11

Q11. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a giv

Accepted Answer

The ratio of the Young's modulus of steel to that of copper is approximately 1.79.

Question 12

Q12. In Fig. 9.11, the stress-strain curves for two different types of rubber are shown.
(a) Which of the two rubbers has a higher Young’s modulus?
(b) Which of the two rubbers is str

Accepted Answer

To answer these questions, we need to interpret the stress-strain curves. Let's assume a typical representation where Curve A has a steeper initial slope and a higher breaking stress compared to Curve B, which might show a larger maximum strain before breaking.

(a) **Which of the two rubbers has a higher Young’s modulus?**
Young's modulus is represented by the slope of the stress-strain curve in the elastic region (initial linear portion). A steeper slope indicates a higher Young's modulus. Therefore, the rubber with the **steeper initial slope** has a higher Young's modulus.

(b) **Which of the two rubbers is stronger?**
Strength of a material is determined by the maximum stress it can withstand before breaking (i.e., its breaking stress or tensile strength). The rubber that can sustain a **higher stress before fracture** is stronger.

(c) **Which of the two rubbers is more elastic?**
In physics, a material is considered more elastic if it experiences less strain for a given stress, or equivalently, has a higher Young's modulus. It implies a greater resistance to deformation and a more perfect return to its original shape. Therefore, the rubber with the **higher Young's modulus (steeper initial slope)** is considered more elastic in the scientific sense. (Note: In common language, 'elastic' often refers to the ability to stretch a lot, but in physics, it refers to the ability to regain original shape perfectly and resist deformation.)

Question 13

Q13. What is the work done in stretching a wire of length L and cross-sectional area A by an amount l if Young’s modulus of the material is Y?

Accepted Answer

The work done in stretching a wire is stored as elastic potential energy in the wire. To calculate this, we consider the work done by a variable force as the wire is stretched incrementally. The final expression for the work done is given by:

Work Done (W) = (1/2) × (Stress × Strain) × Volume

Or, in terms of the given parameters:

W = (1/2) × Y × (l/L)² × A × L = (1/2) × Y × (l²A / L)

Question 14

Q14. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a giv

Accepted Answer

The ratio of the Young’s modulus of steel to that of copper is approximately 1.79.

Question 15

Q15. Figure 9.11 shows the stress-strain curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

Accepted Answer

(a) Young's modulus for the material is approximately 7.0 × 10¹⁰ N/m².
(b) The approximate yield strength for this material is 3.0 × 10⁸ N/m².

Question 16

Q16. 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 the greater Young’s modulus?
(b) Which mate

Accepted Answer

(a) Material A has the greater Young's modulus.
(b) Material A is stronger.

Question 17

Q17. Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretc

Accepted Answer

(a) False. The Young's modulus of steel is much greater than that of rubber.
(b) True. The stretching of a coil (helical spring) involves a twisting or shearing deformation of its wire, which is governed by the shear modulus of the material.

Question 18

Q18. A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10⁻² cm² is stretched, well within its elastic limit, by a hanging mass of 10 kg. Calculate the extension produced

Accepted Answer

The extension produced in the mild steel wire is 9.8 × 10⁻⁴ m.

Question 19

Q19. How much should the pressure on a litre of water be changed to compress it by 0.10%? Bulk modulus of water = 2.2 × 10⁹ N m⁻².

Accepted Answer

The pressure on a litre of water should be changed by 2.2 × 10⁶ N m⁻² to compress it by 0.10%.

Question 20

Q20. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to compress materials to very high pressures. Explain why a diamond anvil has a shape with

Accepted Answer

Diamond anvils are designed with a broad base and a pointed apex to achieve extremely high pressures for material compression. This design leverages the fundamental principle of pressure, which is defined as force per unit area (P = F/A).

**Explanation:**
1.  **Broad Base:** The broad base of the diamond anvil serves to distribute the large force applied to it over a relatively larger area. This helps in two ways:
    *   **Stability:** It provides a stable foundation for the anvil, preventing it from toppling or deforming under the immense forces applied during compression.
    *   **Strength:** It allows the anvil to withstand the large external forces without fracturing, as the stress is spread out over a larger region where the force is initially applied.

2.  **Pointed Apex:** The pointed apex is the crucial part for achieving high pressure. When the force from the broad base is transmitted to the pointed apex, this force is concentrated over an extremely small area at the tip. According to the formula P = F/A, if the area (A) is very small, even a moderate force (F) can result in an enormous pressure (P).

**In summary:** The broad base provides stability and allows for the application of a large force, while the pointed apex concentrates this force onto a minuscule area, thereby generating the extremely high pressures required to compress materials, often to conditions found deep within planets or for synthesizing new materials.

Question 21

Q21. A steel wire has a length of 4.7 m and cross-sectional area of 3.0 × 10⁻⁵ m². The wire stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0

Accepted Answer

The ratio of the Young’s modulus of steel to that of copper is approximately 1.79.

Chapter 9: Mechanical Properties of Solids