Given below are two statements : One is labeled as Assertion (A) and the other is labeled as Reason (R). Assertion (A) : Clothes containing oil grease statins cannot be cleaned by water wash. Reason (R) : Because the angle of contact between the oil/grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below.

Both (A) and(R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true but (R) is not the correct explanation of (A)

NCERT Solutions for Class 11 Physics Ch 8 Mechanical Properties of Solids

Q: 9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Mass of the big structure, M = 50000 kg = 50000 ×9.8N = 4.9 ×105 N Inner radius of the column, r = 30 cm = 0.3 m Outer radius of the column, R = 60 cm = 0.6 m Young’s modulus of steel, Y = 2 ×1011 Pa Total force exerted on 4 columns, F = 4.9 ×105 N Force exerted on single column, f = F/4 = 1.225 ×105 N Cross sectional area of each column, A = π (R2-r2) = π (0.62-0.32) = 0.848 m2 Stress in each column = f/A Young’s modulus, Y = Stress / Strain, Strain = Stress / Y = f/ (A ×Y) = 1.225×1050.848×2×1011=7.22×10-7 m

Q: 9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

Let us assume the depth = h, pressure at depth, Ph = 80 atm = 80 ×1.013×105 Pa Density of water at the surface, ρs = 1.03 ×103 kg/ m3 Let density of water at depth h be ρh Let Vs be the volume at the surface and Vh be the volume at depth h and ΔV be the change in volume. Let m be the mass of water. ΔV = Vs-Vh From the relation m = ρV, we get ΔV = m ( 1ρs - 1ρh ) = ρsVs ( 1ρs - 1ρh ) ΔVVs = 1- ρsρh ……(i) Bulk modulus of water, k = V1ΔPΔV = VsΔPΔV ΔVVs = ΔPk ……..(ii) Bulk modulus of water, k = 2.1 ×109 Pa Hence ΔVVs=80×1.013×1052.1×109 = 3.86 ×10-3 From equation (i) ρsρh = 0.99614 ρh=1.03×103/0.99614 = 1.034 ×103 kg/ m3

Q: 9.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Area of cross-section, A = 15.2 mm ×19.1mm = 15.2 ×19.1×10-6m2=2.9×10-4m2 Force, F = 44500 N Stress, F/A = (44500/ 2.9×10-4) N/ m2 Modulus of elasticity, η = Stress / Strains, Strains = Stress / η For copper, η = 42 ×109 N/m2 Strains = (44500/ 2.9×10-4)/ ( 42 ×109) = 3.65 ×10-3

Updated on Jul 17, 2025 15:54 IST

By Pallavi Pathak, Assistant Manager Content

Mechanical Properties of Solids Class 11 Physics is about solid's properties, which states that solid bodies are not perfectly rigid. Solid has a definite shape and size. The solid's shape and size can be changed by applying force. Even for a rigid steel bar, it can be deformed by applying a large external force. Mechanical properties of fluids class 11 NCERT solutions cover the concept of elasticity, plasticity, stress and strain, Hooke's law, elastic moduli, stress-strain curve, and applications of elastic behavior of materials.
Physics Class 11 Mechanical Properties of Solids PDF is also available for students. The well-structured solutions given in the PDF will help students to improve their problem-solving skills and to get high scores in the CBSE Board exam and other competitive exams.
Students preparing for CBSE board and NEET, and JEE exams can access the complete Class 11 Physics Notes and get the solved examples, weightage information, important topics, and chapter-wise free PDFs.

NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids: Key Concepts, Weightage

The Mechanical Properties of Fluids Class 11 Physics is a crucial chapter for students planning to appear in the competitive exams later. In NEET, it has moderate weightage, and its questions also appear in the JEE Mains. The key topics on which students should focus are stress and strain, Hooke's law, elasticity, and elastic potential energy.
See below the topics covered in this chapter:

Exercise	Topics Covered
8.1	Introduction
8.2	Stress and Strain
8.3	Hooke's Law
8.4	Stress-Strain Curve
8.5	Elastic Moduli
8.6	Applications of Elastic Behaviour of Materials

Mechanical Properties of Fluids Weightage in NEET, JEE Mains

Exam	Number of Questions
NEET	1-2 questions
JEE Main	one question

NCERT Physics Class11th Solution PDF for Mechanical Properties of Solids

Students must download the Physics Class 11 Mechanical Properties of Solids PDF from the link given below. It covers all the NCERT textbook questions of this chapter. The solutions help students to deepen their understanding of concepts and boost their confidence to score high in the CBSE and other competitive exams.

Download Here: NCERT Solution for Class XI Physics Chapter Mechanical Properties of Solids PDF

NCERT Physics Class11th Mechanical Properties of Solids Solutions

Q.9.1 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 of 4.0 × 10^–5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Ans.9.1

Length of the steel wire, $L_{1}$ = 4.7 m

Area of cross-section of the steel wire, $A_{1}$ = 3.0 $\times 10^{- 5}$ m2

Length of the copper wire, $L_{2}$ = 3.5 m

Area of cross-section of the copper wire, $A_{2}$ = 4.0 $\times 10^{- 5}$ m2

Change in length, $Δ {L =}_{} L_{1} - L_{2}$

Let the force applied = F

Young’s modulus in steel wire,

$Y_{1}$ = $\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{Δ L_{}}$ …..(1)

Young’s modulus in copper wire,

$Y_{2}$ = $\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{Δ L_{}}$ …….(2)

The ratio of Young’s modulus

$\frac{Y_{1}}{Y_{2}}$ = $\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{Δ L_{}} \times \frac{A_{2}}{F_{2}} \times \frac{Δ L_{}}{L_{2}}$ = $\frac{L_{1}}{A_{1}} \times \frac{A_{2}}{L_{2}}$ = $\frac{4.7 \times 4 \times 10^{- 5}}{3 \times 10^{- 5} \times 3.5}$ = $\frac{18.8}{10.5} = 1.79$

Q.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?

Ans.9.2

From the given graph, for the value stress 150 $\times 10^{6}$ N/ $m^{2}$ , the strain is 0.002

Young’s modulus = $\frac{150 \times 10^{6}}{0.002}$ = 7.5 $\times 10^{10}$ N/ $m^{2}$

Yield strength is the maxium strength the material can withstand in elastic limit. From the graph, the yield strength is 300 $\times 10^{6}$ or 3 $\times 10^{8} N / m^{2}$

Q.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 of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

Ans.9.3

Material A has greater Young’s modulus.

Material A is the strongest as it can withstand more strain than material B without fracture.

Q. 9.4 Read the following two statements below carefully and state, with reasons, if it is true

or false.

The Young’s modulus of rubber is greater than that of steel

The stretching of a coil is determined by its shear modulus

Ans.9.4

For a given stress, the strain in rubber is more than it is in steel, hence the Young’s modulus of rubber is lesser than in steel. So the statement is False.

Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

= 2.2 $\times 10^{- 4} m^{3}$

Commonly asked questions

9.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Diameter of wires, d = 0.25 cm, radius, r = 0.125 cm

Cross-sectional area, $A_{1}$ = $A_{2}$ = $π \times r^{2} = 0.049 {c m}^{2}$ = 4.908 $\times 10^{- 6}$ $m^{2}$

Length of the steel wire, $L_{1} = 1.5 m$ , length of the brass wire, $L_{2} = 1.0 m$

Change in length of the steel wire $= ? L_{1},$ Change in length of the copper wire $= ? L_{2}$

Total force exerted on the steel wire, $F_{1}$ = ( 4+6) kg = 10 kg = 98 N

Young’s modulus of steel , $Y_{1}$ = $\frac{\frac{F_{1}}{A_{1}}}{\frac{? L_{1}}{L_{1}}}$ = 2.0 $\times 10^{11}$ Pa

$? L_{1}$ = $\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{Y_{1}} = \frac{98 \times 1.5}{4.908 \times 10^{- 6} \times 2.0 \times 10^{11}}$ = 1.497 $\times 10^{- 4}$ m

Similarly for brass wire, $F_{2}$ = 6 kg = 58.8 N, $Y_{2}$ = $0.91 \times 10^{11} P a$

$? L_{2}$ = $\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{Y_{2}} = \frac{58.8 \times 1.0}{4.908 \times 10^{- 6} \times 0.91 \times 10^{11}}$ = 1.316 $\times 10^{- 4}$ m

9.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Mass of the big structure, M = 50000 kg = 50000 $\times 9.8 N$ = 4.9 $\times 10^{5}$ N

Inner radius of the column, r = 30 cm = 0.3 m

Outer radius of the column, R = 60 cm = 0.6 m

Young’s modulus of steel, Y = 2 $\times 10^{11}$ Pa

Total force exerted on 4 columns, F = 4.9 $\times 10^{5}$ N

Force exerted on single column, f = F/4 = 1.225 $\times 10^{5}$ N

Cross sectional area of each column, A = $π (R^{2} - r^{2})$ = $π ({0.6}^{2} - {0.3}^{2})$ = 0.848 $m^{2}$

Stress in each column = f/A

Young’s modulus, Y = Stress / Strain, Strain = Stress / Y = f/ (A $\times Y)$ = $\frac{1.225 \times 10^{5}}{0.848 \times 2 \times 10^{11}} = 7.22 \times 10^{- 7}$ m

9.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.

Length of an edge of the solid copper cube, l = 10 cm = 0.1 m

Volume of the copper cube = 1 $\times 10^{- 3}$ $m^{3}$

Hydraulic pressure, p = 7.0 $\times 10^{6}$ Pa

Bulk modulus of copper, k = 140 $\times 10^{9}$ Pa

From the relation k = $V \frac{Δ P}{Δ V}$ we get $Δ V$ = $V \frac{Δ P}{k}$ = $\frac{1 \times 10^{- 3} \times 7.0 \times 10^{6}}{140 \times 10^{9}}$ = 5 $\times 10^{- 8}$ $m^{3}$

9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

Let us assume the depth = h, pressure at depth, $P_{h}$ = 80 atm = 80 $\times 1.013 \times 10^{5}$ Pa

Density of water at the surface, $ρ_{s}$ = 1.03 $\times 10^{3}$ kg/ $m^{3}$

Let density of water at depth h be $ρ_{h}$

Let $V_{s}$ be the volume at the surface and $V_{h}$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_{s} - V_{h}$ From the relation m = $ρ V,$ we get

ΔV = m ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ ) = $ρ_{s} V_{s}$ ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ )

$\frac{Δ V}{V_{s}}$ = 1- $\frac{ρ_{s}}{ρ_{h}}$ ……(i)

Bulk modulus of water, k = $V_{1} \frac{Δ P}{Δ V}$ = $V_{s} \frac{Δ P}{Δ V}$

$\frac{Δ V}{V_{s}}$ = $\frac{Δ P}{k}$ ……..(ii)

Bulk modulus of water, k = 2.1 $\times 10^{9}$ Pa

Hence $\frac{Δ V}{V_{s}} = \frac{80 \times 1.013 \times 10^{5}}{2.1 \times 10^{9}}$ = 3.86 $\times 10^{- 3}$

From equation (i) $\frac{ρ_{s}}{ρ_{h}}$ = 0.99614

$ρ_{h} = 1.03 \times 10^{3} / 0.99614$ = 1.034 $\times 10^{3}$ kg/ $m^{3}$

9.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Area of cross-section, A = 15.2 mm $\times 19.1 m m$ = 15.2 $\times 19.1 \times 10^{- 6} m^{2} = 2.9 \times 10^{- 4} m^{2}$

Force, F = 44500 N

Stress, F/A = (44500/ $2.9 \times 10^{- 4})$ N/ $m^{2}$

Modulus of elasticity, $η$ = Stress / Strains, Strains = Stress / $η$

For copper, $η$ = 42 $\times 10^{9}$ $N / m^{2}$

Strains = (44500/ $2.9 \times 10^{- 4}) / ($ 42 $\times 10^{9})$ = 3.65 $\times 10^{- 3}$

9.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Edge of the aluminium cube, L = 10 cm = 0.1 m, Area A = 0.01 $m^{2}$

Mass attached, m = 100 kg = 100 $\times$ 9.8 = 980 N = Applied force F

Shear modulus $η$ = 25 GPa = 25 $\times 10^{9} P a$

Shear modulus $η$ = Shear stress / Shear strain = $\frac{\frac{F}{A}}{\frac{? L}{L}}$ , $? L = \frac{F \times L}{η \times A}$ = $\frac{980 \times 0.1}{25 \times 10^{9} \times 0.01}$ = 3.92 $\times 10^{- 7} m$

9.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N m–2, what is the maximum load the cable can support?

Area of cross-section, A = $π {1.5}^{2}$ ${c m}^{2}$ = 7.07 $\times 10^{- 4} m^{2}$

Maximum stress = Maximum load / cross sectional area

Maximum load = Maximum stress $\times$ cross sectional area = 108 $\times$ 7.07 $\times 10^{- 4}$ = 7.07 $\times 10^{4}$ N

9.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, $ω$ = 2 rev/s

Cross sectional area of the wire, A = 0.065 ${c m}^{2}$ = 0.065 $\times 10^{- 4} m^{2}$

Let Δl be the elongation of the wire

When the mass is placed at the position of the vertical circle, the total force on the mass is

F = mg + ml $ω^{2}$ = 14.5 $\times (9.81 + 1 \times 2^{2})$ = 200.25 N

Young’s modulus for steel, $Y_{s}$ = Stress / Strain = 2 $\times 10^{11}$ Pa

Stress = F/A = 200.25/0.065 $\times 10^{- 4}$

Strain = (Δl/l) = (Δl/1) = Δl

Δl = (200.25/0.065 $\times 10^{- 4}) /$ 2 $\times 10^{11}$ = 1.54 $\times 10^{- 4}$ m

9.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 liter, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 liter. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Initial volume, $V_{1}$ = 100 lit = 100 $\times 10^{- 3}$ $m^{3}$

Final volume, $V_{2}$ = 100.5 lit = 100.5 $\times 10^{- 3}$ $m^{3}$

Increase in volume, ΔV = $V_{2} - V_{1}$ = 0.5 $\times 10^{- 3}$ $m^{3}$

Increase in pressure, ΔP = 100 atm = 100 $\times$ 1.013 $\times 10^{5} P a$

Bulk modulus, k = $V_{1} \frac{Δ P}{Δ V}$ = $\frac{100 \times 10^{- 3} \times 100 \times 1.013 \times 10^{5}}{0.5 \times 10^{- 3}}$ Pa = 2.026 $\times 10^{9}$ Pa

Bulk modulus of air = 1.0 $\times 10^{5} P a$

(Bulk modulus of water / Bulk modulus of air) = (2.026 $\times 10^{9}) /$ 1.0 $\times 10^{5}$ = 2.026 $\times 10^{4}$

This higher ratio is attributed to the higher compressibility of air than water.

9.4 Read the following two statements below carefully and state, with reasons, if it is true or false.

The Young’s modulus of rubber is greater than that of steel

The stretching of a coil is determined by its shear modulus

For a given stress, the strain in rubber is more than it is in steel, hence the Young's modulus of rubber is lesser than in steel. So the statement is False.

Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

= 2.2 $* 10^{- 4} m^{3}$

9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Water pressure at the bottom, p = 1.1 $\times 10^{8}$ Pa

Initial volume of the steel ball, V = 0.32 $m^{3}$

Bulk modulus of the steel, B = 1.6 $\times 10^{11}$ N/ $m^{2}$

The ball falls at the bottom of the Pacific ocean which is 11 km beneath the surface

Let the change in volume of the ball on reaching the bottom of the trench be ΔV

We know, bulk modulus, B = $\frac{p}{\frac{Δ V}{V}}$ or ΔV = $\frac{p V}{B}$

ΔV = $\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}$ = 2.2 $\times 10^{- 4} m^{3}$

9.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Hydraulic pressure exerted on glass slab, p = 10 atm = 10 $\times 1.1013 \times 10^{5}$ Pa

Bulk modulus of glass, k = 37 $\times 10^{9}$ N $m^{- 2}$

From the relation k = $V \frac{Δ P}{Δ V}$ , we get $\frac{Δ V}{V}$ = $\frac{Δ P}{k}$ = $\frac{10 \times 1.1013 \times 10^{5}}{37 \times 10^{9}}$ = 2.976 $\times 10^{- 5}$

9.16 How much should the pressure on a liter of water be changed to compress it by 0.10%?

Volume of water, V = 1 liter. If the water is compressed by 10%, then

ΔV = 0.10% of V= (0.1/100) $\times 1$ = 1 $\times 10^{- 3}$

Bulk modulus of water, k = 2.2 $\times 10^{9}$ N/ $m^{2}$

From the relation, k = $V \frac{Δ P}{Δ V}$ , we get ΔP = k $\times \frac{Δ V}{V}$ = 2.2 $\times 10^{9} \times$ 1 $\times 10^{- 3} \frac{}{1}$ = 2.2 $\times 10^{6}$ N/ $m^{2}$

9.20 Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa? Assume that each rivet is to carry one quarter of the load.

Diameter of the metal strip, d = 6.0 mm = 6 $\times 10^{- 3}$ m

Radius, r = d/2 = 3 $\times 10^{- 3}$ m

Maximum shearing stress = 6.9 $\times 10^{7}$ Pa = $\frac{M a x i m u m f o r c e}{A r e a}$

Maximum force = Maximum stress $\times A r e a$

= 6.9 $\times 10^{7} \times π \times r^{2}$ = 6.9 $\times 10^{7} \times π \times {(3 \times 10^{- 3})}^{2}$ = 1950.93 N

Since each rivet carries 1/4th of load,

Maximum tension on each rivet = 4 $\times 1950.93$ N = 7803.72 N

9.1 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 of 4.0 × 10^–5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Length of the steel wire, $L_{1}$ = 4.7 m

Area of cross-section of the steel wire, $A_{1}$ = 3.0 $\times 10^{- 5}$ m2

Length of the copper wire, $L_{2}$ = 3.5 m

Area of cross-section of the copper wire, $A_{2}$ = 4.0 $\times 10^{- 5}$ m2

Change in length, $Δ {L =}_{} L_{1} - L_{2}$

Let the force applied = F

Young’s modulus in steel wire,

$Y_{1}$ = $\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{Δ L_{}}$ …..(1)

Young’s modulus in copper wire,

$Y_{2}$ = $\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{Δ L_{}}$ …….(2)

The ratio of Young’s modulus

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?

From the given graph, for the value stress 150 $* 10^{6}$ N/ $m^{2}$ , the strain is 0.002

Young's modulus = $\frac{150 * 10^{6}}{0.002}$ = 7.5 $* 10^{10}$ N/ $m^{2}$

Yield strength is the maxium strength the material can withstand in elastic limit. From the graph, the yield strength is 300 $* 10^{6}$ or 3 $* 10^{8} N / m^{2}$

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 of the materials has the greater Young’s modulus?

(b) Which of the two is the stronger material?

Material A has greater Young's modulus.

Material A is the strongest as it can withstand more strain than material B without fracture.

9.10 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 ratios of their diameters if each is to have the same tension.

It is given that the tension, F in each wire is same. Since the wire is of same length, Strain also will be same.

If $d_{c}$ is the diameter of copper wire and Young’s modulus of copper $Y_{c}$ = 110 $\times 10^{9} P a$ and strain is s, then $Y_{c}$ = $\frac{F}{\frac{π}{4} {d_{c}}^{2}}$ , $d_{c} = \sqrt{\frac{4 F}{π \times Y_{c}}}$

Similarly, if $d_{i}$ is the diameter of iron wire and $Y_{i}$ is the Young’s modulus of iron = 190 $\times 10^{9} P a$ , then $d_{i}$ = $\sqrt{\frac{4 F}{π \times Y_{i}}}$

$\frac{d_{c}}{d_{i}}$ = $\sqrt{\frac{Y_{i}}{Y_{c}}}$ = $\sqrt{\frac{190}{110}}$ = 1.314

9.17 Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behavior of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?

Diameter of the cone at the narrow end, d = 0.5 mm = 0.5 $\times 10^{- 3}$ m

Radius, r = d/2 = 0.25 $\times 10^{- 3}$ m

Area, A = $π \times r^{2}$ = 1.96 $\times 10^{- 7}$ $m^{2}$

Compressional force, F = 50000 N

Pressure at the tip of the anvil, p = F/A = 50000/1.96 $\times 10^{- 7}$ Pa = 2.54 $\times 10^{11}$ Pa

9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $\times 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $\times 10^{- 6} m^{2}$

Young’s modulus for steel, $Y_{1}$ = 2 $\times 10^{11}$ N/ $m^{2}$

Young’s modulus for aluminium, $Y_{2}$ = 7 $\times 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1} y$ = $F_{2} (1.05 - y)$

$\frac{F_{1}}{F_{2}}$ = $\frac{(1.05 - y)}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2} = \frac{(1.05 - y)}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young’s modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${S t r a i n}_{s t e e l}$ = $\frac{F_{1}}{a_{1} \times Y_{1}}$

In case of Aluminium wire, ${S t r a i n}_{a l u}$ = $\frac{F_{2}}{a_{2} \times Y_{2}}$

Since both the strains are equal, we get

$\frac{F_{1}}{F_{2}}$ = $\frac{a_{1} \times Y_{1}}{a_{2} \times Y_{2}}$ = $\frac{1 \times 10^{- 6} \times 2 \times 10^{11}}{2 \times 10^{- 6} \times 7 \times 10^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05 - y)}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A

9.19 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, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $\times 10^{- 2}$ ${c m}^{2}$ = 0.5 $\times 10^{- 6} m^{2}$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^{2} + l^{2}}$

From equation (i)

Δl = 2 $\times \sqrt{{0.5}^{2} + l^{2}}$ - 1.0 = 2 $\times 0.5 \times \sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0 = $\sqrt{1 + {(\frac{l}{0.5})}^{2}}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^{2}}{0.5}$

We know, Strain = $\frac{I n c r e a s e i n l e n g t h}{O r i g i n a l l e n g t h}$

Let T be the tension in the wire, then

mg = 2T $\cos ? θ$

From the figure

$\cos ? θ$ = $\frac{O Y}{O X}$ = $\frac{0.5}{\sqrt{{0.5}^{2} + l^{2}}}$ = $\frac{0.5}{0.5 \times \sqrt{1 + {(\frac{l}{0.5})}^{2}}}$

Expanding the expression and eliminating the higher terms:

$\cos ? θ$ = $\frac{l}{0.5 \times \sqrt{1 + {(\frac{1}{2}) (\frac{l}{0.5})}^{2}}}$

Since ( 1+ $\frac{l^{2}}{0.5}$ ) = 1 for small l, $\cos ? θ$ = $\frac{1}{0.5}$

Then T = $\frac{m g}{2 (\frac{l}{0.5})}$ = $\frac{m g \times 0.5}{2 l}$ = $\frac{m g}{4 l}$

Stress = $\frac{T e n s i o n}{A r e a}$ = $\frac{m g}{4 l \times A}$

Young’s modulus = $\frac{S t r e s s}{S t r a i n}$

Y = $\frac{m g \times 0.5}{? l \times A \times l^{2}}$

l= $\sqrt{\frac{m g \times 0.5}{4 Y A}}$

For steel, Y = 2 $\times 10^{11}$ Pa, then

l = $\sqrt{\frac{0.1 \times 9.8 \times 0.5}{4 \times 2 \times 10^{11} \times 0.5 \times 10^{- 6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

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