NCERT Solutions for class-11 Physics Chapter 9 Mechanical Properties of Solid As Per Latest Syllabus

NCERT class-11 Physics Chapter 9 Mechanical Properties of Solid

Summary:

1. Introduction to Mechanical Properties:
   • Mechanical properties of solids refer to their ability to resist deformation under the action of external forces.
   • These properties include elasticity, plasticity, toughness, hardness, brittleness, ductility, and malleability.
     
     
2. Elasticity:
   • Elasticity is the ability of a material to regain its original shape after deformation when the external forces are removed.
   • Hooke’s law describes the relationship between stress (σ) and strain (ϵ) for elastic materials: σ=E⋅ϵ, where E is the Young’s modulus of elasticity.
     
     
3. Young’s Modulus:
   • Young’s modulus (Y) is a measure of the stiffness of a material and is defined as the ratio of stress to strain within the elastic limit.
   • It is expressed in units of pascals (Pa) or newtons per square meter (N/m²).
     
     
4. Shear Modulus:
   • Shear modulus (G) is a measure of a material’s resistance to shear deformation.
   • It quantifies the ratio of shear stress to shear strain and is applicable to materials subjected to perpendicular forces.
     
     
5. Bulk Modulus:
   • Bulk modulus (B) measures a substance’s resistance to uniform compression.
   • It relates the relative change in volume to the applied compressive stress.
     
     
6. Poisson’s Ratio:
   • Poisson’s ratio (σ) is a measure of the ratio of transverse contraction strain to longitudinal extension strain in the direction of an applied force.
   • It is a dimensionless quantity and typically lies between -1 and 0.5.
     
     
7. Plasticity:
   • Plasticity is the ability of a material to undergo permanent deformation without rupture under the action of external forces.
   • Unlike elastic deformation, plastic deformation is irreversible.
     
     
8. Toughness:
   • Toughness is the ability of a material to absorb energy and plastically deform before fracturing.
   • Materials with high toughness can withstand significant impact without breaking.
     
     
9. Hardness:
   • Hardness is a measure of a material’s resistance to indentation or scratching.
   • It depends on the material’s strength, elastic modulus, plasticity, ductility, and strain-hardening characteristics.
     
     
10. Brittleness:
    • Brittleness is the tendency of a material to fracture or fail without significant plastic deformation when subjected to stress.
    • Brittle materials have low toughness and little capacity for energy absorption.
      
      
11. Ductility:
    • Ductility is the ability of a material to undergo significant plastic deformation before fracture under tensile stress.
    • Ductile materials can be drawn into wires or stretched into thin sheets without breaking.
      
      
12. Malleability:
    • Malleability is the ability of a material to undergo deformation under compressive stress, leading to the flattening or spreading of its mass into thin sheets.
    • Malleable materials can be shaped or formed by hammering or rolling.

Questions And Answers:

Question 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?

Solution :
Length of the steel wire, L₁ = 4.7 m
Area of cross-section of the steel wire, A₁ = 3.0 × 10^–5 m²
Length of the copper wire, L₂ = 3.5 m
Area of cross-section of the copper wire, A₂ = 4.0 × 10^–5 m²
Change in length = ΔL₁ = ΔL₂ = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:
Y₁ = (F₁ / A₁) (L₁ / ΔL₁)
= (F / 3 X 10^-5) (4.7 / ΔL)     ….(i)
Young’s modulus of the copper wire:
Y₂ = (F₂ / A₂) (L₂ / ΔL₂)
= (F / 4 × 10^-5) (3.5 / ΔL)     ….(ii)
Dividing (i) by (ii), we get:
Y₁ / Y₂  =  (4.7 × 4 × 10^-5) / (3 × 10^-5 × 3.5)
= 1.79 : 1
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

Question2. 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?

Solution :
(a) It is clear from the given graph that for stress 150 × 10⁶ N/m², strain is 0.002.
∴Young’s modulus, Y = Stress / Strain
= 150 × 10⁶ / 0.002  =  7.5 × 10¹⁰ Nm^-2
Hence, Young’s modulus for the given material is 7.5 ×10¹⁰ N/m².

(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.
It is clear from the given graph that the approximate yield strength of this material is 300 × 10⁶ Nm/² or 3 × 10⁸ N/m².

Question3. 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?

Solution :
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence, Young’s modulus (=stress/strain) is greater for A than that of B.

(b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

Question4. 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 stretching of a coil is determined by its shear modulus.

Solution :
(a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.

(b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elsticity is involved.

Question 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.

Solution :
Elongation of the steel wire = 1.49 × 10^–4 m
Elongation of the brass wire = 1.3 × 10^–4 m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = d/2  =  0.125 cm
Length of the steel wire, L₁ = 1.5 m
Length of the brass wire, L₂ = 1.0 m
Total force exerted on the steel wire:
F₁ = (4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:
Y₁ = (F₁/A₁) / (ΔL₁ / L₁)
Where,
ΔL₁ = Change in the length of the steel wire
A₁ = Area of cross-section of the steel wire = πr₁²
Young’s modulus of steel, Y₁ = 2.0 × 10¹¹ Pa
∴ ΔL₁ = F₁  × L₁ / (A₁  × Y₁)
= (98  × 1.5) /[ π(0.125  × 10^-2)²  × 2  × 10¹¹]   =   1.49  × 10^-4 m

Total force on the brass wire:
F₂ = 6 × 9.8 = 58.8 N
Young’s modulus for brass:
Y₂ = (F₂/A₂)/ (ΔL₂ / L₂)
Where,
ΔL₂ = Change in the length of the brass wire
A₁ = Area of cross-section of the brass wire = πr₁²
∴ ΔL₂ = F₂  × L₂ / (A₂  × Y₂)
= (58.8 X 1) / [ (π  × (0.125  × 10^-2)²  × (0.91  × 10¹¹)] = 1.3  × 10^-4 m
Elongation of the steel wire = 1.49 × 10^–4 m
Elongation of the brass wire = 1.3 × 10^–4 m.

Question 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?

Solution :
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 10⁹ Pa
Shear modulus, η = Shear stress / Shear strain  =  (F/A) / (L/ΔL)
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m²
ΔL = Vertical deflection of the cube
∴ ΔL = FL / Aη
= 980 × 0.1/ [ 10^-2 × (25 × 10⁹) ]
= 3.92 × 10^–7 m
The vertical deflection of this face of the cube is 3.92 ×10^–7 m.

Question 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 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.< /p>

Solution :
Mass of the big structure, M = 50,000 kg
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, F = Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = 50000 × 9.8 / 4  =  122500 N
Young’s modulus, Y = Stress / Strain
Strain = (F/A) / Y
Where,
Area, A = π (R² – r²) = π ((0.6)² – (0.3)²)
Strain = 122500/ [ π ((0.6)² – (0.3)²) × 2 × 10¹¹ ]  =  7.22 × 10^-7
Hence, the compressional strain of each column is 7.22 × 10^–7.

Question8. 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?

Solution :
Length of the piece of copper, l = 19.1 mm = 19.1 × 10^–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10^–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10^–3 × 15.2 × 10^–3
= 2.9 × 10^–4 m²
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 10⁹ N/m²
Modulus of elasticity, η = Stress / Strain
= (F/A) / Strain
∴ Strain = F / Aη
= 44500/ (2.9 × 10^-4 × 42 × 10⁹)
= 3.65 × 10^–3.

Question 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?

Solution :
Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 10⁸ N m^–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π (0.015)²
= 7.065 × 10⁴ N
Hence, the cable can support the maximum load of 7.065 × 10⁴ N.

Question 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 ratio of their diameters if each is to have the same tension.

Solution :
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:
Y = Stress / Strain
= (F/A)/ Strain  =  (4F/πd²)/ Strain     ….(i)
Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ (1/d²)
Young’s modulus for iron, Y₁ = 190 × 10⁹ Pa
Diameter of the iron wire = d₁
Young’s modulus for copper, Y₂ = 120 × 10⁹ Pa
Diameter of the copper wire = d₂
Therefore, the ratio of their diameters is given as:

Question 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.

Solution :
Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm² = 0.065 × 10^-4 m²
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω²
= 14.5 × 9.8 + 14.5 × 1 × (12.56)²
= 2429.53 N
Young’s modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 10¹¹ Pa
∆l = 2429.53 × 1 / (0.065 × 10^-4 × 2 × 1011)   =   1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10^–3 m.

Question 12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Solution :
Initial volume, V₁ = 100.0l = 100.0 × 10 ^–3 m³
Final volume, V₂ = 100.5 l = 100.5 ×10 ^–3 m³
Increase in volume, ΔV = V₂ – V₁ = 0.5 × 10^–3 m³
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 10⁵ Pa
Bulk modulus = Δp/ (ΔV/V₁)  =  Δp × V₁ / ΔV
= 100 × 1.013 × 10⁵ × 100 × 10^-3 / (0.5 × 10^-3)
= 2.026 × 10⁹ Pa
Bulk modulus of air = 1 × 10⁵ Pa
∴ Bulk modulus of water / Bulk modulus of air  =  2.026 × 10⁹/ (1 × 10⁵)  =  2.026 × 10⁴
This ratio is very high because air is more compressible than water.

Question13. 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?

Solution :
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10⁵ Pa
Density of water at the surface, ρ₁ = 1.03 × 10³ kg m^–3
Let ρ₂ be the density of water at the depth h.
Let V₁ be the volume of water of mass m at the surface.
Let V₂ be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.
ΔV = V₁ – V₂
= m [ (1/ρ₁) – (1/ρ₂) ]
∴ Volumetric strain = ΔV / V₁
= m [ (1/ρ₁) – (1/ρ₂) ] × (ρ₁ / m)
ΔV / V₁ = 1 – (ρ₁/ρ₂)     ……(i)
Bulk modulus, B = pV₁ / ΔV
ΔV / V₁ = p / B
Compressibility of water = (1/B) = 45.8 × 10^-11 Pa^-1
∴ ΔV / V₁ = 80 × 1.013 × 10⁵ × 45.8 × 10^-11  =  3.71 × 10^-3    ….(ii)
For equations (i) and (ii), we get:
1 – (ρ₁/ρ₂)   =   3.71 × 10^-3
ρ₂ = 1.03 × 10³/ [ 1 – (3.71 × 10^-3) ]
= 1.034 × 10³ kg m^-3
Therefore, the density of water at the given depth (h) is 1.034 × 10³ kg m^–3.

Question14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Solution :
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 10⁵ Pa
Bulk modulus of glass, B = 37 × 10⁹ Nm^–2
Bulk modulus, B = p / (∆V/V)
Where,
∆V/V = Fractional change in volume
∴ ∆V/V = p / B
= 10 × 1.013 × 105 / (37 × 109)
= 2.73 × 10^-5
Hence, the fractional change in the volume of the glass slab is 2.73 × 10^–5.

Question15. 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.

Solution :
Length of an edge of the solid copper cube, l = 10 cm = 0.1 m
Hydraulic pressure, p = 7.0 × 10⁶ Pa
Bulk modulus of copper, B = 140 × 10⁹ Pa
Bulk modulus, B = p / (∆V/V)
Where,
∆V/V = Volumetric strain
ΔV = Change in volume
V = Original volume.
ΔV = pV / B
Original volume of the cube, V = l³
∴ ΔV = pl³ / B
= 7 × 10⁶ × (0.1)³ / (140 × 10⁹)
= 5 × 10^-8 m³  =  5 × 10^-2 cm^-3
Therefore, the volume contraction of the solid copper cube is 5 × 10^–2 cm^–3.

Question16. How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution :
Volume of water, V = 1 L
It is given that water is to be compressed by 0.10%.
∴ Fractional change, ∆V / V = 0.1 / (100 × 1)  =  10^-3
Bulk modulus, B = ρ / (∆V/V)
ρ = B × (∆V/V)
Bulk modulus of water, B = 2.2 × 10⁹ Nm^-2
ρ = 2.2 × 10⁹ × 10^-3  =  2.2 × 10⁶ Nm^-2
Therefore, the pressure on water should be 2.2 ×10⁶ Nm^–2.

Question17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour 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?

Solution :
Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10^–3 m
Radius, r = d/2  =  0.25 × 10^-3 m
Compressional force, F = 50000 N
Pressure at the tip of the anvil:
P = Force / Area  =  50000/ π(0.25 × 10^-3)²
= 2.55  × 10¹¹ Pa
Therefore, the pressure at the tip of the anvil is 2.55 × 10¹¹ Pa.

Question 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.

Solution :
Cross-sectional area of wire A, a₁ = 1.0 mm² = 1.0 × 10^–6 m²
Cross-sectional area of wire B, a₂ = 2.0 mm² = 2.0 × 10^–6 m²
Young’s modulus for steel, Y₁ = 2 × 10¹¹ Nm^–2
Young’s modulus for aluminium, Y₂ = 7.0 ×10¹⁰ Nm^–2

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then:
F₁ / a₁  =  F₂ / a₂
Where,
F₁ = Force exerted on the steel wire
F₂ = Force exerted on the aluminum wire
F₁ / F₂ = a₁ / a₂  =  1 / 2    ….(i)
The situation is shown in the following figure:

Taking torque about the point of suspension, we have:
F₁y = F₂ (1.05 – y)
F₁ / F₂ = (1.05 – y) / y    ……(ii)
Using equations (i) and (ii), we can write:
(1.05 – y) / y  = 1 / 2
2(1.05 – y)  =  y
y = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young’s modulus = Stress / Strain
Strain = Stress / Young’s modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F₁/a₁)/ Y₁  =  (F₂/a₂)/ Y₂
F₁ / F₂ = a₁Y₁ / a₂Y₂
a₁ / a₂ = 1/2
F₁ / F₂ = (1 / 2) (2 × 10¹¹ / 7 × 10¹⁰)  =  10 / 7      …….(iii)
Taking torque about the point where mass m, is suspended at a distance y₁ from the side where wire A attached, we get:
F₁y₁ = F₂ (1.05 – y₁)
F₁ / F₂  =  (1.05 – y₁) / y₁    ….(iii)
Using equations (iii) and (iv), we get:
(1.05 – y₁) / y₁  =  10 / 7
7(1.05 – y₁)  =  10y₁
y₁ = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.

Question 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.

Solution :

From the above figure,

Let x be the depression at the mid point

i.e. CD = x.

In fig.,

AC= CB = l = 0.5 m ;

m = 100 g = 0.100 Kg

AD= BD = (l² + x²)^1/2

Increase in length, ∆l = AD + DB – AB = 2AD – AB

Question 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.

Solution :
Diameter of the metal strip, d = 6.0 mm = 6.0 × 10^–3 m
Radius, r = d/2 = 3 × 10^-3 m
Maximum shearing stress = 6.9 × 10⁷ Pa
Maximum stress = Manimum load or force / Area
Maximum force = Maximum stress × Area
= 6.9 × 10⁷ × π × (r) ²
= 6.9 × 10⁷ × π × (3 ×10^–3)²
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N.

Question 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?

Solution :
Water pressure at the bottom, p = 1.1 × 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = p / (∆V/V)
∆V  =  B / pV
= 1.1 × 10⁸ × 0.32 / (1.6 × 10¹¹ )  =  2.2 × 10^-4 m³
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10^–4 m³.