Solution-Class-11-Science-Physics-Chapter-6-Mechanical Properties of Solids-Maharashtra Board

(A) deformation

(B) yield point

(B) 3 m³

(C) hardness

(A) toughness

Elasticity is the property of a body or material which enables it to regain its original shape and size after removal of the deforming force.

A change in the shape or size or both of a body due to an external force is called deformation. [size ≡ volume]

The SI unit of stress is the Pascal or the newton per square metre (Pa ≡ N/m²). The dimensions of stress are [L^-1M¹T^-2].

Strain is defined as the ratio of change in the physical dimensions of a stressed body to its original dimensions.

A perfectly rigid body cannot be deformed by any external force. Hence, there is no question of stress, strain and Young's modulus in this case.

After extensive use, bridges acquire wear and tear and lose their elastic qualities. Hence, they become unsafe.

To produce shear stress in a body, the applied force or its component should be tangential to the surface of the body.

Hooke's law holds good up to a certain point in deformation, where the applied force is minimal. In an elastic deformation, the atoms/molecules, of the body shift only a small fraction of an interatomic/intermolecular spacing in response to deforming forces and the internal restoring force is proportional to the strain produced by the applied force.

Within the elastic limit, the ratio of lateral strain to the linear strain is called Poisson’s ratio.

A material which can be elastically stretched to a large value of strain is called an elastomer.

In a material like vulcanized rubber, the material does not regain its original state immediately on removal of the applied force. Strain lags behind stress. This behaviour is called elastic hysteresis.

• Hardest material : Diamond
• Softest material : Plasticine clay

The property which resists the relative motion between two surfaces in contact is called friction.

Up to a certain limit, the force of static friction changes with the applied force such that the two forces are equal in magnitude and opposite in direction, keeping the body at rest Hence, the force of static friction is known as ‘self- adjusting force.

The coefficient of friction depends on the normal reaction between the surfaces in contact, the materials and nature of the surfaces.

• Polishing the bearings
• Using a lubricant
• Using ball bearings
• Giving a body an aerodynamic shape
• Using antifriction metals.

Friction between two bodies in contact when one body rolls over the other, is called rolling friction. In this case, the point of contact between the surfaces keeps changing.

Lubricating moving parts with oil creates a thin oil film between them, reducing contact friction significantly. Hence, substituting contact friction with fluid friction greatly reduces the contact friction between the moving parts. Petroleum oils and greases are common lubricants.

Examples of solid lubricants : Graphite, molybdenum disulphide and talc.

(i) The limiting force of static friction (f_s) is directly proportional to the magnitude of the normal reaction (N) between the two surfaces in contact: f_s ∝ N.

(ii) The limiting force of static friction is independent of the apparent geometrical area between the surfaces in contact, so long as the normal reaction remains the same.

(iii) The limiting force of static friction depends upon materials in contact and the nature of their surfaces.

• The magnitude of the force of kinetic friction (f_k) is directly proportional to the magnitude of the normal reaction (N) between the two surfaces in contact.

f_s ∝ N.

• The force of kinetic friction is independent of shape and apparent area of the surfaces in contact.
• The force of kinetic friction depends upon the nature and material of the surfaces in contact.
• The magnitude of the force of kinetic friction is independent of the relative velocity between the object and the surface provided that the relative velocity is neither too large nor too small.

It is due to friction that,

• locomotion, such as walking or running, is possible
• machines work.
• a wheel can roll on a surface.
• we can hold objects.

Disadvantages of friction :

• Friction between moving parts of a machine or that between the tyres and road decreases the speed.
• The work done by frictional force is dissipated as heat. Thus, only a part of the input energy is available for useful work. Fuel is wasted in order to overcome the frictional force and maintain the desired speed.
• Friction leads to wear and tear of the surfaces in contact which are in relative motion.
• Friction produces noise.

A substance is said to be brittle if fracture occurs soon after the elastic limit.

Examples : glass and ceramics.

• Stress is defined as the net internal restoring force per unit area of the surface of a body over which the deforming force is applied.
• Strain is defined as the ratio of change in the physical dimensions of a stressed body to its original dimensions.

Types of stresses and corresponding strains :

• Normal or longitudinal stress, which changes the length of a body. Longitudinal strain is the resulting fractional change in length of the body.
• Volume stress, which changes the volume of a body without producing a change in its shape. Volume strain is the resulting fractional change in volume of the body.
• Shear stress, which changes the shape of a body. The corresponding strain is called the shear strain or shear.

Determination of Young's modulus of the material of a long and thin wire :

Apparatus :

• The experimental wire ‘B’ is suspended from a rigid wall-support along
• with a compensating or reference wire ‘A’ of the same length, material and cross section.
• A main scale graduated in millimetre is attached to the wire ‘A’.
• A Vernier scale attached to the wire ‘B’ can move alongside the main scale.

Procedure :

(1) The length L and diameter d of the experimental wire are measured. The radius r (=\(\frac{1}{2}d\)) of the wire is calculated.

(2) Wire B is initially loaded with 1 kg to 2 kg to keep it taut and free from kinks. This load is treated as a zero-load. The main scale and Vernier scale readings are noted as those for zero-load.

(3) The stretching load Mg on wire B is increased by 0.5 kg at a time. The scale readings are noted for each increase in load, after allowing sufficient time for the wire to elongate.

(4) After 6-8 such readings, the wire is unloaded 0.5 kg at a time, and the corresponding scale readings are noted. Within the limits of experimental error, the readings for the same load—while loading and unloading—should be the same.

(5) The average elongation per unit mass, ΔL/M is found.

(6) Young's modulus of the material of the wire is calculated using the formula.

Y = \(\frac{longitudinal\,stress}{longitude\,strain} = \frac{Mg/πr^2}{ΔL/L}=\frac{gL}{πr^2(ΔL/M)}\)

Expression for strain energy per unit volume of the material of a wire :

Consider a uniform wire of length L and area of cross section A is suspended from a rigid support and stretched by applying a load Mg at the other end.

The internal restoring force gradually and uniformly increases from O to F (=Mg). During the time the wire stretches, at some instant, let the magnitude of the applied force be f and the elongation produced be x.

When fully stretched, x becomes l.

Young's modulus = \(\frac{\text{longitudinal stress}}{\text{longitudinal strain}}=\frac{f/A}{x/L}=\frac{fL}{Ax}\) 

∴ f = \(\frac{YAx}{L}\)     ….(1)

The work done by the force f, against the internal restoring force, in stretching the wire through a further infinitesimal length dx is

dW = fdx = \(\frac{YAx}{L}dx\)  

The total work done in producing the elongation ΔL is

W = ∫ dW = \(\int_{0}^{ΔL}\frac{YAx}{L}dx\)= \(\frac{YA}{L}\int_{0}^{ΔL}x\,dx\)

= \(\frac{YA}{L}[\frac{x^2}{2}]_{0}^{ΔL}\)=\(\frac{YAΔL^2}{2L}\)=\(\frac{1}{2}(\frac{YAΔL}{L})ΔL\)

But Y = \(\frac{F/A}{ΔL/L}=\frac{Mg/A}{ΔL/L}=\frac{MgL}{AΔL}\)

∴ Maximum restoring force, F = Mg = \(\frac{YAΔL}{L}\)

Hence W = \(\frac{1}{2}FΔL=\frac{1}{2}MgΔL\)  ….(2)

Thus, the total work done in stretching the wire = \(\frac{1}{2}\)stretching force x elongation.

The work done by the applied force in stretching the wire is stored in the stretched (or strained) wire as the potential energy which is then called the strain energy.

Thus, strain energy = \(\frac{1}{2}FΔL\)  ….(3)

∴ Strain energy per unit volume of the stretched wire  = \(\frac{\frac{1}{2}FΔL}{AL}\) = \(\frac{1}{2}(\frac{F}{A})(\frac{ΔL}{L})\)

= \(\frac{1}{2}\)(tensile stress).(tensile strain)  ……(4)

= \(\frac{1}{2Y}\)(tensile stress)²    .....\(['.'Y=\frac{tensile\,stress}{tensile\,strain}]\)

Also, strain energy per unit volume

=  \(\frac{1}{2}\)(tensile stress).(tensile strain = \(\frac{1}{2}Y\)(tensile strain)²

The property which resists the relative motion between two surfaces in contact is called friction.

The coefficient of static friction (μ_s) is defined as the ratio of the limiting force of

friction (f_s) to the normal reaction (N) between the two surfaces in contact : μ_s = \(\frac{f_s}{N}\)

The coefficient of kinetic friction (μ_k) is defined as the ratio of the force of kinetic friction (f_k) to the normal reaction (N) between the two surfaces in contact.

μ_{k = \(\frac{f_k}{N}\)}

Hooke's law: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it.

Fig. Graph of tensile stress against tensile strain for a metal wire up to the breaking point

Given : m = 50 kg, g = 9.8 m/s₂, f_s = 50N

f_s = μ_s N = μ_smg

μs = \(\frac{f_s}{mg}=\frac{50}{50×9.8}\) = 0.102

0.102 is the coefficient of static friction.

Given : m = 37 kg, μ _s = 0.3, g = 9.8 m/s²,

Required force f_s = μ_smg = 0.3 x 37 x 9.8 = 108.8 N

108.8 N is the required force.

Given = m = 37kg, f_s = 68.5 N, f_k = 43 N, g = 9.8 m/s²

(a) μ_s = \(\frac{f_s}{mg}=\frac{68.5}{37×9.8}\)  = 0.1889

0.1889 is the coefficient of static friction.

(b) μ_k = \(\frac{f_s}{mg}=\frac{43}{37×9.8}\) =  = 0.1186

0.1186 is the coefficient of kinetic friction

Given : Δ L = 4 mm, L₂ = \(\frac{L_1}{2}\), d₂ = 2d₁ , r₂ = 2r₁,

same load F, same Y

Y = \(\frac{FL}{πr^2ΔL}\)  ∴ ΔL = \(\frac{FL}{πr^2Y}\)

ΔL₁ = \(\frac{FL_1}{πr_1^2Y}\)  and ΔL₂ = \(\frac{FL_2}{πr_2^2Y}\)

∴ \(\frac{ΔL_1}{ΔL_2}=(\frac{L_2}{L_1})(\frac{r_1}{r_2})^2\) = \((\frac{1}{2})(\frac{1}{2})^2\) = \(\frac 18\)

ΔL₂ = \(\frac{ΔL_1}{8}=\frac{4}{8}\) = 0.5mm

0.5 mm is the change in the length of the second wire.

Given : L = 2m, F = 100 N,

A = 2.25 x 10^-8 m²,

Y = 2 x 10¹¹ N/m²

W = \(\frac{1}{2}\)FΔL, Y = \(\frac{FL}{AΔL}\)  ∴ ΔL = \(\frac{FL}{AY}\)

W = \(\frac{1}{2}\)F x \(\frac{FL}{AY}\) = \(\frac{F^2L}{2AY}\)

= \(\frac{(100)^2(2)}{(2)(2.25×10^{-8})(2×10^{11})}\)

= \(\frac{20}{4×2.25}\) = 2.222 J

The work done in stretching the wire is 2.222 J.

Given : v = 0.31 m³, p = 2 x 10⁷ N/m², K = 6.1 x 10¹⁰ N/m²

K = \(-\frac{p}{ΔV/V}\) = \(-\frac{pV}{ΔV}\)

ΔV = \(-\frac{pV}{K}\) = \(-\frac{(2×10^7)(0.31)}{6.1×10^{10}}\)

= \(-\frac{0.62}{6.1}×10^{-3}\)

= −0.1016 X 10⁻³

= −1.016 x 10^-4 m³

The change (decrease) in the volume of the sphere is −1.016 x 10^-4 m³ .

Given : L = 1.5 m, d = 0.60 mm, Δ L = 6.3 mm = 6.3 x 10^-3m, Y = 2.1 x 10¹¹ N/m²

r = \(\frac{d}{2}\) = 0.30mm = 3 x 10^-4 m

Y= \(\frac{FL}{AΔL}\)  ∴ F = \(\frac{YAΔL}{L}\) = \(\frac{Yπr^2ΔL}{L}\)

∴ F = \(\frac{2.1×10^{11}×3.14(3×10^{-4})^2×6.3×10^{-3}}{1.5}\)

= 2.1 x 3.142 x 6 x 6.3 = 249.4 N

The applied force is 249.4 N.

Given : L_T = L_S, A_T = A_S, Δ L_T + Δ L_S = 3.25 mm, Y_T and Y_S are not given, F same

Y = \(\frac{FL}{AΔL}\),   ∴ ΔL = \(\frac{FL}{AY}\)

ΔL_T = \(\frac{FL_T}{AY_T}\) and ΔL_S = \(\frac{FL_S}{AY_S}\)

\(\frac{ΔL_T}{ΔL_S}\) = \(\frac{Y_S}{Y_T}\)  (as L_T = L_S and A_T = A_S) …(1)

ΔL_T + ΔL_S = 3.25mm                    …(2)

We have two equations and two unknown quantities (ΔL_T and ΔL_S) if the ratio \(\frac{Y_S}{Y_T}\) is known. Since the ratio is not given, ΔL_T and ΔL_S cannot be determined.

Given : A = 1.2 mm² = 1.2 x 10^-6 m², F = 120 N, lateral strain = 0.1455 mm = 1.455 x 10^-4 m, Y_S = 2 x 10¹¹ N/m²

Y = \(\frac{Fy}{AΔx}\)

∴ \(\frac{Δx}{y}\) = \(\frac{F}{AY}=\frac{120}{1.2×10^{-6}×2×10^{11}}\)

= \(\frac{100}{2×10^5}\) = 50 x 10^-5 = 5 x 10^-4 m

This is the linear strain produced in the wire.

Poisson’s ratio, σ = \(\frac{lateral\,strain}{linear\,strain}\) = \(\frac{1.455×10^{-4}}{5×10^{-4}}\) = 0.291

Given : L = 125 m, r = 1 mm = 1 x 10^-3 m, L+ Δ L = 125.25 m, F = 800 N

Δ L= 0.25 m

Y = \(\frac{FL}{AΔL}\) = \(\frac{FL}{πr^2ΔL}\) = 

= \(\frac{800×125}{3.14×(1×10^{-3})^2×0.25}\) = \(\frac{40}{3.14}\)x 10¹⁰ = 12.73 x 10¹⁰ = 1.273 x 10¹¹ N/m²

Value of Young's modulus for the material of the wire = 1.273 x 10¹¹ N/m²

Given : L₁ = 30 cm, L₂ = 32 cm

The strain produced in the rubber band = \(\frac{L_2-L_1}{L}\) = \(\frac{32-30}{30}\) 

= 0.06667 = 6.667 x 10^-2

Given : L = 50 m, A = 0.01 cm² = 1 x 10^-6m², M = 100 kg, g = 9.8 m/s²

The stress in the wire = \(\frac{F}{A}\) = \(\frac{Mg}{A}\) = \(\frac{100×9.8}{1×10^{-6}}\) = 9.8 x 10⁸ N/m²

Given = L = 50m, ΔL = 2.5 cm = 2.5 x 10^-2 m

Strain = \(\frac{ΔL}{L}=\frac{2.5×10^{-2}}{50}\) = 5 x 10^-4

The strain produced in wire is 5 x 10^-4