# 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids

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## Karnataka 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids

Introduction

Question 1.
What is elasticity?
The property of a body by virtue of which it tends to regain its original shape and size when the applied forces are removed is known as elasticity.

Question 2.
What are elastic materials?
The materials which regain their shape and size when external forces are removed are known as elastic materials.

Question 3.
Mention few examples for nearly elastic materials.
Quartz, phosphor – bronze, etc.

Question 4.
What are deforming forces?
Answer: The forces that change the shape or size of the body are called deforming forces.

Question 5.
What is plasticity?
The property of a body by virtue of which it does not have a tendency to regain its original shape and size when the applied forces are removed is known as plasticity.

Question 6.
What are plastic materials?
The materials which do not have a tendency to regain its original shape and get permanently deformed even when external forces are removed are called plastic materials.

Question 7.
Given an example for plastic material.
Lump of putty or mud.

Question 8.
Which is more elastic: rubber or steel?
Steel is more elastic than rubber.

Elastic Behavior of Solids

Question 1.
Explain the elastic behaviour of solid on the basic of intermolecular forces?
In a solid, each atom or molecule is surrounded by neighbouring atoms or molecules. They are held together by interatomic or intermolecular forces and are in stable equilibrium position. When a solid is deformed, the atoms and molecules are displaced from their equilibrium position causing a change in interatomic or intermolecular distances. When the deforming force is removed, the interatomic forces tend to bring the atoms and molecules to their original position. Thus the body regains its original shape and size.

Stress and Strain

Question 1.
What are restoring forces?
When a body is subjected to external forces, due to property of elasticity, internal forces are developed within the body and tend to regain its original shape and size. These internal forces are called restoring forces.

Question 2.
What is the cause for the elasticity?
Elasticity of a body is due to intermolecular forces between the molecules of the body.

Question 3.
Define stress.
The restoring force per unit area of the body is called stress.

Question 4.
Mention the expression for stress.
Stress = $$\frac{\text { force }}{\text { area }}=\frac{F}{A}$$

Question 5.
Mention the SI unit of stress.
The SI unit of stress is Nm-2 or pascal (Pa).

Question 6.
Give the dimensions of stress.
Dimensions of stress is [ML-1T-2 ]

Question 7.
What is longitudinal stress?
The stress which tends to change the length of the body is called longitudinal stress.
Longitudinal stress = $$\frac{\text { normal force }}{\text { area }}$$

Question 8.
Define tensile stress.
When a cylinder is stretched by two equal forces applied normal to its cross-sectional area, then restoring force per unit area is called tensile stress.

Question 9.
Define compressive stress.
If a cylinder is compressed under the action of applied forces, the restoring force per unit area is called compressive stress.

Question 10.
Define strain.
Strain is defined as ratio of change in dimension to the original dimension.

Question 11.
Define longitudinal strain.
Longitudinal strain is defined as ratio of change in length to original length of the body (cylinder).
Longitudinal strain =$$\frac{\text { Change in length }}{\text { Original length }}$$ = (∆L/L)

Question 12.
What are the dimensions of strain?
Strain has no dimensions.

Question 13.
Strain is a unitless, dimensional quantity. Give reason.
Strain is the ratio of same physical quantity.
Hence it is unitless and dimensionless quantity.

Question 14.
Explain tangential stress or shearing stress.
If two equal and opposite deforming forces are applied parallel to the cross¬sectional area of the cylinder there is relative displacement between the opposite faces of the cylinder. The restoring force per unit area developed due to the applied tangential force is known as tangential or shearing stress.

Question 15.
Define tangential stress or tangential stress.
The restoring force per unit area developed due to the applied tangential force is known as tangential stress or shearing stress.

Question 16.
Define shearing strain.
Shearing strain is defined as the ratio of relative displacement of the faces Ax of the cylinder to the length of
the cylinder.
Shearing strain = $$\frac{\Delta x}{L}$$
= tan θ
for small angle tan θ = θ
∴ θ = ( $$\frac{\Delta x}{L}$$ )

Question 17.
Explain. What is hydraulic stress?
Consider a solid sphere placed in a fluid under high pressure. The sphere is compressed by the fluid uniformly on all sides. The force applied by the fluid acts in perpendicular to every point on the surface of sphere. Now the sphere is said to be under hydraulic compression. Hence the volume of the sphere decreases without any change in its shape.
The sphere develops internal forces and is equal and opposite to the force applied by the fluid. The internal restoring force per unit area is called hydraulic stress.

Question 18.
Define hydraulic stress.
The stress which tends to change the volume of a body without any change in its geometrical shape is called hydraulic stress.

Question 19.
What is the relation between the hydraulic stress and hydraulic pressure?
The hydraulic stress is equal in magnitude to the hydraulic pressure.

Question 20.
What is volume strain?
The strain produced by hydraulic pressure is called volume strain.

Question 21.
Define volume strain.
Volume strain is defined as ratio of change in volume (AV) to the original volume (V).
Volume strain = $$\frac{\text { Change in volume }}{\text { Original volume }}=\frac{\Delta V}{V}$$

Hooke’s Law

Question 1.
State Hooke’s law.
For small deformation stress is directly proportional to strain.

Question 2.
State and explain Hookes’ law.
For small deformation stress is directly proportional to strain.
Stress α strain
Stress = k × strain
where k is a constant called modulus of elasticity.

Stress Strain Curve

Question 1.
Draw the typical stress – strain graph for a metal. Explain the significance of the different regions in the graph.
The typical stress strain graph for a metal is as shown.
OA → elastic range A proportional limit
B → yield point
CD → plastic range
D → breaking point
E → fracture point
OF → permanent set

Consider a wire of uniform cross section stretched by applied force. The applied force is gradually increased in small steps and change in the length is noted. The graph of stress versus strain is as shown:
The study of stress strain graph gives the following details.

1. In the region between O and A, the curve is linear. In this region Hooke’s law is obeyed. The body regains its original dimension when applied force is removed. The body behaves like elastic body.
2. In the region A to B stress and strain are not proportional. But still the body regains its original dimension when the load is removed. The point B is called yield point or elastic limit, the corresponding stress is called yield strength.
3. If the load is increased further, the stress developed exceeds the yield strength. The strain increases rapidly even for a small change in the stress. This is represented by BD in the graph.
4. For a point C between B and D, the body does not regain its original dimensions even if the load is removed. In this case, even when the stress is zero, the strain is not zero. The material is said to have permanent set. The deformation is said to be plastic deformation.
5. The point D is the ultimate tensile strength of the material. Beyond this point, additional strain is produced even by a reduced applied force and the fracture occurs at E.
6. If the ultimate strength and fracture points D and E are close, the material is said to be brittle. If they are far apart, the material is said to ductile.

Question 2.
Define elastic limit. (Yield strength)
The maximum stress below which Hooke’s law is applicable is called elastic limit or yield strength.

Question 3.
Define ultimate tensile strength.
The minimum strength needed to cause the fracture of the material .is known as ultimate tensile strength.

Question 4.
When does a body wire is said to be permanent set?
When a wire is stretched too much, that the strain is permanent even when the stress is zero. Then the wire is said be permanent set.

Question 5.
What is fracture point?
The stretched wire breaks for a certain applied stress is called fracture point.

Question 6.
What are elastomers?
Substances which can be stretched to cause large strains are called elastomers.
They can be pulled to several times its original length and still returns to its original shape.

Question 7.
Do the elastomers obey Hooke’s law.
No, elastomers do not obey Hooke’s law.

Question 8.
Mention few examples for elastomers.
Tissue of aorta, rubber etc.

Question 9.
Draw the stress – strain graph for aorta (elastomers)

Elastic Moduli

Question 1.
Define modulus of elasticity.
The ratio stress to strain is called modulus of elasticity.
stress
Modulus of elasticity =$$\frac{\text { stress }}{\text { strain }}$$
strain

Question 2.
Is the modulus of elasticity depends on the dimensions of the body?
No, modulus of elasticity does not depend on dimensions of the body.

Question 3.
Is the modulus of elasticity has the same dimensions as that of stress?
Yes, the modulus of elasticity and stress has same dimensions.

Question 4.
Mention the expression for the Young’s modulus of elasticity.

Young’s Modulus

Question 1.
Is the magnitude of the strain produced is same for both tensile and compressive stress?
Yes magnitude of strain produced is same for both tensile and compressive stress.

Question 2.
Define Young’s modulus of elasticity.
Young’s modulus is defined as ratio of tensile or compressive stress to the longitudinal strain.
Young s modulus =$$\frac{\text { tensile stress }}{\text { longitudinal strain }}=\frac{\sigma}{\varepsilon}$$

Question 3.
Dimensions of Young’s modulus is same as stress. Is it true?
Yes, the dimensions of Young’s modulus is same as stress.

Question 4.
Mention the dimensions of Young’s modulus of elasticity.
Nm-2 or pascal (Pa)

Question 5.
Explain why steel is more elastic than metals like copper, brass or aluminium?
A force of 2000 N is required to increase the length of a thin steel wire of cross sectional area 0.1 cm2 by 0.1%. The force required to produce the same strain in aluminium, brass and copper are 690 N, 900 N and 1100 N respectively. Hence steel is more elastic.

Question 6.
Explain why steel is preferred over other metals in heavy duty machines and in structural designs.
The Young’s modulus of steel is very large than any other metals and is highly elastic. A large force is required even for a small strain. Hence steel is preferred in heavy duty machines and in structural designs.

Question 7.
Springs are made of steel and not copper. Why?
Steel is more elastic than copper. Therefore, the extension produced is less in steel than copper for same applied force. Steel spring can sustain larger applied force. Hence steel is preferred than copper for making springs.

Question 8.
Explain the experiment to determine the Young’s modulus of the material of the wire.
The experimental arrangement consists of two long straight wires of same length and equal radius suspended side by side from a fixed rigid support. A wire A, called reference wire carries a millimeter main scale M and a pan to place weight. The wire B, called sectional area also carries a pan in which known weights can be placed. A vernier scale V is attached to a pointer at the bottom of the experimental wire B, and the main scale M is fixed to the reference wire A.
Both the reference and experimental wires are given an initial load to keep the wire straight and the vernier reading is noted.

Now the experimental wire is gradually loaded with more weights and the new readings of the vernier is noted.
The difference between two vernier readings gives the elongation produced in the wire.
Let r and L be the initial radius and length of the experimental wire, then area of cross section of the wire is πr2
Let M is the mass that produces an elongation ∆L.

Share Modulus

Question 1.
Define shear modulus or modulus of rigidity.
The ratio of shearing stress to the corresponding shearing strain is called shear modulus or modulus of rigidity.

Question 2.
Mention the expression for shear modulus

Question 3.
Mention the unit of shear modulus.
The unit shear modulus is Nm-2 or Pa.

Question 4.
What is the relation between shear modulus (G) and Young’s modulus (Y)?
Generally shear modulus (G) is less than Young’s modulus (Y).
For most materials, G = $$\frac{Y}{3}$$

Bulk Modulus

Question 1.
Define bulk modulus.
The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus.

Question 2.
Define volume strain.
Volume strain is defined as ratio of change in volume to the original volume.

Question 3.
Mention the expression for bulk modulus.

Question 4.
Mention the unit of bulk modulus.
Unit of bulk modulus is Nm-2 or Pa.

Question 5.
Is the unit of bulk modulus is same as unit of pressure?
Yes unit of bulk modulus is same as unit of pressure and is equal to Nm-2 or Pa.

Question 6.
What is the significance of negative sign in the expression for bulk modulus?
The negative sign indicates that as the pressure increases, the volume decreases.

Question 7.
What is compressibility?
The reciprocal of the bulk modulus is called compressibility.

Question 8.
Define compressibility.
Compressibility is defined as the fractional change in volume per unit increase in pressure.

Question 9.
Mention the units of compressibility.
Unit of compressibility is N-1m2

Question 10.
Gases are about million times more compressible than solids. Explain.
The bulk modulus of solids is much greater than gases. The reciprocal of bulk modulus is compressibility. Therefore compressibility of gases is much greater than solids. Gases are more compressible than solids.

Applications of Elastic Behaviour of Materials

Question 1.
What is elastic after effect?
The delay in regaining the original state by a body after the removal of the deforming force is called elastic, after effect.
Example: A glass fibre will take hours to return to its original state.

Question 2.
What are the factors affecting elasticity?

• Effect of hammering and rolling
• Effect of annealing
• Effect of the presence of impurities
• Effect of temperature

Question 3.
The beams used in construction of bridges have a cross section of the type I. Why?
This section provides a large load bearing surface and enough depth to prevent bending.
This shape reduces the weight of the beam without sacrificing the strength and hence reduces the cost.

Question 4.
What is meant by buckling?
The bending of beam under a load is called buckling.

Question 5.
A crane which is used to lift heavy loads is provided with rope made of number of wires braided together rather than single thick wire.
A single thick wire is a rigid rod. Instead number of wires braided together is flexible and has more strength.

Question 6.
Show that the pillars or columns of the bridges and buildings have distributed shape at the ends. Why?
The pillars with a distributed shape at the ends support more load than pillars with rounded ends.

Question 7.
Why the maximum height of the mountain on earth is 10 km?
Let h be the height of the mountain.
The force per unit area due to the weight of the mountain = ρgh.
The elastic limit for the rock = 30 x 107 Nm-2 density of the rock = 3 x 103 kgm3
.’. 30 x 107 =ρgh
h = $$\frac{30 \times 10^{7}}{3 \times 10^{3} \times 10}$$
Maximum height h = 10 x 103 = 10 km

Poisson’s Ratio

Question 1.
What happens to diameter (area of cross section) of the wire when it is stretched?
Diameter (area of cross section) of the wire decreases.

Question 2.
What is lateral strain?
The strain perpendicular to the applied force is called lateral strain.

Question 3.
Define Poisson’s ratio.
The ratio of the lateral strain to the longitudinal strain is called Poisson’s ratio.

Question 4.
Mention the expression for lateral strain.
Lateral strain = $$\frac{\Delta d}{d}$$
Where d → is the original diameter of the wire, ∆d is the contraction of the diameter under the stress.

Question 5.
Mention the expression for Poisson’s ratio.
Poisson ratio = $$\frac{\text { Lateral strain }}{\text { Longitudical strain }}$$

Question 6.
What is the unit and dimensions of Poisson’s ratio?
Poisson’s ratio is a unitless and dimensionless quality. It is a pure number.

Elastic Potential Energy in a Stretched Wire

Question 1.
What is elastic potential energy of a wire?
When a wire under a tensile stress, work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy.

Question 2.
Obtain an expression for elastic potential energy in a stretched wire.
Consider a wire of length L and area of a cross section and is subjected to a deforming force F along its length. Let / be the increase in the length (elongation) of the wire.

If dl is the small elongation, work done, dW = Fdl
Similarly, total work done when the wire elongates by l is

W =$$\frac{1}{2}$$ x Young’s modulus x (strain)2 x volume of the wire
But Y = $$\frac{\text { stress }}{\text { strain }}$$
∴ W = $$\frac{1}{2}$$ x stress x strain x volume of the wire
This work done is stored in the wire in the form elastic potential energy
∴ W=U
∴ U = $$\frac{1}{2}$$ x stress x strain x volume of the wire
$$\frac{U}{\text { Volume of the wire }}$$ U, potential energy per unit volume
U = $$\frac{1}{2}$$ stress x strain
U = $$\frac{1}{2}$$ σε

Numerical Problems

Question 1.
A steel wire of length 4.7 m and cross sectional area 3 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross sectional area of 4 × 10-5 m2 under the given load. What is the ratio of the Young’s modulus of steel to that of copper?
For steel
F1 = F, L1 = 4.7 m, A1 = 3 × 10-5m2, ∆L1 = ∆l
For copper F2 = F.L2= 3.5 m, A2 = 4 × 10-5m2, ∆L2 = ∆l

Question 2.
The given figure shows the strain – stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?

Young’s modulus = Stress/Strain
Y = $$\frac{300 \times 10^{6}}{0.004}$$
= 75000 × 106 =7-5 × 1010Nm-2
(b) Approximate yield strength = maximum stress within elastic limit
∴ yield strength = 300 × 106
= 3 × 108 Nm-2

Question 3.
The stress-strain graphs for materials A and B are as shown.

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?
(a) Since the slope A is greater than B, Young’s modulus of A is greater.
(b) Strength of a material is determined by the amount of stress required to cause fracture. Hence A is stronger than the material B.

Question 4.
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.
(a) False
For a given stress, strain produced is more in rubber than steel. Further Young’s modulus ∝ 1/strain
Hence Young’s modulus of rubber is less than that of steel.

(b) True

Since stretching of a coil changes the shape of the body, the stretching of a coil is determined by its shear modulus.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are load as shown in the diagram. The unloaded length of steel wire 1.5 m and that of brass wire is 1 m. Compute the elongations of the steel and the brass wires. (Young’s modulus of steel = 200 × 109 Nm-2 ; Young’s modulus of brass = 94 × 109 Nm-2)
For steel wire:

Question 6.
The edge of an aluminum 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?
L = 10cm = 0-lm
A = L2 =0.12 =001 m2
F = 100 × 9.8 = 980 N
G = 25 GPa = 25 × 109 Pa

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and the 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.
F = 50,000 × 9.8 N
Area of cross section of each column = π{r12 – r22) = π{(60 × 10-2)2 – (30 × 10-2)2]
= π{3600-900} × 10-4=8483.4 × 10-4m2
Area of cross section of 4 columns
A = 4 × 8483 × 10-4m2
A = 3 393 m2

Question 8.
A piece of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain. (Shear modulus of Cu = 42 GPa)
A = 15.2 mm× 19.1 mm = 15.2 × 19.2 × 10-6 m2
F = 44,500 N
(G = 42 GPa)

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 Nm-2 , what is the maximum load the cable can support?
r = 1.5 × 10-2 m
A = πr2 = π × (1.5 × 10-2 )2 = 7.0695 × 10-4 m2
Maximum stress P = 108 Nm-2

∴ Maximum load the cable can support
F = maximum stress × A
= 108 × 7.0695 × 10-4
= 7-0695 × 104N =7.07 × 104N

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 copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
The three wires are under same tension, the extension due to the rigid bar is also same. Further the strain in each wire is also same.

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m is virled 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.
m = 14.5 kg, L = 1 m
ω = 2 rev/s = 2 × 2πrad s-1 = 4n rad s-1
A = 0.065 cm2 = 0.065 × 10-4 m2 = 6.5 × 10-6 m3
∆L = ?
When the wire is at the bottom of the circle
F = mg + mrω2
= m(g + rω2) {∵ mrω2 reaction due to centripetal force}
=m(g + rω2) = 14.5[9.8 +1 × (4π)2] = 2432 N

Question 12.
Compute the bulk modulus of water from the following data: Initial volume 100.0 litre. Pressure increase = 100 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.
y,=100 litre = 100 × 103m3, V2 =100-5 litre = 100.5 × 10-3m3
∆V =V2 -V1 = (100.5 – 100) × 10-3 =0.5 × 10-3m3 ∆P = 100 × 1.013 × 105Pa = 101.3 × 105 Pa
For water

The above ratio implies that air is more compressible than water.

Question 13.
What is the density of water at a depth where the pressure is 80.0 atm. Given that its density at the surface is 1.03 × 103 kg m-3. (Compressibility of water = 45.8 × 10-11 N-1 m2)
Increase in the density dp is given by
dp = K × P × ρ
where K is the compressibility
P is the pressure
ρ is the density at the surface
dp = (45.8 × 10-11) × (80 × 1.013 × 105) × (1.03 x 103) =3.823 kg m-3
Density at a depth where the pressure at 80 atm is given by
= p + dρ =1.03 × 103 + 3.823 = 1 034 × 103kg m-3

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm (bulk modulus of glass 37 GPa)