TS Inter 1st Year Physics Study Material Chapter 10 Mechanical Properties of Solids

Telangana TSBIE TS Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 10th Lesson Mechanical Properties of Solids

Very Short Answer Type Questions

Question 1.
State Hooke’s law of elasticity.
Answer:
Hooke’s law states that within elastic limit stress is proportional to strain.

This constant is known as elastic modulus of the body.

Question 2.
State the units and dimensions of stress.
Answer:

Dimensional formula ML^-1 T^-2

Question 3.
State the units and dimensions of modulus of elasticity.
Answer:

Nm^-2 (or) pascal
Dimensional formula ML^-1 T^-2

Question 4.
State the units and dimensions of Young’s modulus.
Answer:

Dimensional formula ML^-1T²

Question 5.
State the units and dimensions of modulus of rigidity.
Answer:
Modulus of rigidity,

(or) pascal
Dimensional formula ML^-1T².

Question 6.
State the units and dimensions of Bulk modulus.
Answer:

unit is Nm^-2 (or) pascal
Dimensional formula ML^-1 T^-2

Question 7.
State the examples of nearly perfect elastic and plastic bodies.
Answer:
There is no perfectly elastic body. But behaviour of Quartz fiber is very nearer to perfectly elastic body.

Real bodies are not perfectly plastic, but behaviour of wet clay, butter etc., can be taken as examples for perfectly plastic bodies.

Short Answer Questions

Question 1.
Define Hooke’s law of elasticity, proportionality, permanent set, and breaking stress.
Answer:
Hooke’s Law :
It states that within elastic limit, stress is proportional to strain.

This constant is called elastic constant (E).

Proportionality limit:
When load is increased the elongation of the wire will also increases. The maximum load upto which the elongation is directly proportional to the load is called proportionality limit (A). The graph drawn between load and extension. It is a straight line OA’.

Permanent set :
If the load on the wire is increased beyond elastic limit, the elongation is not pro portional to load. On removal of the load the wire cannot regain its original length. The length of the wire increases permanently. In figure permanent set is given by OP. This is called permanent set.

Breaking stress :
If the load is increased beyond yield point the elongation is very rapid, even for small changes in load and wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Question 2.
Define modulus of elasticity, stress, strain and Poisson’s ratio.
Answer:
1) Stress:
Restoring force acting on unit area is called stress.

Unit: N/m² (or) pascal;
Dimensional formula: ML^-1T^-2.

2) Strain :
The change in dimension per unit original dimension of a body is called strain.

It is a ratio, so no units and dimensional formula.

3) Modulus of elasticity :
∴ From Hooke’s Law Stress x Strain or

The ratio of stress to strain is called modulus of elasticity.
Unit: N/m².
Dimensional formula: ML^-1T^-2.

4) Poisson’s ratio :
It is defined as the ratio of lateral contraction strain to longitudinal elongation strain.
Poisson’s ratio

It is a ratio, so no units and dimensional formula.

Question 3.
Define Young’s modulus, Bulk modulus, and Shear modulus.
Answer:
1) Young’s Modulus Y:
Within elastic limit, the ratio of longitudinal stress to longitudinal strain is called Young’s modulus

2. Bulk Modulus (B) :
Within elastic limit, the ratio of volumetric stress to volumetric strain is called Bulk modulus

3) Shear Modulus or Rigidity Modulus (G):
Within elastic limit, the ratio of tangential or shearing stress to shearing strain is called Shear modulus or Rigidity modulus.

Question 4.
Define stress and explain the types of stress. [AP War. 19; TS War. 16]
Answer:
Stress:
When a body is subjected to a deforming force, then restoring forces will develop inside the body. These restoring forces will oppose any sort of change in its original shape. The restoring force per unit area of the surface is called stress.

Stress:
Stress is defined as force applied per unit area.

D.F. = ML^-1T^-2 ; Unit: N/m² (or) Pascal.

Types of stress :
It is of three types. They are : 1) Longitudinal stress 2) Tangential stress (or) Shear stress 3) Volumetric stress.

Longitudinal stress:
If the force applied on a body is along its lengthwise direction then it is called longitudinal stress. It produces deformation in length.

Tangential stress (or) Shear stress:
Force applied per unit area parallel to the surface of a body trying to displace the upper layers of the body is called shearing stress.

(Parallel to the surface layers)

Volumetric stress:
If force is applied on all the sides of a body or on the volume of a body then it is called volumetric stress.

Question 5.
Define strain and explain the types of strain.
Answer:
Strain:
Strain is defined as deformation produced per unit dimension. It is a ratio. So no units.

Types of strain :
Strain is of three types. They are: 1) Longitudinal strain 2) Tangential strain or Shear strain 3) Volumetric strain.

Longitudinal strain:
The ratio of elongation to original length along length wise direction is defined as longitudinal strain.

Shearing strain :
If the force applied on a body produces a change in shape only it is called shearing force. The angle through which a plane originally perpendicular to the fixed surface shifts due to the application of shearing stress is called shearing strain or simply shear (θ).

Question 6.
Define strain energy and derive the equation for the same. [TS Mar. ’19; May ’18; AP Mar. ’14; May ’14]
Answer:
Strain energy :
The energy developed in (string) a body when it is strained is called strain energy.

Let a force F be applied on lower end of wire, fixed at the upper end. Let the extension be dl.
∴ Work done = dW = Fdl
Total work done in stretching it from 0

Question 7.
Explain why steel is preferred to copper, brass, aluminium in heavy-duty machines and in structural designs.
Answer:
For metals Young’s moduli are large. Therefore, these materials require a large force to produce small change in length. To increase the length of a thin steel wire of 0.1 cm² cross-sectional area by 0.1 %, a force of 2000 N is required. The force required to produce the same strain in aluminium, brass, and copper wire having the same cross-sectional area are 690 N, 900N, and 1100 N respectively. It means that steel is more elastic than copper, brass, and aluminium. It is for this reason that steel is preferred in heavy duty machines and in structural designs.

Question 8.
Describe the behaviour of a wire under gradually increasing load. [AP Mar. ’18, ’17, ’16, ’15, ’13, May ’16, ’13; TS Mar. 18, 17, 15; May 17,16; June 15]
Answer:
Behaviour of a wire under increasing load:
Let a wire is suspended at one end and loads are attached to the other end. When loads are gradually increased the following changes are noticed.

1) Proportionality limit (A) :
When load is increased the elongation of the wire gradually increases. The maximum load upto which the elongation is directly pro-portional to the load is called proportionality limit (A). The graph drawn between load and extension is a straight line. So point A is called proportionality limit. In this region Hooke’s Law is obeyed.

2) Elastic limit (B):
If the load is increased above the proportionality limit the elongation is not proportional to the load. Hooke’s law is not obeyed. But it exhibits elasticity which means that it regains the original length if load is removed. The maximum load on the wire upto which it exhibits elasticity is called elastic limit (B in the graph).

3) Permanent set (C) :
If the load on the wire is increased beyond elastic limit say upto C, the elongation is not proportional to load. On removal of the load, the wire does not regain its original length. Length of wire increases perma-nently. In figure permanent set is given by OP. So OP is called permanent set.

4) Point of ultimate tensile strength (D) :
If the load is further increased, upto D’ then strain increases rapidly even though there is no increase in stress.

At this stage the restoring forces seems to be subdued to then deforming forces. Elongation without increase in load is called creeping. This behaviour of metal is called yielding.

5) Fracture point (E) :
If the load is increased beyond Yield point the elongation is very rapid, even for small changes in load the wire becomes thinner and breaks. This is shown as E. The breaking force per unit area is called breaking stress.

Question 9.
Two identical solid balls, one of ivory and the other of wet clay are dropped from the same height on to the floor. Which one will rise to a greater height after striking the floor and why?
Answer:
We know that ivory ball is more elastic than wet-clay ball. Therefore, the ivory ball will tend to regain its original shape in a very short time after the collision. Due to it, there will be large energy and momentum transfer to the ivory ball in comparison to the wet-clay ball. As a result of it, the ivory ball will raise higher after the collision.

Question 10.
While constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends. Why?
Answer:
Use of pillars or columns is very common in buildings and bridges. A pillar with roun-ded-ends supports less load than that with a distributed shape at the ends. Hence, for this reason, while constructing buildings and bridges a pillar with distributed ends is preferred to a pillar with rounded ends.

Question 11.
Explain why the maximum height of a mountain on earth is approximately 10 km?
Answer:
A mountain base is not under uniform compression and this provides some shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow.

At the bottom of a mountain of height ‘h’, the force per unit area due to the weight of the mountain is hpg where p is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction and the sides of the mountain are free. There is a shear component, approximately hpg itself. Now the elastic limit for a typical rock is 3 × 10⁷ Nm^-2. Equating this to hpg with ρ = 3 × 10³ kg m^-3 gives
h = \(\frac{30\times10^{-7}}{3\times10^3\times10}\) = 10 km
Hence, the maximum height of a mountain on earth is approximately 10 km.

Question 12.
Explain the concept of Elastic Potential Energy in a stretched wire and hence obtain the expression for it. [AP May ’ 18, 17; AP June 15]
Answer:
When a wire is put under a tensile stress, work is done against the inter atomic forces. This work is stored in the form of “Elastic potential energy.”

Expression to elastic potential energy:
To stretch a wire, force is applied. As a result it elongates. So the force applied is useful to do some work. This work is stored in it as potential energy. When the deforming force is removed, this energy is liberated as heat. The energy developed in a body (string) when it is strained is called strain energy.

Let a force F be applied on a wire fixed at the upper end. Let the extension be dl.
∴ Work done = dW = Fdl
Total work done in stretching from

Long Answer Questions

Question 1.
Define Hooke’s law of elasticity and describe an experiment to determine the Young’s modulus of the material of a wire.
Answer:
Hooke’s Law :
Within elastic limit, stress is directly proportional to strain.
strain ∝ stress

where E constant called modulus of elasticity of the material of a body.

Determination of Young’s modulus of a wire:
The apparatus used to find Young s modulus of a wire consists of two long wires A and B of same length made with same material are used. These two wires are suspended from a rigid support and a vernier scale V’ is attached to them. Wire A is connected to the main scale (M). A fixed load is connected to this cord to keep tension in the wire. This is called reference wire. The second Wire B’ is connected to vernier scale V’. Adjustable load hanger is connected to this wire. This is called experimental wire.

Procedure :
Let a load M₁ is attached to the weight hanger at vernier. Main scale reading (M.S.R) and vernier scale reading (V.S.R) are noted. Weights are gradually increased in the steps of \(\frac{1}{2}\) kg upto a maximum load of say 3 kg. Every time M.S.R and V.S.R are noted. They are placed in tabular form.

Now loads are gradually decreased in steps of \(\frac{1}{2}\) kg. While decreasing M.S.R and V.S.R are noted for every load, values are posted in tabular form.

Let 1^st reading with mass M₁ is e₁ and 2^nd reading with mass M₂ is e₂.

Change in load M = M₂ – M₁
elongation e = e₂ – e₁
‘M’ and ‘e’ values are calculated and a graph is plotted.

Force on the wire = mg
Area of cross section of the wire = πr²
(r = radius of the wire)
Elongation = e
Original length = l

A graph is between load (m) and elongation ‘e’, is straight line passing through the origin. The slope of the graph (tan θ) gives (\(\frac{m}{e}\)). The value is substituted in the above equation to find Young’s modulus of the material of the wire.

Precautions:

1. The load applied should be much smaller than elastic limit.
2. Reading is noted only after the air bubble is brought to centre of spirit level.

Problems

Question 1.
A copper wire of 1mm diameter is stretched by applying a force of 10 N. Find the stress in the wire.
Solution:
Diameter, d = 1mm
∴ radius, r = 0.5 mm = 0.5 × 10^-3m
Force, F = 10N

Question 2.
A tungsten wire of length 20cm is stretched by 0.1cm. Find the strain on the wire.
Solution:
Length of wire, l = 20cm = 0.2m
elongation, e = 0.1cm = 1 × 10^-3m

Question 3.
If an iron wire is stretched by 1%, what is the strain on the wire?
Solution:

Question 4.
A brass wire of diameter 1mm and of length 2m is stretched by applying a force of 20N. If the increase in length is 0.51mm, find i) the stress, ii) the strain and iii) the Young’s modulus of the wire.
Solution:
Length of wire, l = 2m;
L diameter, d = 1mm = 10^-3 m
Force, F = 20N;
Increase in length, e = 0.51mm

Question 5.
A copper wire and an aluminium wire have lengths in the ratio 3: 2, diameters in the ratio 2 : 3 and forces applied in the ratio 4: 5. Find the ratio of increase in length of the two wires. (Y_Cu = 1.1 × 10¹¹ Nm^-2, Y_Al = 0.7 × 10¹¹Nm^-2)
Solution:
Ratio of lengths, l₁ : l₂ = 3 : 2;
Ratio of diameters, d₁ : d₂ = 2 : 3
Ratio of forces, F₁ : F₂ = 4 : 5
Y₁ = Y of copper = 1.1 × 10¹¹
Y₂ = Y of Aluminium = 0.7 × 10¹¹;
Ratio of elongation, e₁ : e₂ = ?

Question 6.
A brass wire of cross-sectional area 2mm² is suspended from a rigid support and a body of volume 100cm³ is attached to its other end. If the decrease in the length of the wire is 0.11mm, when the body is completely immersed in water, find the natural length of the wire.
(Y_brass = 0.91 × 10¹¹ Nm^-2, ρ_water = 10³kgm^-3)
Solution:
Area of cross section, A = 2mm = 2 × 10^-6 m²
Volume of body, V = 100 cc = 100 × 10^-6 m³
Decrease in length, e’ = 0.11mm = 0.11 × 10^-3m
Youngs modulus of brass, Y = 0.91 × 10¹¹ N/m²
Density of water, ρ = 1000 kg / m³
Use e’ = \(\frac{V\rho gl}{AY}\)

Question 7.
There are two wires of same material. Their radii and lengths are both in the ratio 1 : 2. If the extensions produced are equal, what is the ratio of the loads?
Solution:
Ratio of lengths, l₁ : l₂ = 1 : 2
Ratio of radii, r₁ : r₂ = 1 : 2
Extensions produced are equal ⇒ e₁ = e₂;
Made of same material ⇒ Y₁ = Y₂
Ratio of loads m₁ : m₂ = ?

Question 8.
Two wires of different material have same lengths and areas of cross-section. What is the ratio of their increase in length when forces applied are the same? (Y₁ = 0.90 × 10¹¹ Nm^-2, Y₂ = 3.60 × 10¹¹ Nm^-2.)
Solution:
Lengths are same ⇒ l₁ = l₂ ;
Area of cross sections are same, A₁ = A₂
Y₁ = 0.9 × 10¹¹ N/m²
Y₂ = 3.60 × 10¹¹ N/m²

Question 9.
A metal wire of length 2.5m and area of cross-section 1.5 × 10^-6 m² is stretched through 2mm. If its Young’s modulus is 1.25 × 10¹¹ Nm^-2, find the tension in the wire.
Solution:
Length of wire, l = 2.5m
Y = 1.25 × 10¹¹N/m²
Area of cross section, A = 1.5 × 10^-6 m²
Elongation, e = 2 m.m = 2 × 10^-3m
Tension, T = mg = F = ?

Question 10.
An aluminium wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 1.35 mm, find the ratio of the (i) stress in the two wires and 0Q strain in the two wires. (Y_Al = 0.7 × 10¹¹ Nm^-2, Y_steel = 2 × 10¹¹m^-2)
Solution:
i) Length is same ⇒ l₁ = l₂
Area is same ⇒ A₁ = A₂
In composite wire same load will act on both wires.
∴ Ratio of stress = 1 : 1

ii) Total elongation, e = 1.35mm =e_Al + e_s
Young’s modulus of aluminium = 7 × 10¹⁰ N/m²
Y of steel = 2 × 10¹¹ N/m²
Elongation, e = \(\frac{Fl}{AY}\)
But F, l and A are same

∴ Ratio of strains in the wires is 20 : 7.

Question 11.
A 2 cm cube of some substance has its upper face displaced by 0.15cm due to a tangential force of 0.3 N while keeping the lower face fixed. Calculate the rigidity modulus of the substance.
Solution:
Side of cube, a = 2.0 cm = 2 × 10^-2 m
Area, A = 4 × 10^-4 m²
Displacement of upper layer = 0.15cm
= 0.15 × 10^-2m
Tangential force, F = 0.30N

Question 12.
A spherical ball of volume 1000 cm³ is subjected to a pressure of 10 atmosphere. The change in volume is 10^-8 cm³. If the ball is made of iron, find its bulk modulus. (1 atmosphere = 1 × 10⁵ Nm^-2)
Solution:
Volume of ball, V = 1000 cm³ = 10^-3 m³
(∵ 1M³ = 10⁶ cm³)
Pressure, P = 10 atmospheres
= 10 × 10⁵ pa ( v 1 atm = 10⁵ pascal)
Change in volume, ∆V = 10^-8 cm³

Question 13.
A copper cube of side of length 1 cm is subjected to a pressure of 100 atmosphere. Find the change in its volume if the bulk modulus of copper is 1.4 × 10¹¹ Nm^-2 (1 atm = 1 × 10⁵ Nm^-2).
Solution:
Side of cube, a’ = 1cm = 10^-2m
∴ Volume of cube = 10^-6m
Pressure, P = 100 atm = 100 × 10⁵ = 10⁷ pa
Bulk modulus, K = 1.4 × 10¹¹ N/m²;

Question 14.
Determine the pressure required to reduce the given volume of water by 2%. Bulk modulus of water is 2.2 × 10⁹ Nm^-2
Solution:

Question 15.
A steel wire of length 20 cm is stretched to increase its length by 0.2 cm. Find the lateral strain in the wire if the Poisson’s’ ratio for steel is 0.19.
Solution:
Length of wire, l = 20, cm = 0.20m,
Poisson’s ratio, σ =0.19
Increase in length, ∆l = 0.2 cm = 2 × 10^-3m
lateral strain = ?
Lateral strain = σ × longitudinal strain ‘e’