NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

Answer:

Young’s modulus ,Y =Stress/Strain
=150 x 10⁶/0.002
= 150 x 10⁶/2 x 10^-3
= 75 x 10⁹ Nm^-2
=75 x 10¹⁰ Nm^-2

(a)Yield strength of a material is the maximum stress that the material can sustain and retain its elastic property. From graph, the approximate yield strength of the given material
= 300 x 10⁶ Nm^-2
= 3 x 10⁸ Nm^-2 .

Q3. The stress-strain graphs for materials A and B are shown in the figure below. 

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?

Answer:

Young’s modulus = Stress/Strain

(a) From the graphs we can see that for the given strain, stress for A is greater than that of B. Therefore, Young’s modulus of A is greater than B.
(b) Young’s modulus is also a measure of the strength of the material. Young’s modulus is greater for A, therefore material A is stronger than B.

Q4. 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.

Answer.

( a ) True. Stretching a coil does not change its length, only its shape is altered and this involves shear modulus.

( b ) False. This is because, for the same value of stress, there is more strain in rubber than in steel. And as Young Modulus is an inverse of strain, it is greater in steel.

Q5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 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. [Young’s modulus of steel is 2.0 x 10¹¹ Pa. (1 Pa = 1 N m²)]

Answer:

Diameter of the two wires, d =0.25m
Radius of the wires, r= d/2 =0.125cm
Unloaded length of the steel wire, l₁ =1.5m
Unloaded length of the brass wire, l₂ =1.0m

Force exerted on the steel wire:
F₁ =(4+6)g=10×9.8=98N

Cross-section area of the steel wire, a₁ =πr₁²

Change in length of the steel wire = Δl₁

Young’s modulus for steel= 2.0 x 10¹¹ Pa

= 1.49 x 10^-4 m

Force of the brass wire ,F₂ = 6 x 9.8 = 58. 8 N

Cross-section area of the brass wire, a₂ =πr₂²

Change in length of the brass wire = Δl₂

Young’s modulus of the brass wire = 0.91 x 10¹¹ Pa

= 1.3 x 10^-4 m

Elongation of the steel wire is 1.49 x 10^-4 m and that of brass is 1.3 x 10^-4 m.

Q6. 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?

Answer:

Edge of the aluminium cube, L = 10 cm =10/100= 0.1 m
Area of each face, A = (0.1)² = 0.01 m²

Mass attached to the opposite face of the cube = 100 kg

Tangential force acting on the face, F = 100 kg = 100 x 9.8 = 980 N

Shear modulus, η = Tangential stress/Shearing strain

Shearing strain = Tangential stress/ Shear modulus

= F/Aη = 980/(0.01 x 25 x 10⁹) = 3.92 x 10^-6

Since, Shearing strain = Lateral strain/Side of the cube

Lateral strain = Shearing strain x Side of the cube = 3.92 x 10^-6 x 0.1

= 3.92 x 10^-7 m ≈ 4 x 10^-7 m

Q7. Four identical hollow cylindrical columns of mild steel support a big structure of a 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.

Answer:

Mass of the big structure, M = 50,000 kg

Total force exerted on the four columns= total weight of the structure=50000×9.8N

The compressional force on each column = Mg/4 = (50000×9.8)/4 N= 122500 N
Therefore, Stress = 122500 N
Young’s modulus of steel, Y=2×10¹¹ Pa

Young’s modulus, Y= Stress/Strain

Strain = Young’s modulus/Stress

Strain = (F/A)/Y

Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Where,
Area, A=π(R²−r²)=π((0.6)²−(0.3)²) = 0.27 π m²

Strain =122500/[0.27 x 3.14×2×10¹¹]=7.22×10⁻⁷

Hence, the compressional strain of each column is 7.22×10⁻⁷.

Q8. 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?

Answer:

Area of the copper piece, A=19.1×10⁻³ ×15.2×10⁻³=2.9×10⁻⁴m²

Tension force applied on the piece of copper, F=44,500 N

Modulus of elasticity of copper, Y=42×10⁹ Nm⁻²

Modulus of elasticity (Y) = Stress / Strain

=(F/A) / Strain
Strain = F/(YA)
= 44500/(2.9×10⁻⁴×42×10⁹)
= 3.65×10⁻³

Q9. 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?

Answer:

Radius of the steel cable, r=1.5 cm = 0.015 m

Cross-sectional area of the cable = πr²= 3.14 x ( 0.015)²

= 7.06 x 10^-4 m

Maximum stress allowed on the steel cable = 10⁸ N/m²

Maximum load the cable can support = Maximum stress × Area of cross-section

= 10⁸ x 7.06 x 10^-4

= 7.065×10⁴ N
Hence, the cable can support the maximum load of 7.065×10⁴ N.

Q10. 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.

Answer.

As the tension on the wires is the same, the extension of each wire will also be the same. Now, as the length of the wires is the same, the strain on them will also be equal.

Now, we know :

Y = Stress / Strain
= (F/A) / Strain  =  (4F/πd²) / Strain     . . . . . . . . . . ( 1 )
Where,
A = Area of cross-section
F = Tension force
d = Diameter of the wire
We can conclude from equation ( 1 ) that Y ∝ (1/d²)
We know that Young’s modulus for iron, Y₁ = 190 × 10⁹ Pa
Let the diameter of the iron wire = d₁
Also, Young’s modulus for copper, Y₂ = 120 × 10⁹ Pa
let the diameter of the copper wire = d₂
Thus, the ratio of their diameters can be given as :

= $sqrt{frac{190×10^{9}}{120×10^{9}}}$ =1 : 25 : 1

Q11. 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.

Answer: 

Mass, m = 14.5 kg

Length of the steel wire, l = 1 m

Angular velocity, v = 2 rev/s

Cross-sectional area of the wire, A = 0.065 x 10^-4 m²

Total pulling force on the steel wire when the mass is at the lowest point of the vertical circle, F = mg + mr ω²

= 14.5×9.8+14.5×1×(12.56)²
=2429.53 N

Young’s modulus = Stress / Strain
$Y=frac{F}{A}frac{l}{Delta l}$

Δl=(2429.53×1)/(0.065×10⁻⁴) × (2×10¹¹) = 1.87×10⁻³m
Hence, the elongation of the wire when the mass is at the lowest is 1.87×10⁻³m.

Q12. 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.

Answer:

Initial volume, V₁ =100.0 litre =100.0×10⁻³ m³
Final volume, V₂ =100.5 litre =100.5×10⁻³ m³

Change in the volume, ΔV=V₂ −V₁ =0.5×10⁻³ m³

Pressure increase, p=100.0atm=100×1.013×10⁵Pa

= 101.3 x 10⁵Pa
Bulk modulus of water= p/(ΔV/V₁)=pV₁/ΔV
= 101.3×10⁵×100×10⁻³/(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

The intermolecular force in the liquids is much larger than air as the distance between the molecules is much lesser in liquid than in air. Therefore, at the same temperature, strain for water is much more than air.

Q13. 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?

Answer.

Let the depth be the alphabet ‘d’.

Given,
Pressure at the given depth, p = 60.0 atm = 60 × 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 d.
V₁ be the volume of water of mass m at the surface.
Then, let V₂ be the volume of water of mass m at the depth h
and ΔV is 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)        . .  . . .  . . ( 1 )
We know, Bulk modulus, B = pV₁ / ΔV
=> ΔV / V₁ = p / B

Compressibility of water = ( 1/B ) = 45.8 × 10^-11 Pa^-1
∴ ΔV / V₁ = 60 × 1.013 × 10⁵ × 45.8 × 10^-11  =  2.78 × 10^-3    . . . . . ( 2 )

Using equation ( 1 ) and equation ( 2 ), we get:
1 – (ρ₁/ρ₂)   =   2.78 × 10 ^-3
ρ₂ = 1.03 × 10³ / [ 1 – (2.78 × 10^-3) ]
= 1.032 × 10³ kg m^-3
Therefore, at the depth d water has  a density of 1.034 × 10³ kg m^–3.

Q14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer.

Given,

Pressure acting on the glass plate, p = 10 atm = 10 × 1.013 × 10⁵ Pa

We know,

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 ^-4
Therefore, the fractional change in the volume of the glass plate is 2.73 × 10^–4.

Q15. 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.

Answer:

Side of the copper cube, a = 10 cm

Therefore, Volume of the copper cube, V = a³ = 10^-3 m³

hydraulic pressure, p = 7.0 x 10⁶ Pa

Bulk modulus of copper B = 140 G Pa = 140 x 10⁹ Pa.

Bulk modulus, K=−P/(ΔV/V) ​

We get the value of volume contraction as, ΔV = – PV/K
=−(7×10⁶ ×0.001)/(140×10⁹)

=−0.05×10⁻⁶m³

Q16.How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answer:

Volume of water, V=1 litre

Water should be compressed by  0.10%
The fractional change in volume, △V/V=(0.1/100)×1=10⁻³

Bulk modulus, B =P/(△V/V) = PV/△V
P=B×(△V/V)

Bulk modulus of water, B = 2.2×10⁹ Nm ⁻²

Pressure on water, P=2.2×10⁹ ×10⁻³ =2.2×10⁶ Pa

Q17. Anvils made of single crystals of diamond, with the shape as shown in the figure, are used to investigate the 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?

Answer:

Flat faces at the narrow end of the anvil have a diameter, d=0.50mm=0.5×10⁻³ m
Radius, r=d/2=0.25×10⁻³ m
Compressional force, F=50000N
Pressure at the tip of the anvil:
P=Force/Area

Area = πr² = 3.14 x (0.25×10⁻³ )² = 0.1925 x 10^-6m²

Pressure at the tip of the anvil = F/A

= 50000/0.1925 x 10^-6
=2.59×10¹¹ Pa

Q18. 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 the figure. 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.

Answer.

Given,

Cross-sectional area of wire A, a₁ = 1.0 mm² = 1.0 × 10^–6 m²
Cross-sectional area of wire B, a₂ = 2 mm² = 2 × 10^–6 m²

We know, Young’s modulus for steel, Y₁ = 2 × 10¹¹ Nm^–2
Young’s modulus for aluminium, Y₂ = 7.0 ×10¹⁰ Nm^–2

( i ) Let a mass m be hung on the stick at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
Now it is given that the two wires have equal stresses ;
F₁ / a₁  =  F₂ / a₂

Where,
F₁ = Force acting on wire A
and F₂ = Force acting on wire B
F₁ / F₂ = a₁ / a₂  =  1 / 2    . . . . . . . . . . . . ( 1 )
The above situation can be represented as :

Moment of forces about the point of suspension, we have:
F₁y = F₂ (1.5 – y)
F₁ / F₂ = (1.5 – y) / y    . . . . . . . . . . ( 2 )
Using equation ( 1 ) and equation ( 2 ), we can write:
(1.5 – y) / y  = 1 / 2
2 (1.5 – y)  =  y
y = 1 m

Therefore, the  mass needs to be hung at a distance of 1m from the end where wire A is attached in order to produce equal stress in the two wires.

( ii ) We know,

Young’s modulus = Stress / Strain
=> Strain = Stress / Young’s modulus  =  ( F/a)/ Y
It is given that the strain in the two wires is equal :
( F₁/a₁) / Y₁  =  ( F₂/a₂) / Y₂
F₁ / F₂ = a₁Y₁ / a₂Y₂
a₁ / a₂ = 1 / 2
F₁ / F₂ = (1 / 2) (2 × 10¹¹ / 7 × 10¹⁰)  =  10 / 7      . . . . . . . . . . . ( 3 )

Let the mass m be hung on the stick at a distance y₁ from the end where the steel wire is attached in order to produce equal strain

Taking the moment of force about the point where mass m is suspended :

F₁y₁ = F₂ (1.5 – y₁)
F₁ / F₂  =  (1.5 – y₁) / y₁                                    . . . . . . . . . . . ( 4 )

From  equations ( 3 ) and ( 4 ), we get:
(1.05 – y₁) / y₁  =  10 / 7
7(1.05 – y₁)  =  10y₁
y₁ = 0.432 m

Therefore, the mass needs to be hung at a distance of 0.432 m from the end where wire A is attached in order to produce equal strain in the two wires.

Q19. 9 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.

Answer.

Given,

Water pressure at the bottom, p = 1000 atm = 1000 x 1.013 x 10⁵ Pa

p = 1.01 x 10⁸ Pa
Initial volume of the steel ball, V = 0.30 m³

We know, bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2

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  =   pV / B
=  [ 1.01 × 10⁸ × 0.30 ] / (1.6 × 10¹¹ )  =  1.89 × 10^-4 m³
Hence, volume of the ball changes by 1.89 × 10^-4 m³ on reaching the bottom of the trench.

Q20. 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.

Answer: 

Diameter of the metal strips= 6mm = 6 x  10^-3 m

Radius, r = 3 x 10^-3 m;
Shearing stress on the rivet= 6.9 x 10⁷ Pa
Maximum load or force on a rivet
= Maximum stress x cross-sectional area
= 6.9 x 10⁷ x 3.14 x (3 x 10^-3)² N = 1950 N
Maximum tension = 4 x 1950 N = 7800 N

Q21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the 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 the bottom?

Answer:

Water pressure at the bottom of the trench, p=1.1×10 ⁸ Pa
Initial volume of the steel ball, V=0.32m³
Bulk modulus of steel, B=1.6×10¹¹ Nm⁻²

The ball falls at the bottom of the trench which is nearly 11 km beneath the surface of the water.

The volume change of the ball after reaching the bottom of the trench is  △V
Bulk modulus, B=p/(△V/V) = pV/△V
△V= pV/B
=(1.1×10⁸×0.32)/(1.6×10¹¹)
= 0.352 ×10⁸/1.6×10¹¹

= 0.22 x 10^-3m³

The change in volume of the ball on reaching the bottom of the trench is 0.22 x 10^-3m³

Q22.  A mild steel wire of cross-sectional area 0.60 x 10 ^-2 cm² and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.

Answer.

Let YZ be the mild steel wire of length 2l = 2m and cross sectional area A = 0.60 x 10 ^-2 cm² . Let  the mass of m = 100 g = 0.1 kg be hung from the midpoint O, as shown in the figure. And let x be the depression at the midpoint  i.e OD

From the figure;

ZO =YO = l = 1 m ;

M = 0.1 KG

ZD = YD =  (l² + x²)^1/2

Increase in length, ∆l = YD + DZ – ZY

= 2YD – YZ                           ( As DZ = YD)

= 2(l² + x²)^1/2 – 2l

∆l = 2l( x²/2l² ) = x² / l

Therefore, longitudinal strain = ∆l / 2l = x²/2l²  . . . . . . . . ( i )

If T is the tension in the wires, then in equilibrium 2Tcosθ = 2mg

Or,          T = mg / 2cos θ

= [ mg (l² + x²)^1/2] / 2x =mgl / 2x

Therefore, Stress = T / A = mgl / 2Ax   . . . . . . . . . . . . ( ii )

= $frac{mgl^{3}}{2Ax^{3}}$

x = $l[frac{mg}{YA}]^{frac{1}{3}}$ = $1[frac{0.1 imes 10}{20 imes 10^{11} imes 0.6 imes 10^{-6}}]^{frac{1}{3}}$

= 9.41 x 10^-3 m.

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids is one of the most important chapters for the students of Class 11. Questions from Mechanical Properties of Solids are asked in most Class 11 examinations. Students who aim to score good marks in Class 11 examination and entrance examinations should try to solve NCERT questions given at the end of the chapter. Solving NCERT questions will help you to understand the chapter in a better way.

Finding out solid’s mechanical properties is one of the core concepts in physics. Here you will be covering topics such as Young’s modulus of copper and steel. We will be deriving information such as yield strength from a plotted graph and even compare between stress-strain relation graph and find out the values in Young’s modulus and strength in this chapter.

Check out NCERT Solutions for class 11 Physics for more information.

Class 11 Physics NCERT Solutions for  Chapter 9 Mechanical Properties of Solids

Have you ever stretched a coil, do you want to know what factors determine the stretch of a coil. Find out why Young’s modulus is greater in steel than in rubber. We will be finding the compression strain of each column of a cylinder and we are finding the ratio of the diameter of three wires if tension is the same in all of them. We will be finding the density of water when the pressure is at the bottom and we will also find the fractional change in the volume of a glass plate when pressure is applied. We will be calculating the pressure on a litre of water if it is compressed; along with that, you will be seeing what will happen when a ball is dropped into the Marina Trench, the deepest point in the planet’s ocean. Similar to this, we will be seeing many other examples in this chapter whose sole purpose is to make you understand the concepts clearly and if practised effectively, they will give you excellent results.

Subtopics of Class 11 Physics Chapter 9 Mechanical Properties of Solids

1. Introduction
2. Elastic behaviour of solids
3. Stress and strain
4. Hooke’s law
5. Stress-strain curve
6. Elastic Moduli
7. Applications of elastic behaviour of materials.

NCERT Solutions for Class 11 Physics Chapter 9 is prepared by the subject experts by verifying different textbooks, previous year question papers and sample papers. In order to get a good score in Class 11 examination and entrance exams, it is very important for the students to study these solutions repeatedly. CoolGyan’S provide you with the finest of study materials, notes, sample papers, MCQs (multiple choice questions), short and long answer questions, exemplary problems and worksheets. These will assist you in equipping better to face Class 11 examinations and all-important engineering and medical entrance examinations. These are prepared as per the latest CBSE syllabus 2020-21.