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

NCERT Solutions Class 11 Physics Chapter 9 Mechanical Properties of Solids is a foremost resource that helps your understanding of the topics and helps your Class 11 and entrance exam preparation. It is very important to make notes to understand the topic thoroughly. This solution will assist you in preparing notes through its exemplar problems, worksheets, HOTS (high order thinking skills) and questions from previous year question papers. It is important to solve the questions to get well versed with the topic.

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 of the Class 11 examination. Students who want 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.

## 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 even we will 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.

### 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.

### Important questions of Class 11 Physics Chapter 9 Mechanical Properties of Solids

Q1. 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?

Given,

Length of the steel wire, L1 = 5 m
The cross-sectional area of the steel wire, A1 = 3.0 × 10–5 m2
Length of the copper wire, L2 = 4 m
Cross-sectional area of the copper wire, A2 = 4.0 × 10–5 m2

Change in length = ΔL1 = ΔL2 = ΔL
Let the force being applied in both the situations  = F

We know, Young’s modulus of the steel wire :
Y1 = (F1 / A1) (L1 / ΔL1)
= (F / 3 X 10-5) (5 / ΔL)     . . . . . . . . . . ( 1 )

Also, Young’s modulus of the copper wire:
Y2 = (F2 / A2) (L2 / ΔL2)
= (F / 4 × 10-5) (3 / ΔL)     . . . . . . . . . . .( 2 )

Dividing equation ( 1 ) by equation ( 2 ), we get :
Y1 / Y2  =  (5 × 4 × 10-5) / (3 × 10-5 × 4)
= 1.6 : 1
The ratio of Young’s modulus of steel to Young’s modulus of copper is 1.6: 1

Q2.  The figure shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material? ( i ) It can be seen from the graph that the approximate yield strength of this material is 300 × 106Nm/2or 3 × 108 N/m2.

( ii ) It is observed from the given graph that for strain  0.001, stress is 75 × 106N/m2.
∴ We know, Young’s modulus, Y = Stress / Strain
= 75 × 106 / 0.001  =  7.5 × 1010 Nm-2

Q3. The stress-strain graphs for materials X and Y are 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 ) Comparing the two graphs we can infer that the stress on X is greater than that on Y for the same values of strain. Therefore, Young’s Modulus ( stress/strain ) is greater for X.

( b ) As X’s Young’s modulus is higher, it is the stronger material among the two. For strength is the measure of stress a material can handle before breaking.

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.

( 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 the figure. 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. Given,

Diameter of the wire, d = 0.25 m
Hence, the radius of the wires, r = d/2  =  0.125 cm
Length of the steel wire, L1 = 1.8 m
Length of the brass wire, L2 = 1.0 m

Total force exerted on the steel wire:

F1 = (4 + 8) g = 12 × 9.8 = 117.6 N
We know, 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

We know ,Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
∴ ΔL1 = F1  × L1 / (A1  × Y1)
= (117.6  × 1.8) / [ π(0.125  × 10-2)2  × 2  × 1011]   =   2.15 × 10-4m

Total force on the brass wire:
F2 = 8 × 9.8 = 78.4 N

Young’s modulus for brass:

Y2 = 0.91 x 1011 Pa
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)
= (78.4 X 1) / [ π  × (0.125  × 10-2)2  × (0.91  × 1011) ] = 1.75  × 10-4 m
Therefore, Elongation of the steel wire = 2.15 × 10-4m, and
Elongation of the brass wire = 1.75  × 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?

Given,

Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa

Edge of the cube, L = 8 cm = 0.08 m

The mass attached to the cube, m = 50 kg

Shear modulus, η = Shear stress / Shear strain  =  (F/A ) / ( L /ΔL)

Where,
F = Applied force = mg = 50 × 9.8  =  490 N
A = Area of one face of the cube = 0.08 × 0.08 = 0.0064 m2
ΔL= Vertical deflection of the cube
∴ ΔL = FL / Aη
= 490 × 0.08 / [ 0.0064× (25 × 109) ]
= 2.45 x 10-7 m
Therefore the vertical deflection of this face of the cube is 2.45 ×10–7 m.

Q7. 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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column

A.Given,

Mass of the building, M = 50,000 kg

Outer radius of the column, R = 60 cm = 0.6 m

Inner radius of the column, r = 30 cm = 0.3 m

Young’s modulus of steel, Y = 2 × 1011 Pa

We know,

Total force exerted, F = Mg = 50000 × 9.8 N = 490000 N
Stress = Force exerted on a single column = 490000 / 4  =  122500 N
Also, Young’s modulus, Y = Stress / Strain

Strain = (F/A) / Y
Where,
Area, A = π (R2 – r2)

= π ((0.6)2 – (0.4)2)

= 0.628
Strain = 61250 / [ 0.628× 2 × 1011 ]  =  4.87 × 10-7
Therefore, the compressional strain of each column is 4.87 × 10–7.

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.

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, Y= 120 × 109 Pa
let the diameter of the copper wire = d2
Thus, the ratio of their diameters can be given as :

$\frac{d_{1}}{d_{2}} = \sqrt{\frac{Y_{1}}{Y_{2}}}$

= $\sqrt{\frac{190×10^{9}}{120 x 10^{9}}}$ =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.

A.Given,

Mass, m = 14.5 kg
Length of the 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 = 0.060 cm2 = 0.06 × 10-4m2
Let Δl be the increase in the wire’s length when the body is at the lower most point.
When the body is at the lowest point of the vertical circle , the force on the body is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2513.304 N

We know, Young’s modulus = Stress / Strain
Y = ( F/A ) / ( ∆l / l )
∴ ∆l = Fl / AY
Also, young’s modulus for steel = 2 × 1011 Pa
=> ∆l = ( 2513.304 × 1) / (0.06 × 10-4 × 2 × 1011)   =   2.09 × 10-3 m

Therefore, the increase in the wire is 2.09 × 10–3 m.

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

Given,

P = 100 atmosphere

= 100 x 1.013 x 105 Pa

Final volume, V= 100.5 l = 100.5 ×10 –3 m3

Initial volume, V= 100.0 l = 100.0 × 10 –3 m3

Increase in volume, ΔV = V2 – V= 0.5 × 10–3 m3

Bulk modulus = Δp / ( ΔV / V1 )  =  Δ p × V1 / ΔV
= [ 100 × 1.013 × 105 × 100 × 10-3 ] / (0.5 × 10-3)
= 2.026 × 109 Pa

We know ,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 large because air has more intermolecular space thus it is more compressible than water.

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?

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, ρ= 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 – (ρ12)        . .  . . .  . . ( 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 – (ρ12)   =   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.

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.

A.Given,

Hydraulic pressure, p = 7.0 × 106 Pa

Edge length of the cube, l = 10 cm = 0.10 m

Bulk modulus of copper, B = 140 × 109 Pa

We know, bulk modulus, B = p / (∆V/V )
Where,
V = Original volume = V = l 3
ΔV = Change in volume
.∆V/V = Volumetric strain.
ΔV = pV / B

∴ ΔV = pl3 / B
= [ 8 × 106 × (0.05)3 ] / (140 × 109)
= 7.142 × 10-9 m3  =  7.142 × 10 -3 cm-3
Hence, the volume contraction of the solid copper cube is 7.142 × 10  –3 cm–3.

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

Given, volume of water, V= 1 L
And water needs to be compressed by 0.10%.
∴ Fractional change, ∆V / V = 0.10 / (100 × 1)  =  1.0 x 10-3
We know,

Bulk modulus, B = ρ / (∆V/V)
=> ρ = B × (∆V/V)

We know, bulk modulus of water, B = 2.2 × 109 Nm-2
=> ρ = 2.2 × 109 × 1.0 x 10-3  =  3.3 × 106 Nm-2
Thus, a pressure of  3.3 ×106 Nm–2 should be applied on the water.

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? Given,

Diameter 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
Therefore the pressure at the tip of the anvil:
P = Force / Area  =  50000 / π(0.25 × 10-3)2
= 4.07  × 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. Given,

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

F1y= 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.

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.

Given,

Diameter of the metal bar, d = 6.0 mm = 6.0 × 10–3 m
Radius, r = d/2 = 2.5 × 10-3 m

Maximum shearing stress = 6.9 × 107 Pa

We know,

Maximum stress =  Maximum force or tension / Area
=> Maximum force = Maximum stress × Area
= 6.9 × 107 × π × (2.5) 2
= 6.9 × 107 × π × (2.5 ×10–3)2
= 1354.125 N
Since each rivet carries ¼  of the load.
∴ Maximum tension on each rivet = 4 × 1354.125 = 5416.5 N.

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. 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 )

$Y = \frac{stress}{strain} = \frac{mgl}{2Ax} * \frac{2l^{2}}{x^{2}}$

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

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

= 9.41 x 10-3 m.

 Also Access NCERT Exemplar for Class 11 Physics Chapter 9 CBSE Notes for Class 11 Physics Chapter 9

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. BYJU’S provide you with 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 examination. These are prepared as per the latest CBSE syllabus 2020-21. Related Links Solution Of Class 10 NCERT Maths Class 9 Science NCERT Solutions PDF NCERT Class 10 Science Solutions NCERT Solution For Class 12 Chemistry NCERT Class 7 Solutions NCERT Solutions Class 8 All Subjects NCERT Questions For Class 9 NCERT Class 10 Solution NCERT Solution Of Maths Of Class 7 NCERT Solutions For Class 8 Maths Free Download NCERT Class 9 Maths Solution Download Class 11 Maths NCERT Solutions