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

## NCERT Solutions For Class 11 Physics Chapter 9 PDF Free Download

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 in this solution 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 11th 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 practiced 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 cross-sectional area 3 x 10– 5mand length of 5m stretches by the same length as a copper wire of cross-sectional area 4.0 x 10-5m2 and length 4m  under a given load. Find the ratio of  Young’s modulus of copper to that of steel?

Ans.

Given,

Length of the steel wire, L1 = 5 m
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 graph given below is the stress-strain curve of a material. Find this material’s ( i )Approximate yield strength (  ii ) Young’s modulus.

Ans. ( 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. Given below are stress-strain curves for two materials X and Y. If both the graphs are drawn to the same scale.

( a ) Identify the material with a greater Young’s Modulus.

( b ) Identify the stronger material.

Ans.

( 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. A state with reasons whether the following statements are true or false.

( a ) Shear modulus determines how much a coil can stretch.

( b ) Rubber has Young’s Modulus greater than that of steel.

Ans.

( 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. A wire made up of steel and brass and having a diameter of 0.30 cm is loaded as depicted in the diagram below. The unloaded length of the brass wire is 1 m and that of the steel wire is 1.8.Find the length of elongations in the brass and steel wires.

Ans. 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. An aluminium cube has an edge 8cm long, while one face is firmly against the wall the other is pressed against a 50 kg body. If the shear modulus of aluminium is 25 GPa. Find the vertical deflection of this face.

Ans.

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. A big building of 25000 kg mass is supported by four follow cylindrical columns made of mild steel. If the outer and inner radii of each column are 60 cm and 40 cm respectively, find the compressional strain of each column. ( Consider the load distribution is in uniform )

Ans.

A.Given,

Mass of the building, M = 25,000 kg

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

Inner radius of the column, r = 40 cm = 0.4 m

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

We know,

Total force exerted, F = Mg = 25000 × 9.8 N = 245000 N
Stress = Force exerted on a single column = 245000 / 4  =  61250 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 column having a mass of 20 kg is supported by three wires each of length 2 m. The wires at the end are copper and the one in the middle is iron. If the tension on the three wires in the same, find the ratio of their diameters.

Ans.

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 15 kg mass is tied to a steel wire of 1m ( unstretched length). It is then spun in vertical circles where its angular velocity is  2 rev/s at the lowermost point. If the wire’s cross-sectional area is 0.060 cm2, find the elongation in the wire when the weight is at the lowest point.

Ans.

A.Given,

Mass, m = 15 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
= 15 × 9.8 + 15 × 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. Using the data provided, find the bulk modulus of water. Also, compare the bulk modulus of water and air (at constant temperature) and explain why the ratio is so large. Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre.

Ans.

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. At a depth where the pressure is 60 atm find the density of water if at the surface it is 1.03 x 103 kg m-3 ?

Ans.

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. Calculate the fractional change in the volume of a glass plate when it is subjected to a pressure of100 atm.

Ans.

Given,

Pressure acting on the glass plate, p = 100 atm = 100 × 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
= [ 100 × 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. A solid copper cube with edges of length 5 cm is subjected to a hydraulic pressure of 8 x 106 Pa. Calculate the volume contraction in it.

Ans.

A.Given,

Hydraulic pressure, p = 8.0 × 106 Pa

Edge length of the cube, l = 5 cm = 0.05 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. Calculate the pressure on a litre of water if it is to be compressed by 0.15%.

Ans.

Given, volume of water,V= 1 L
And water needs to be compressed by 0.15%.
∴ Fractional change, ∆V / V = 0.15 / (100 × 1)  =  1.5 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 × 1.5 x 10-3  =  3.3 × 106 Nm-2
Thus, a pressure of  3.3 ×106 Nm–2 should be applied on the water.

Q17.  A diamond anvil cell is used to create extremely high-pressure environments. The narrow ends of the anvil have a diameter of 0.50 mm and the wide ends are subjected to a compressive force of 80,000 N. Calculate the pressure at the tip of the anvil.

Ans. 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 = 80000 N
Therefore the pressure at the tip of the anvil:
P = Force / Area  =  80000 / π(0.25 × 10-3)2
= 4.07  × 1011 Pa

Q18. A stick of length 1.5 m is supported by a steel wire (wire A) and an aluminum wire (wire B) from its two ends. The wires A and B have a cross-sectional area of 1 mm2 and 2 mm2 respectively. Calculate the location of a point along the stick from where we can hang a body of mass m  that will produce ( i ) equal stresses, and  ( ii ) equal strains in both the wires.  Assume the stick to be weightless.

Ans. 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 aluminum, 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. The deepest known point in our planet’s oceans is the Marina Trench, it is 10,994 m deep and the water pressure at its bottom is 1000 atm. If a steel ball having an initial volume of 0.30 m 3 is dropped into the trench, find the change in the volume of this ball when it hits the bottom.

Ans.

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 metal bars are riveted together at their ends by four rivets, each having a diameter of 5 mm. Calculate the maximum tension the riveted bars can bear if the maximum shearing stress a rivet can take is 6.9 x 107 Pa. Consider that each rivet carries ¼ of the total load.

Ans.

Given,

Diameter of the metal bar, d = 5.0 mm = 5.0 × 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.

Ans. 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

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