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## NCERT Class 11 Physics Textbook Chapter 9 With Answer PDF Free Download

### Chapter 9: Mechanical Properties of Solids

#### 9.1 Introduction

In Chapter 7, we studied the rotation of the bodies and then realised that the motion of a body depends on how mass is distributed within the body.

We restricted ourselves to simpler situations of rigid bodies. A rigid body generally means a hard solid object having a definite shape and size.

But in reality, bodies can be stretched, compressed, and bent.

Even the appreciably rigid steel bar can be deformed when a sufficiently large external force is applied to it.

This means that solid bodies are not perfectly rigid. A solid has a definite shape and size. In order to change (or deform) the shape or size of a body, force is required.

If you stretch a helical spring by gently pulling its ends, the length of the spring increases slightly. When you leave the ends of the spring, it regains its original size and shape.

The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity and the deformation caused is known as elastic deformation.

However, if you apply force to a lump of putty or mud, they have no gross tendency to regain their previous shape, and they get permanently deformed.

Such substances are called plastic and this property is called plasticity. Putty and mud are close to ideal plastics.

#### 9.2 Elastic Behaviour of Solids

We know that in a solid, each atom or molecule is surrounded by neighboring atoms or molecules.

These are bonded together by interatomic or intermolecular forces and stay in a stable equilibrium position.

When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the interatomic (or intermolecular) distances.

When the deforming force is removed, the interatomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size.

#### 9.3 Stress And Strain

When forces are applied to a body in such a manner that the body is still in static equilibrium,

it is deformed to a small or large extent depending upon the nature of the material of the body and the magnitude of the deforming force.

The deformation may not be noticeable visually in many materials but it is there.

When a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but opposite in direction to the applied force.

The restoring force per unit area is known as stress. If F is the force applied normal to the cross-section and A is the area of cross-section of the body, the Magnitude of the stress = F/A (9.1)

The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T –2 ].

There are three ways in which a solid may change its dimensions when an external force acts on it.

These are shown in Fig. 9.2. In Fig.9.2(a), a cylinder is stretched by two equal forces applied normally to its cross-sectional area.

The restoring force per unit area in this case is called tensile stress. If the cylinder is compressed under the action of applied forces, the restoring force per unit area is known as compressive stress.

Author | NCERT |

Language | English |

No. of Pages | 15 |

PDF Size | 1542 KB |

Category | Physics |

Source/Credits | ncert.nic.in |

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

**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=2×10^{11} 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(R^{2}−r^{2})=π((0.6)^{2}−(0.3)^{2}) = 0.27 π m^{2}

Strain =122500/[0.27 x 3.14×2×10^{11}]=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^{−4}m^{2}

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

Modulus of elasticity of copper, Y=42×10^{9 }Nm^{−2}

Modulus of elasticity (Y) = Stress / Strain

=(F/A) / Strain

Strain = F/(YA)

= 44500/(2.9×10^{−4}×42×10^{9})

= 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 10 ^{8} 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 = πr^{2}= 3.14 x ( 0.015)^{2}

= 7.06 x 10^{-4} m

Maximum stress allowed on the steel cable = 10^{8} N/m^{2}

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

= 10^{8} x 7.06 x 10^{-4}

= 7.065×10^{4 }N

Hence, the cable can support the maximum load of 7.065×10^{4} 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/πd^{2}) / 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/d^{2})

We know that Young’s modulus for iron, Y_{1} = 190 × 10^{9} Pa

Let the diameter of the iron wire = d_{1}

Also, Young’s modulus for copper, Y_{2 }= 120 × 10^{9} Pa

let the diameter of the copper wire = d_{2}

Thus, the ratio of their diameters can be given as :\frac{d_{1}}{d_{2}} = \sqrt{\frac{Y_{1}}{Y_{2}}}*d*2*d*1=*Y*2*Y*1

= \sqrt{\frac{190×10^{9}}{120×10^{9}}}120×109190×109 =1 : 25 : 1

Mechanical Properties of Solids Textbook With Solutions PDF Free Download