- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- Chapter 3- Motion in a Straight Line
- Chapter 4- Motion in a plane
- Chapter 5- Laws of motion
- Chapter 6- Work Energy and power
- Chapter 7- System of particles and Rotational Motion
- Chapter 8- Gravitation
- Chapter 10- Mechanical Properties Of Fluids
- Chapter 11- Thermal Properties of matter
- Chapter 12- Thermodynamics
- Chapter 13- Kinetic Theory
- Chapter 14- Oscillations
- Chapter 15- Waves

Chapter 1- Physical World |
Chapter 2- Units and Measurements |
Chapter 3- Motion in a Straight Line |
Chapter 4- Motion in a plane |
Chapter 5- Laws of motion |
Chapter 6- Work Energy and power |
Chapter 7- System of particles and Rotational Motion |
Chapter 8- Gravitation |
Chapter 10- Mechanical Properties Of Fluids |
Chapter 11- Thermal Properties of matter |
Chapter 12- Thermodynamics |
Chapter 13- Kinetic Theory |
Chapter 14- Oscillations |
Chapter 15- Waves |

**Answer
1** :

**Answer
2** :

(a) Young’smodulus of the material (Y) is given by

Y =Stress/Strain

=150 x 10^{6}/0.002

150 x 10^{6}/2 x 10-3

=75 x 10^{9} Nm^{-2}

=75 x 10^{10} Nm^{-2}

(a)Yield strength of a material is defined as the maximum stress it cansustain. From graph, the approximate yield strength of the given material

= 300 x 10^{6} Nm^{-2}

= 3 x 10^{8} Nm^{-2} .

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
3** :

(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus =(Stress /Strain) is greater for A than that of B.

(b) Strength of a material is determined by the amount of stress required to cause fracture. This stress corresponds to the point of fracture. The stress corresponding to the point of fracture in A is more than for B. So, material A is stronger than material B.

Read the ‘allowing 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
4** :

(a) False. The-Young’s modulus is defined as the ratio of stress to the strain within elastic limit. For a given stretching force elongation is more in rubber and quite less in steel. Hence, rubber is less elastic than steel.

(b) True. Stretching of a coil is determined by its shear modulus. When equal and opposite forces are applied at opposite ends of a coil, the distance, as well as shape of helicals of the coil change and it, involves shear modulus

**Answer
5** :

**Answer
6** :

Here, side of cube, L = 10 cm =10/100= 0.1 m

.•. Area of each face, A = (0.1)^{2} = 0.01 m^{2}

**Answer
7** :

**Answer
8** :

**Answer
9** :

**Answer
10** :
Since each wire is to have same tension therefore, each wire has same extension. Moreover, each wire has the same initial length.So, strain is same for each wire.

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