CBSE Class 11 Physics Notes for Mechanical Properties of Solid , Hooke’s Law , Stress and Strain

NCERT Class 11 Physics : If we apply a force to a rubber band, we find that the rubber band stretches. Similarly, if we attach a wire to a support, as shown in figure 1, and sequentially figure 2 stretching an object. Apply forces of magnitude F, 2F, and 3F to the wire, we find that the wire stretches by an amount ∆L, 2∆L, and 3∆L, respectively. (Note that the amount of stretching is greatly exaggerated in the diagram for illustrative purposes.)

Fig. 1 Stretching of an object

The deformation, ∆L, is directly proportional to the magnitude of the applied force F and is written mathematically as

∆L ∝ F

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This aspect of elasticity is true for all solids. It would be tempting to use equation as it stands to formulate a theory of elasticity, but with a little thought, it becomes obvious that although it is correct in its description, it is incomplete.

Fig 2. The deformation is inversely proportioned to the cross-section of wire.

Physics Notes for Class 11

Let us consider two wires, one of cross-sectional area A, and another with twice that area, namely 2A, as shown in figure 2. When we apply a force F to the Figure 2 The deformation is inversely proportional to the cross-sectional area of the wire. first wire, that force is distributed over all the atoms in that cross-sectional area A. If we subject the second wire to the same applied force F, then this same force is distributed over twice as many atoms in the area 2A as it was in the area A. Equivalently we can say that each atom receives only half the force in the area 2A that it received in the area A. Hence, the total stretching of the 2A wire is only 1/2 of what it was in wire A. Thus, the elongation of the wire ∆L is inversely proportional to the cross-sectional area A of the wire, and this is written

∆L ∝ 1

A Note also that the original length of the wire must have something to do with the amount of stretch of the wire. For if a force of magnitude F is applied to two wires of the same cross-sectional area, but one has length L0 and the other has length 2L0, the same force is transmitted to every molecule in the length of the wire. But because there are twice as many molecules to stretch apart in the wire having length 2L0, there is twice the deformation, or 2∆L, as shown in figure 4. We write this as the proportion

∆L ∝ L0

Fig. 3 the deformation is directly proportional to the original length

The results of equations are, of course, also deduced experimentally. The deformation ∆L of the wire is thus directly proportional to the magnitude of the applied force F, inversely proportional to the cross-sectional area A, and directly proportional to the original length of the wire L0. These results can be incorporated into the one proportionality

∆L ∝ FL0/A

which we rewrite in the form

F/A ∝∆L/A L0

The ratio of the magnitude of the applied force to the cross-sectional area of the wire is called the stress acting on the wire, while the ratio of the change in length to the original length of the wire is called the strain of the wire. Above equation is a statement of Hooke’s law of elasticity, which says that in an elastic body the stress is directly proportional to the strain, that is,

stress ∝ strain

The stress is what is applied to the body, while the resulting effect is called the strain. To make an equality out of this proportion, we must introduce a constant of proportionality. This constant depends on the type of material used, since the molecules, and hence the molecular forces of each material, are different. This constant, called Young’s modulus of elasticity is denoted by the letter Y. thus above equation becomes

F/A = Y ∆L/LO

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