JEE Main & Advanced Physics Elasticity Young's Modulus (Y)

It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality.

\[Y=\frac{\text{Normal stress}}{\text{longitudinal strain}}=\frac{F/A}{l/L}=\frac{FL}{Al}\]

If force is applied on a wire of radius r by hanging a weight of mass M, then

\[Y=\frac{MgL}{\pi {{r}^{2}}l}\]

(i) If the length of a wire is doubled,

Then longitudinal strain = \[\frac{\text{change in length(}l\text{)}}{\text{initial length(}L\text{)}}=\frac{\text{final length}-\text{initial length}}{\text{Initial length}}=\frac{2L-L}{L}=1\] \[=\frac{\text{final length}-\text{initial length}}{\text{Initial length}}=\frac{2L-L}{L}=1\]

\[\therefore \] Young's modulus = \[\frac{\text{stress}}{\text{strain}}\]   \[\Rightarrow \]  Y = stress                                                 [As strain = 1]

So young's modulus is numerically equal to the stress which will double the length of a wire.

(ii) Increment in the length of wire \[l=\frac{FL}{\pi {{r}^{2}}Y}\]                                                                                      

\[\left[ \text{As }Y=\frac{FL}{Al} \right]\]

So if same stretching force is applied to different wires of same material, \[l\propto \frac{L}{{{r}^{2}}}\]          [As F and Y are constant]

i.e., greater the ratio \[\frac{L}{{{r}^{2}}}\], greater will be the elongation in the wire.

(iii) Elongation in a wire by its own weight : The weight of the wire Mg act at the centre of gravity of the wire so that length of wire which is stretched will be L/2.

\[\therefore \]  Elongation \[l=\frac{FL}{AY}=\frac{Mg(L/2)}{AY}\]= \[\frac{MgL}{2AY}\]\[=\frac{{{L}^{2}}dg}{2Y}\]                     

[As mass (M) = volume (AL) \[\times \] density (d)]

(iv) Thermal stress : If a rod is fixed between two rigid supports, due to change in temperature its length will change and so it will exert a normal stress (compressive if temperature increases and tensile if temperature decreases) on the supports. This stress is called thermal stress.

As by definition, coefficient of linear expansion \[\alpha =\frac{l}{L\Delta \theta }\]

\[\Rightarrow \] thermal strain \[\frac{l}{L}=\alpha \Delta \theta \]

So  thermal stress \[=Y\alpha \Delta \theta \]                     [As Y = stress/strain]

And tensile or compressive force produced in the body \[=YA\alpha \Delta \theta \]

Note :

• In case of volume expansion Thermal stress \[=K\gamma \Delta \theta \]

Where K = Bulk modulus, \[\gamma =\] coefficient of cubical expansion

(v) Force between the two rods : Two rods of different metals, having the same area of cross section A, are placed end to end between two massive walls as shown in figure. The first rod has a length \[{{L}_{1}},\] coefficient of linear expansion \[{{\alpha }_{1}}\] and young?s modulus \[{{Y}_{1}}\]. The corresponding quantities for second rod are \[{{L}_{2}},\,{{\alpha }_{2}}\] and \[{{Y}_{2}}\]. If the temperature of both the rods is now raised by T degrees.

Increase in length of the composite rod (due to heating) will be equal to \[{{l}_{1}}+{{l}_{2}}=\]\[[{{L}_{1}}{{\alpha }_{1}}+{{L}_{2}}{{\alpha }_{2}}]\,T\]                  \[[\text{As}\,\,l=L\alpha \Delta \theta ]\]

and due to compressive force F from the walls due to elasticity, decrease in length of the composite rod will be equal to

\[\left[ \frac{{{L}_{1}}}{{{Y}_{1}}}+\frac{{{L}_{2}}}{{{Y}_{2}}} \right]\frac{F}{A}\]                                               \[\left[ \text{As }l=\frac{FL}{AY} \right]\]

as the length of the composite rod remains unchanged the increase in length due to heating must be equal to decrease in length due to compression i.e. \[\frac{F}{A}\left[ \frac{{{L}_{1}}}{{{Y}_{1}}}+\frac{{{L}_{2}}}{{{Y}_{2}}} \right]=[{{L}_{1}}{{\alpha }_{1}}+{{L}_{2}}{{\alpha }_{2}}]\,T\]

or   \[F=\frac{A[{{L}_{1}}{{\alpha }_{1}}+{{L}_{2}}{{\alpha }_{2}}]T}{\left[ \frac{{{L}_{1}}}{{{Y}_{1}}}+\frac{{{L}_{2}}}{{{Y}_{2}}} \right]}\]

(vi) Force constant of wire : Force required to produce unit elongation in a wire is called force constant of material of wire. It is denoted by k.

\[\therefore \] \[k=\frac{F}{l}\]                                                                             ...(i)

but from the definition of young's modulus

\[\frac{F}{l}=\frac{YA}{L}\]                                                                                    ...(ii)        

from (i) and (ii) \[k=\frac{YA}{L}\]

It is clear that the value of force constant depends upon the dimension (length and area of cross section) and material of a substance.

(vii) Actual length of the wire : If the actual length of the wire is L, then under the tension \[{{T}_{1}},\]  its length becomes \[{{L}_{1}}\] and under the tension \[{{T}_{2}},\] its length becomes \[{{L}_{2}}\].

\[{{L}_{1}}=L+{{l}_{1}}\] Þ \[{{L}_{1}}=L+\frac{{{T}_{1}}}{k}\]                             ...(i)        

and \[{{L}_{2}}=L+{{l}_{2}}\]\[\Rightarrow \]\[{{L}_{2}}=L+\frac{{{T}_{2}}}{k}\] ...(ii)

From (i) and (ii) we get \[L=\frac{{{L}_{1}}{{T}_{2}}-{{L}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]