Question

The increase in energy of a metal bar of length ‘L’ and cross sectional area ‘A’ when compressed with a load ‘M’ along its length is: (Y= young modulus of the material of metal bar)
A) $\dfrac{{FL}}{{2AY}}$
B) $\dfrac{{{F^2}L}}{{2AY}}$
C) $\dfrac{{FL}}{{AY}}$
D) $\dfrac{{{F^2}{L^2}}}{{2AY}}$

Answer 1

For the above question Young’s modulus concept is as follows:
Young Modulus is given by the formula:
$Y = \dfrac{\text{longitudinal stress}}{{strain}}$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}$ , F is the force, A is the area and l is the length and $\Delta L$ is the change in length.
Energy, U is given by $= \dfrac{1}{2} \times stress \times strain \times volume$
Using the above relations we will calculate the energy of the metal bar.

Complete step by step solution:
Young’s Modulus is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression .Young’s modulus is the ratio of longitudinal stress to strain.SI unit of Young’s modulus is Pascal.

Now let’s derive the equation for the energy of the metal bar
U$= \dfrac{1}{2} \times stress \times strain \times volume$(equation for energy)
Young’s modulus is given by:
$stress = \dfrac{F}{A}$ (F is force and A is the area)
Then, strain is given by:
$strain = \dfrac{{stress}}{Y} = \dfrac{F}{{AY}}$ (Y is young modulus here)
Volume of the metal bar is given: AL.
Therefore, energy of the metal rod is given as:
$U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{{AY}})AL$ .....................1
Now, we will cancel the common terms in equation 1 to bring out the formula for energy.
$\Rightarrow U = \dfrac{1}{2} \times \dfrac{F}{A} \times (\dfrac{F}{Y})L \\\ \Rightarrow U = \dfrac{1}{2} \times \dfrac{{{F^2}L}}{{AY}} \\\$

Note: Application of Young’s Modulus in our day to day life is in designing a bridge ,one has to keep in mind the traffic load, weight of the bridge ,force of wind .Bridge should also not bend too much nor break. Another application is heavy load lifted by the crane, which depends on the weight of the load and the stress and strain applied by the crane because pressure is equal to the force upon the area.