Question

Consider an iron rod of length 1 metre and cross-section 1 $cm^2$ with a Young's modulus of $10^{12} dyne cm^{-2}$. We wish to calculate the force with which the two ends must be pulled to produce an elongation of 1 mm. It is equal to:

Answer 1

Hint: Within the elastic limits, the ratio of longitudinal stress and longitudinal strain for the rod Barbie equal to Young's modulus of elasticity. Longitudinal stress acting on the rod is the amount of force acting on per unit area.

Formula used:
The force for the rod can be written as:
$F = \dfrac{AY \Delta l}{l}$

Complete step by step answer:
Consider any rod with cross-section area A and length l, if you suspend this rod with a weight, then a force will act on the rod which will change its length by an amount depending on the modulus of elasticity of that material.
On application of stress, the strain produced varies linearly till the elastic limit is reached. This is basic Hooke's law. Mathematically, for the case of elongation (linear stretch):
Stress = Y $\times$ Strain ;
where Y is the constant of proportionality.
Now, stress is also force per unit area and strain is the ratio to change in length to original length. So we may write:
$\dfrac{F}{A} = \dfrac{Y \Delta l}{l}$,
which gives us:
$F = \dfrac{AY \Delta l}{l}$ .

For our case, we are given the area of cross-section of the rod, A = 1 $cm^2$, length of the rod, l = 1 m = 100 cm, elongation $\Delta l$ = 1mm = 0.1 cm, Young's modulus Y = $10^{12}$ dyne $cm^{-2}$. Keeping these values we get:
$F = \dfrac{ 1 \times 10^{12} \times 0.1}{ 100} = 10^9$ dyne.

In standard units, we can use the conversion:
$ 1 dyne = 10^{-5} N$
So,
$10^9 dyne = 10^9 \times 10^{-5} N = 10^4 N$.

Therefore the given rod we pulled with a force of $10^4$ N to cause an elongation of 1mm.

Note:
Carefully note the units in this case as some of them are in metres and millimetres. We chose to find force in C.G.S. units first and then convert the obtained force in S.I. units. One could also convert all the given values in SI units first and then find the required force. Although, it has not been asked to specific unit systems but still I converted into S.I. is recommended.