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Question: A metallic wire has a radius of \( \;0.2{\text{ }}mm \) . How much force is required to have an incr...

A metallic wire has a radius of   0.2 mm\;0.2{\text{ }}mm . How much force is required to have an increase of 0.2%0.2\% in its length? (Y = 9.0 × 1010N m−2)\left( {Y{\text{ }} = {\text{ }}9.0{\text{ }} \times {\text{ }}{{10}^{10}}N{\text{ }}{m^{ - 2}}} \right) .

Explanation

Solution

Hint : In this solution, we will use the relation of stress and strain with young’s modulus to determine the force needed. The stress acting on an object is the ratio of force to the cross-section area on which the force is acting.

Formula used: In this solution, we will use the following formula
- Y=FLAΔLY = \dfrac{{FL}}{{A\Delta L}} where YY is the young’s modulus, FF is the force on the wire, AA is the cross-sectional area of the wire.

Complete step by step answer
We’ve been given that a metallic wire has a radius of   0.2 mm\;0.2{\text{ }}mm and we want to find the force that has to be acted on the wire to have a change of 0.2%0.2\% in its length. We know that the young’s modulus of a material is equal to the ratio of the stress that is acting on the body to the stain. We also know that the strain is the ratio of the force that is acting on the wire to the cross-sectional area of the wire. And that strain is the relative change in length of the wire when force is applied to it.
This implies that ΔLL=0.2100=0.002\dfrac{{\Delta L}}{L} = \dfrac{{0.2}}{{100}} = 0.002 . Also, we know that the area of cross-section of the wire is calculated as A=πr2A = \pi {r^2} where r=0.2 mm=0.2×10−3 mr = 0.2\,mm = 0.2 \times {10^{ - 3}}\,m
Then using the formula for young’s modulus, we get
9×1010=Fπ(0.2×10−3)2(0.002)9 \times {10^{10}} = \dfrac{F}{{\pi {{\left( {0.2 \times {{10}^{ - 3}}} \right)}^2}\left( {0.002} \right)}}
Solving for the force, we get
F=22.6 NF = 22.6\,N
Hence the force that should be acting on the wire to make a change in length of 0.2%0.2\% is given F=22.6 NF = 22.6\,N .

Note
The force that will be acting on the wire can be compressive or expansive since we only know the relative change in its length however it is not of consequence while using the formula. While calculating the area of the wire, we can consider it as a circular cross-section and hence the formula of the area of a circle.