
Let L be the length and d be the diameter of the cross-section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of the wire will be maximum?
A. L=200cm , d=0.5mm
B. L=300cm , d=1.0mm
C. L=50cm , d=0.05mm
D. L=100cm , d=0.2mm
Answer
584.7k+ views
HintThe calculations from Young’s modulus depends on the applied force, the type of material and the area of the material. The stress of the medium relates to the ratio of the applied force with respect to the cross-sectional area. Also, the strain considers the change in length of a material with respect to its original length.
Complete step-by-step solution:In such questions we have to check all the given options.
They are asking in which option extension will be maximum
First we will the Young’s relation and try to relate it with given quantities in the question
Young’s relation \[\dfrac{{\Delta l}}{l} = \dfrac{F}{{YA}}\]
Where: \[\Delta l\] change in length
\[Y\] young’s constant for given material
\[l\] is original length
\[F\] is applied force
\[A\] area on which force is applied
From the above expression we try to find some proportionality between the change in length with given quantities.
\[\Delta l = \dfrac{{Fl}}{{YA}}\]
\[\Delta l \propto \dfrac{l}{A}\](as else quantities are same )
And we know \[A \propto {d^2}\]
Therefore, \[\Delta l \propto \dfrac{l}{{{d^2}}}\]
Now from above relation checking every relation
\[A:\;\dfrac{l}{{{d^2}}} = \dfrac{{2000}}{{0.5 \times 0.5}} = 8000\]
\[B:\;\dfrac{l}{{{d^2}}} = \dfrac{{3000}}{{1.0 \times 1.0}} = 3000\]
\[C:\;\dfrac{l}{{{d^2}}} = \dfrac{{50}}{{0.05 \times 0.05}} = 20000\]
\[D:\;\dfrac{l}{{{d^2}}} = \dfrac{{100}}{{0.2 \times 0.2}} = 2500\]
Hence option C is correct.
Note: Keep your units consistent when performing this calculation. Basically Young’s modulus is stress/strain in other words. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material in the linear elasticity regime of a uniaxial deformation. Young’s modulus of hard things is high like concrete and soft things are low.
Complete step-by-step solution:In such questions we have to check all the given options.
They are asking in which option extension will be maximum
First we will the Young’s relation and try to relate it with given quantities in the question
Young’s relation \[\dfrac{{\Delta l}}{l} = \dfrac{F}{{YA}}\]
Where: \[\Delta l\] change in length
\[Y\] young’s constant for given material
\[l\] is original length
\[F\] is applied force
\[A\] area on which force is applied
From the above expression we try to find some proportionality between the change in length with given quantities.
\[\Delta l = \dfrac{{Fl}}{{YA}}\]
\[\Delta l \propto \dfrac{l}{A}\](as else quantities are same )
And we know \[A \propto {d^2}\]
Therefore, \[\Delta l \propto \dfrac{l}{{{d^2}}}\]
Now from above relation checking every relation
\[A:\;\dfrac{l}{{{d^2}}} = \dfrac{{2000}}{{0.5 \times 0.5}} = 8000\]
\[B:\;\dfrac{l}{{{d^2}}} = \dfrac{{3000}}{{1.0 \times 1.0}} = 3000\]
\[C:\;\dfrac{l}{{{d^2}}} = \dfrac{{50}}{{0.05 \times 0.05}} = 20000\]
\[D:\;\dfrac{l}{{{d^2}}} = \dfrac{{100}}{{0.2 \times 0.2}} = 2500\]
Hence option C is correct.
Note: Keep your units consistent when performing this calculation. Basically Young’s modulus is stress/strain in other words. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material in the linear elasticity regime of a uniaxial deformation. Young’s modulus of hard things is high like concrete and soft things are low.
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