Qstnid -Iv-4-a Steel Wire And A Copper Wire Of Equal Length And Equal Cross-sectional Area Are Joined End To End And The Combination Is Subjected To A Tension. Find The Ratio Of - 14: Some Mechanical Properties Of Matter

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NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 5

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#4
A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of
Ans : Given:
Young's modulus of steel = 2 × 10¹¹ N m^-2
Young's modulus of copper = 1.3 × 10 11 N m^-2
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
As per the question:
`` {L}_{\,\mathrm{\,steel\,}}={L}_{\,\mathrm{\,Cu\,}}``
`` {A}_{\,\mathrm{\,steel\,}}={A}_{\,\mathrm{\,Cu\,}}``
`` {F}_{\,\mathrm{\,Cu\,}}={F}_{\,\mathrm{\,Steel\,}}``
`` ``
Here: L_steel and L_Cu denote the lengths of steel and copper wires, respectively.
A_steeland A_Cu denote the cross-sectional areas of steel and copper wires, respectively.
F_steel and F_Cu denote the tension of steel and cooper wires, respectively.
`` \left(\,\mathrm{\,a\,}\right)\frac{\,\mathrm{\,Stress\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Stress\,}\,\mathrm{\,of\,}\,\mathrm{\,Steel\,}}=\frac{{F}_{\,\mathrm{\,Cu\,}}}{{A}_{\,\mathrm{\,Cu\,}}}\frac{{A}_{\,\mathrm{\,Steel\,}}}{{F}_{\,\mathrm{\,Steel\,}}}=1``
(b)
`` \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\displaystyle \frac{∆{L}_{\,\mathrm{\,Steel\,}}}{{L}_{\,\mathrm{\,Steel\,}}}}}{{\displaystyle \frac{∆{L}_{\,\mathrm{\,cu\,}}}{{L}_{\,\mathrm{\,cu\,}}}}}=\frac{{F}_{\,\mathrm{\,Steel\,}}{L}_{\,\mathrm{\,Steel\,}}{A}_{cu}{Y}_{cu}}{{A}_{\,\mathrm{\,Steel\,}}{Y}_{\,\mathrm{\,Steel\,}}{F}_{cu}{L}_{cu}}``
`` \left(\,\mathrm{\,Using\,}\frac{∆L}{L}=\frac{F}{AY}\right)``
`` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,cu\,}}}{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,Steel\,}}}=\frac{1.3\times {10}^{11}}{2\times {10}^{11}}``
`` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,C\,}\,\mathrm{\,u\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{13}{20}``
`` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}=\frac{20}{13}``
Hence, the required ratio is 20 : 13.
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