A) \[(2.5{{l}_{2}}-1.5{{l}_{1}})m\]
B) \[(6{{l}_{2}}-1.5{{l}_{1}})m\]
C) \[(3{{l}_{2}}-2{{l}_{1}})m\]
D) \[(3.5{{l}_{2}}-2.5{{l}_{1}})m\]
E) \[(2.5{{l}_{2}}+1.5{{l}_{1}})m\]
Correct Answer: A
Solution :
Let the original unstretched length be\[l.\] \[Y=\frac{Stress}{Strain}=\frac{T/A}{\Delta l/l}=\frac{T}{A}\times \frac{l}{\Delta l}\] Now\[Y=\frac{4}{A}\frac{l}{({{l}_{1}}-l)}=\frac{6}{A}\frac{l}{({{l}_{2}}-l)}\] \[=\frac{9}{A}\frac{l}{({{l}_{3}}-l)}\] \[\therefore \] \[4({{l}_{3}}-l)=9({{l}_{1}}-l)\] \[\Rightarrow \] \[4{{l}_{3}}+5l=9{{l}_{1}}\] ?.. (i) Again, \[6({{l}_{3}}-l)=9({{l}_{2}}-l)\] \[\Rightarrow \] \[2{{l}_{3}}+l=3{{l}_{2}}\] ...(ii) Solve Eqs. (i) and (ii), we obtain \[{{l}_{3}}=(2.5{{l}_{2}}-1.5{{l}_{1}})\]
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