Mechanical Properties of Solids

JEE Main level, numerical answer, medium

32 of 118

Question 32 of 118

The length of a wire becomes l1{l}_{1} and l2{l}_{2} when 100N100N and 120N120N tension are applied respectively. If 10l2=11l110{l}_{2}=11{l}_{1}, then the natural length of wire will be 1xl1\dfrac{1}{x}{l}_{1}. Here the value of xx is

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Answer: 2

Young's modulus is given by Y=FA∆llY=\dfrac{\dfrac{F}{A}}{\dfrac{∆l}{l}}
⇒FA=Y(Δll)⇒Δl=FlAY\Rightarrow \dfrac{F}{A}=Y\left(\dfrac{\Delta l}{l}\right) \Rightarrow \Delta l=\dfrac{Fl}{AY}
⇒l+Δl=l1⇒(F1LAY+L)=L(F1AY+1)=l1...(i)\Rightarrow l+\Delta l={l}_{1} \Rightarrow \left(\dfrac{{F}_{1}L}{AY}+L\right)=L\left(\dfrac{{F}_{1}}{AY}+1\right)={l}_{1}...(i)
Similarly,
l2=(F2AY+1)L{l}_{2}=\left(\dfrac{{F}_{2}}{AY}+1\right)L
⇒10l2=11l1\Rightarrow 10{l}_{2}=11{l}_{1}
It is given that 10l1=11l210{l}_{1}=11{l}_{2}
⇒10×L[F2AY+1]=11L[F1AY+1]\Rightarrow 10\times L\left[\dfrac{{F}_{2}}{AY}+1\right]=11L\left[\dfrac{{F}_{1}}{AY}+1\right]
⇒10F2AY+10=11F1AY+11\Rightarrow \dfrac{10{F}_{2}}{AY}+10=\dfrac{11{F}_{1}}{AY}+11
Substituting F1=100N,F2=120N{F}_{1}=100N,{F}_{2}=120N
⇒1200AY−1100AY=1⇒100AY=1\Rightarrow \dfrac{1200}{AY}-\dfrac{1100}{AY}=1 \Rightarrow \dfrac{100}{AY}=1
⇒1AY=1100\Rightarrow \dfrac{1}{AY}=\dfrac{1}{100}
Using equation (i),
L=l1(F1AY+1)=l1100100+1=l12L=\dfrac{{l}_{1}}{\left(\dfrac{{F}_{1}}{AY}+1\right)}=\dfrac{{l}_{1}}{\dfrac{100}{100}+1}=\dfrac{{l}_{1}}{2}

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