Mechanical Properties of Solids

JEE Main level, numerical answer, medium

29 of 118

Question 29 of 118

Two metallic wires PP and QQ have same volume and are made up of same material. If their area of cross sections are in the ratio 4:14:1 and force F1{F}_{1} is applied to PP, an extension of ∆l∆l is produced. The force which is required to produce same extension in QQ is F2{F}_{2}. The value of F1F2\dfrac{{F}_{1}}{{F}_{2}} is ______.

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Answer: 16

The formula for Young's modulus is given by
Y=FA∆ll=FlA∆l\begin{matrix}Y & = & \dfrac{\dfrac{F}{A}}{\dfrac{∆l}{l}} \\ & = & \dfrac{Fl}{A∆l}\end{matrix}
From above equation, it follows that
Δl=FlAY...(1)\Delta l=\dfrac{Fl}{AY}...\left(1\right)
Again, volume of the wire is given by
V=Al⇒l=VAV=Al \Rightarrow l=\dfrac{V}{A}
Hence, from equation (1),
∆l=FVA2Y...(2)∆l=\dfrac{FV}{{A}^{2}Y}...\left(2\right)
As, YY and VV is the same for both the wires,
∆l∝FA2∆l\propto \dfrac{F}{{A}^{2}}
Hence, for two wires, it can be written that
∆l1∆l2=F1A12×A22F2...(3)\dfrac{∆{l}_{1}}{∆{l}_{2}}=\dfrac{{F}_{1}}{{A}_{1}^{2}}\times \dfrac{{A}_{2}^{2}}{{F}_{2}}...\left(3\right)
As ∆l1=∆l2∆{l}_{1}=∆{l}_{2}, from equation (3), it follows that
F1A22=F2A12⇒F1F2=A12A22=(41)2=16{F}_{1}{A}_{2}^{2}={F}_{2}{A}_{1}^{2} \Rightarrow \dfrac{{F}_{1}}{{F}_{2}}=\dfrac{{A}_{1}^{2}}{{A}_{2}^{2}} ={\left(\dfrac{4}{1}\right)}^{2} =16

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