Verified Solution • 2025–26 Session moderate QJEEM24Ph143f: JEE Main 2024 Physics - Class 11 JEEM24Ph143f Solution: Two metallic wires and have same volume and are made up of same … Two metallic wires PPP and QQQ have same volume and are made up of same material. If their area of cross sections are in the ratio 4:14: 14:1 and force F1F_1F1 is applied to PPP , an extension of Δl\Delta lΔl is produced. The force which is required to produce same extension in QQQ is F2\mathrm{F}_2F2 . The value of F1F2\frac{F_1}{F_2}F2F1 is _________. Step-by-Step Verified Solution Y= Stress Strain =F/AΔℓ/ℓ=FℓAΔℓ\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}Y= Strain Stress =Δℓ/ℓF/A=AΔℓFℓ amp;Δℓ=FℓAYamp;V=Aℓ⇒ℓ=VAamp;Δℓ=FVA2Y\begin{aligned}& \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\& \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \\& \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}}\end{aligned}amp;Δℓ=AYFℓamp;V=Aℓ⇒ℓ=AVamp;Δℓ=A2YFV Y &VY ~\& VY &V is same for both the wires amp;Δℓ∝FA2amp;Δℓ1Δℓ2=F1 A12×A22 F2amp;Δℓ1=Δℓ2amp; F1 A22=F2 A12amp;F1 F2=A12 A22=(41)2=16\begin{aligned}& \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \\& \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \\& \Delta \ell_1=\Delta \ell_2 \\& \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \\& \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16\end{aligned}amp;Δℓ∝A2Famp;Δℓ2Δℓ1= A12F1× F2A22amp;Δℓ1=Δℓ2amp; F1 A22=F2 A12amp; F2F1= A22A12=(14)2=16 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring