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QJEEM23Pha661: JEE Main 2023 Physics - Class 11 JEEM23Pha661 Solution: A 100 m long wire having cross-sectional area and Young's…

A 100 m long wire having cross-sectional area 6.25×10−4 m2\mathrm{6.25\times10^{-4}~m^2} and Young's modulus is 1010 Nm−2\mathrm{10^{10}~Nm^{-2}} is subjected to a load of 250 N, then the elongation in the wire will be :

Multiple Choice Options

  • A 6.25×10−6 m\mathrm{6.25\times10^{-6}~m}
  • B 4×10−3 m\mathrm{4\times10^{-3}~m}
  • C 4×10−4 m\mathrm{4\times10^{-4}~m}
  • D 6.25×10−3 m\mathrm{6.25\times10^{-3}~m}

Step-by-Step Verified Solution

Elongation in wire δ=FℓAY\delta=\frac{\mathrm{F} \ell}{\mathrm{AY}}

amp;δ=250×1006.25×10−4×1010amp;δ=4×10−3 m\begin{aligned}& \delta=\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} \\\\& \delta=4 \times 10^{-3} \mathrm{~m}\end{aligned}

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