A wire of length 100 cm and area of cross-section $2 \mathrm{~mm}^2$ is stretched by two forces of each 440…

A wire of length 100 cm and area of cross-section $2 \mathrm{~mm}^2$ is stretched by two forces of each 440 N applied at the ends of the wire in opposite directions along the length of the wire. If the elongation of the wire is 2 mm , the Young's modulus of the material of the wire is
  1. $4.4 \times 10^{11} \mathrm{Nm}^{-2}$
  2. $1.1 \times 10^{11} \mathrm{Nm}^{-2}$
  3. $2.2 \times 10^{11} \mathrm{Nm}^{-2}$
  4. $3.3 \times 10^{11} \mathrm{Nm}^{-2}$

Solution

$1=100 \mathrm{~cm}=100 \times 10^{-2} \mathrm{~m}, \mathrm{~A}=2 \mathrm{~mm}^2=2 \times 10^{-6} \mathrm{~m}^2$ $\mathrm{F}=440 \mathrm{~N}, \delta \mathrm{l}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$ $\therefore$ Young's modulus of the wire is $\mathrm{Y}=\frac{\mathrm{Fl}}{\mathrm{~A} . \delta \mathrm{l}}=\frac{440 \times 100 \times 10^{-2}}{2 \times 10^{-6} \times 2 \times 10^{-3}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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