A metal wire with circular cross section and length one metre is pulled with tensile force of $1000…

A metal wire with circular cross section and length one metre is pulled with tensile force of $1000 \mathrm{~N}$ on each side. For the wire to be stretched not more than $0.25 \mathrm{~cm}$, the minimum diameter of the wire required is (Young's modulus of the metal $=10^{11} \mathrm{~Pa}$, take $\sqrt{\pi}=1.77$ )
  1. $1.13 \mathrm{~mm}$
  2. $2.26 \mathrm{~mm}$
  3. $4.12 \mathrm{~mm}$
  4. $3.1 \mathrm{~mm}$

Solution

Tensile force, $\mathrm{F}=1000 \mathrm{~N}$ Length, $L=1 \mathrm{~m}$ Change in length, $\Delta \mathrm{L}=0.25 \times 10^{-2} \mathrm{~m}$ Young modulus of the metal, $Y=10^{11} \mathrm{pa}$ $ \begin{aligned} & \mathrm{Y}=\frac{\text { stress }}{\text { Strain }}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}} \\ & \mathrm{r}^2=\frac{\mathrm{FL}}{\mathrm{Y} \Delta \mathrm{L}} \\ & =\frac{1000 \times 1}{10^{11} \times 0.25 \times 10^{-2} \times 3.14}=\frac{1}{25 \pi \times 10^4} \end{aligned} $ $ r=\frac{1}{5 \times 10^2 \times 1.77} $ Diameter, $\mathrm{d}=2 \mathrm{r}=\frac{2}{5 \times 1.77} \times 10^{-2}=2.26 \mathrm{~mm}$

Asked in: AP EAMCET 2023 (19 May Shift 1)

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