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Mechanical Properties of Solids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Solids5 Marks
Question
A rubber string 10 m long is suspended from a rigid support at its one end. Calculate the extension in the string due to its own weight. The density of rubber is $1.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ and Young's modulus for the rubber is $5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$. The breaking stress for a metal is $7.8 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. Calculate the maximum length of the wire made of this metal which may be suspended without breaking. The density of metal $=7.8 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$.
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Answer


$\text{l}=10\text{m},\rho=15\times10^3\text{kg}\ \text{m}^3$
$\text{Y}=5\times10^6\text{N}\text{m}^2$ We know, $\text{Y}=\frac{\text{F.l}}{\text{A}\Delta\text{l}}$ Efficient force = Mg Consider a small length dy at a distance y from free end. The length above this, (l - y) wil experience a force of, $\text{F}_\text{dy}=\frac{\text{M}}{\text{l}}(\text{dy}).\text{gdy}$
$\therefore$ Extension dI $=\frac{\text{F}.\text{l}}{\text{AY}}$
$\Rightarrow\text{dl}=\frac{(\text{l}-\text{y})}{\text{AY}}.\frac{\text{M}}{\text{l}}\text{gdy}$
$=\frac{\text{Mg}}{\text{lAY}}(\text{l}-\text{y})\text{dy}$ Net extension due to its own weight} $=\int\text{dl}$
$=\frac{\text{Mg}}{\text{AYl}}\int(\text{l}-\text{y})$
$=\frac{\text{Mg}}{\text{lAY}}\Big[\text{ly}-\frac{\text{y}^2}{2}\Big]^\text{l}_0=\frac{\text{Mgl}}{2\text{AY}}$ Net extension $=\frac{\text{Mgl}}{2}=\frac{\text{Mgl}^2}{2\text{YV}}=\frac{\rho\text{gl}^2}{2\text{Y}}$ Extcnrion of rubbcer string, $=\frac{1.5\times10^3\times10\times10^2}{2\times5\times10^6}=0.15\text{m}$ Breaking stress for a metal = $7.8 \times 10^9N/ m^2$ Density = $7.8 \times 10^3kg/ m^3$. Stress $=\frac{\text{Force}}{\text{Area}}=\frac{\text{Mg}}{\text{A}}=\frac{\text{Mgl}}{\text{Al}}=\frac{\text{Mgl}}{\text{Volume}}\rho\text{gl}$ If $\rho\text{gl}>$ Breaking stress, the wire will break.
$\therefore\text{l}\leq\frac{7.8\times10^9}{\rho\text{g}},\text{l}\leq\frac{7.8\times10^9}{7.8\times10^3\times10}$ i.e., $\text{l}\leq10^5\text{m}$ Maximum length of wire = $10^5m$.

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