A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m^–3 (Young’s modules Y=2×10¹¹ Nm^–2). (b) If the yield strength of steel is 2.5×10⁸ Nm^–2, what is the maximum weight that can be hung at the lower end of the wire?

This is a long answer type question as classified in NCERT Exemplar

When a small element of length dx is considered at x from the load x=0

(a) letT(x) and T(x+dx) are tensions on the two cross sections a distance dx apart then

T(x+dx)+T(x)=dmg= $\mu$ dxg

dT= $\mu$ gx+C

at x=0 T(0)=mg

C=mg

T(x)= $\mu$ gx+Mg

Let length dx at x increases by dr then

Young’s modulus Y= stress/strain

$\frac{\frac{T\left(x\right)}{A}}{\frac{dr}{dx}}=Y$

$\frac{dr}{dx}=\frac{T\left(x\right)}{YA}$

r= $\frac{1}{YA}{\int }_0^L(\mu \mathrm{g}\mathrm{x}+\mathrm{M}\mathrm{g})dx$

= $\frac{1}{YA}\left[\frac{\mu g{x}^2}{2}+Mgx\right]$ ₀^L

r= $\frac{1}{2\times {10}^{11}\times \pi \times {10}^{-6}}\left[\frac{\pi \times 786\times {10}^{-3}\times 10\times 10}{2}+25\times 10\times 10\right]$

 

so r = 4 $\times {10}^{-3}m$

(b) tension will be maximum at x=L

T= $\mu gL$ +Mg=(m+M)g

The force = (yield strength) area= $250\times \pi N$

(m+M)g= 250 $\pi$

Mg $\approx 250\pi$

M= 25 $\pi \approx 75kg$

<div><div><p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>When a small element of length dx is considered at x from the load x=0</p><p><strong>(a)</strong> letT(x) and T(x+dx) are tensions on the two cross sections a distance dx apart then</p><p>T(x+dx)+T(x)=dmg= <span contenteditable="false"> <math> <mi>μ</mi> </math> </span>dxg</p><p>dT=<span contenteditable="false"> <math> <mi>μ</mi> </math> </span>gx+C</p><p>at x=0 T(0)=mg</p><p>C=mg</p><p>T(x)=<span contenteditable="false"> <math> <mi>μ</mi> </math> </span> gx+Mg</p><p>Let length dx at x increases by dr then</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745489095phpHgxCWB_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745489095phpHgxCWB.jpeg" alt="" width="122" height="188"></picture></div><div><p>Young’s modulus Y= stress/strain</p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>T</mi> <mfenced separators="|"> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfrac> <mo>=</mo> <mi>Y</mi> </math> </span></p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>T</mi> <mfenced separators="|"> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mi>Y</mi> <mi>A</mi> </mrow> </mrow> </mfrac> </math> </span></p><p>r=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>Y</mi> <mi>A</mi> </mrow> </mrow> </mfrac> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>L</mi> </mrow> </msubsup> <mrow> <mrow> <mo>(</mo> <mi> </mi> <mi>μ</mi> <mi mathvariant="normal">g</mi> <mi mathvariant="normal">x</mi> <mo>+</mo> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">g</mi> <mo>)</mo> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </math> </span></p><p>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>Y</mi> <mi>A</mi> </mrow> </mrow> </mfrac> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>μ</mi> <mi>g</mi> <msup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mi>M</mi> <mi>g</mi> <mi>x</mi> </mrow> </mrow> </mfenced> </math> </span> ₀^L</p><p>r=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>11</mn> </mrow> </mrow> </msup> <mo>×</mo> <mi>π</mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>π</mi> <mo>×</mo> <mn>786</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>10</mn> <mo>×</mo> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mn>25</mn> <mo>×</mo> <mn>10</mn> <mo>×</mo> <mn>10</mn> </mrow> </mrow> </mfenced> </math> </span></p><p>&nbsp;</p><p>so r = 4<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> </span></p><p><strong>(b) </strong>tension will be maximum at x=L</p><p>T=<span contenteditable="false"> <math> <mi>μ</mi> <mi>g</mi> <mi>L</mi> </math> </span>+Mg=(m+M)g</p><p>The force = (yield strength) area=<span contenteditable="false"> <math> <mn>250</mn> <mo>×</mo> <mi>π</mi> <mi>N</mi> </math> </span></p><p>(m+M)g= 250<span contenteditable="false"> <math> <mi>π</mi> </math> </span></p><p>Mg<span contenteditable="false"> <math> <mo>≈</mo> <mn>250</mn> <mi>π</mi> </math> </span></p><p>M= 25<span contenteditable="false"> <math> <mi>π</mi> <mo>≈</mo> <mn>75</mn> <mi>k</mi> <mi>g</mi> </math> </span></p></div></div></div></div>

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