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In the question number 46, if the mass M is hung the free end of the wire, then the extension produc in the wire is

Answer» <html><body><p>`(mugL^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) + MgL)/(2YA)`<br/>`(2mugL^(2) + MgL)/(YA)`<br/>`(mugL^(2) + 2MgL)/(2YA)`<br/>`(mugL^(2) + MgL)/(YA)`</p>Solution :Consider a <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> element oflengthdx at a distance x fromthe load as shown in thewire. <br/><a href="https://interviewquestions.tuteehub.com/tag/tensionin-3102866" style="font-weight:bold;" target="_blank" title="Click to know more about TENSIONIN">TENSIONIN</a> the wire at a distance x from <a href="https://interviewquestions.tuteehub.com/tag/thelower-3191402" style="font-weight:bold;" target="_blank" title="Click to know more about THELOWER">THELOWER</a> end is`T(x) = mugx + Mg` <br/>Let dl be increase in lengthof the elemenet <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XI_C09_E01_047_S01.png" width="80%"/><br/> Then `Y= (T(x)//A)/(dl//dx)` <br/>`dl =(T(x)dx)/(YA) = ((mugx+Mg)/(YA))dx ""("<a href="https://interviewquestions.tuteehub.com/tag/using-7379753" style="font-weight:bold;" target="_blank" title="Click to know more about USING">USING</a> (i)")` <br/>`therefore` Totalextensionproduced in thewire is `l = underset(0)overset(L)int ((mugx+Mg)/(YA))dx = (1)/(Y) [(mugx)/(2) +Mgx]_(0)^(L) = (1)/(YA) [(mugL^(2))/(L)+Mgl] =(mugx^(2)+2MgL)/(2YA)`</body></html>


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