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When a mass M hangs from a spring of length l, it stretches the spring by a distance x. Now the spring is cut in two parts of length l//3"and" 2l//3, and two springs thus formed are connected to a straight rod of mass M which is horizontal in the configuration shown in figure . Find the stretch in each of the spring

Answer» <html><body><p></p>Solution :AS it given that the mass M <a href="https://interviewquestions.tuteehub.com/tag/stretches-1229372" style="font-weight:bold;" target="_blank" title="Click to know more about STRETCHES">STRETCHES</a> the original spring by a distance x,<br/> we have `kx=Mg "(or)"k=(Mg)/x`….(1)<br/> The new force <a href="https://interviewquestions.tuteehub.com/tag/constants-20556" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANTS">CONSTANTS</a> of the two springs can be given by using equation <br/> `k_(1)=(k_(eq)l)/l_(1) "and" k_(2)=(k_(eq)l)/l_(2)`as,`k_(1)=3k "and" k_(2)=(3k)/2`<br/>Let we take the <a href="https://interviewquestions.tuteehub.com/tag/stretch-1229345" style="font-weight:bold;" target="_blank" title="Click to know more about STRETCH">STRETCH</a> in the two springs be `x_(1)` and <br/>`x_(2)` we have for the equilibrium of the rod <br/> `k_(1)x_(1)+k_(2)x_(2)=Mg`(or)`2kx_(1)+(3k)/2x_(2)=Mg`<br/>From equation (1), we have `x_(1)+x_(2)/2=x/3`...(2)<br/>As the rod is horizontal and in <a href="https://interviewquestions.tuteehub.com/tag/static-1226026" style="font-weight:bold;" target="_blank" title="Click to know more about STATIC">STATIC</a> equlibrium , we have net torque acting on the rod about any point on it must be <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> . Thus we have torque on it above end A are <br/> `k_(2)x_(2)L=Mg l/2`<br/>(or)`x_(2)=(Mg)/(2k_(2))=(Mg)/(3k)=x/3`<br/> Using this value of `x_(2)` in equation (2)<br/> we have `x_(1)=x/6`<br/> * This can also be directly obtained by using torque zero about point B on rod as <br/> `k_(1)x_(1)L=Mg L/2`<br/> (or) `x_(1)=(Mg)/(2k_(1))=(Mg)/(6k)=x/6`<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C06_SLV_011_S01.png" width="80%"/></body></html>


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