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A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load , the net elongation is found to be 0.70 mm. Obtain the load applied.

Answer» <html><body><p></p>Solution :The copper and steel wires are under same tensile <a href="https://interviewquestions.tuteehub.com/tag/stress-1229213" style="font-weight:bold;" target="_blank" title="Click to know more about STRESS">STRESS</a> because they have the same <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> (equal to the load W) and the same area of cross- section A. We have stress = strain `xx` Young.s modulus. Therefore <br/> `W//A=Y_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> )xx(DeltaL_(C )//L_(C ))=Y_(S)xx(DeltaL_(S)//L_(S))` <br/> `DeltaL_(c )//DeltaL_(s)=(Y_(s)//Y_(c ))xx(L_(c )//L_(s))` <br/> `=(2.0xx10^(11)//1.1xx10^(11))xx(2.2//1.6)=2.5"".....(1)` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> elongation is given to be <br/> `DeltaL_(c )+DeltaL_(s)=7.0xx10^(-4)m""....(2)` <br/> Solving the above equations (1) &amp; (2). <br/> `DeltaL_(c )=5.0xx10^(-4)m and DeltaL_(s)=2.0xx10^(-4)m`. <br/> Therefore`W=(AxxY_(c )xxDeltaL_(c ))//L_(c )` <br/> `=pi(1.5xx10^(-3))^(2)<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>[(5.0xx10^(-4)xx1.1xx10^(11))//2.2]` <br/> `=1.8xx10^(2)N`</body></html>


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