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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. Young's modulus of copper Y _(C) =1.1 xx 10 ^(11) Nm ^(-2) Young's modulus of steel Y _(S) =2.0 xx 10 ^(11) Nm^(-2).

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let C copper and S for steel <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> radius of wire, <br/> `r = (3.0 xx 10 ^(-3))/( 2) = 1.5 xx 10 ^(-3) m` <br/> Length of copper wire `L _(C)=2.2 m` <br/> Length of steel wire `L _(S) = 1.6 m` <br/> Total increase in length for both wire, <br/> `Delta L _(C) + Delta L _(S) = 0.7 xx 10 ^(-3) m` <br/> Young.s modulus `=("Strees")/("Strain") = ("Stress")/((Delta L)/(L))` <br/> `therefore Y xx (Delta L )/(L)=` Stress [`because ` Strain `= (Delta)/(L)]` <br/> Stress on both <a href="https://interviewquestions.tuteehub.com/tag/thw-3217424" style="font-weight:bold;" target="_blank" title="Click to know more about THW">THW</a> wire is same, <br/> `therefore Y _(C) xx ( Delta L _(C))/( L _(C)) = Y _(S) xx ( Delta L _(S))/(L _(S))` <br/> `therefore (Delta L _(C))/( Delta L _(S)) = (Y_(S))/( Y _(C)) xx (L _(C))/( L _(S)) = (2.0 xx 10 ^(11))/(1.1 xx 10 ^(11)) xx (2.2)/(1.6)` <br/> `therefore (Delta L _(C))/(Delta L _(S))=2.5` <br/> Expanding `(Delta L _(C) + Delta L _(S))/( Delta L _(S)) = (2.5 + 1.0)/(1.0)` <br/> `therefore (0.7 xx 10 ^(-3))/( Delta L _(S))= 3.5` <br/> `therefore Delta L _(S) = (0.7 xx 10 ^(-3))/( 3.5) = 1/5 xx 10 ^(-3)` <br/> `= 2 xx 10 ^(-4) m` <br/> Now, `Delta L _(C) + Delta L _(S) =0.7 xx 10 ^(-3)` <br/> `therefore Delta L _(C) = 7 xx 10 ^(-4) -2 xx 10 ^(-4)` <br/> `= 5 xx 10 ^(-4) m` <br/> `therefore Y = ("Stress")/("Strain")` <br/> `therefore ` Stress `= Y xx` Strain <br/> `therefore F/A =Y xx` Strain <br/> `therefore F = Y xx` Strain ` xx A` <br/> `therefore F = Y _(S) xx (Delta L _(S))/(L _(S)) xx pi r ^(2)` <br/> `=2.0 xx 10 ^(11) xx(2.0 xx 10 ^(-4))/( 1.6) xx 3.14 xx (1.5 xx 10 ^(-3)) ^(2)=17.6625 xx 10 ^(1)` <br/> `therefore ~~1.8 xx 10 ^(2) <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>` where, F is appied load `F = Y_(C)xx (Delta L _(C))/(L _(C)) xx pi r ^(2)` From this formula load canbe calculate.<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XI_P2_C09_E02_004_S01.png" width="80%"/></body></html>


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