A `20kg` load is suspended from the lower end of a wire `10cm` long and `1mm^(2)` in cross-sectional area. The upper half of the wire is made of iron and the lower half with aluminium. The total elongation in the wire is `(Y_("iron") = 20xx10^(10) N//m^(2), Y_(Al) = 7xx10^(10) N//m^(2))`A. `18.9xx10^(-3)m`B. `17.8xx10^(-3)m`C. `1.78xx10^(-3)m`D. `1.89xx10^(-4)m`

A `20kg` load is suspended from the lower end of a wire `10cm` long an

1.	A `20kg` load is suspended from the lower end of a wire `10cm` long and `1mm^(2)` in cross-sectional area. The upper half of the wire is made of iron and the lower half with aluminium. The total elongation in the wire is `(Y_("iron") = 20xx10^(10) N//m^(2), Y_(Al) = 7xx10^(10) N//m^(2))`A. `18.9xx10^(-3)m`B. `17.8xx10^(-3)m`C. `1.78xx10^(-3)m`D. `1.89xx10^(-4)m`
Answer» Correct Answer - D `e = e_(1) + e_(2), e = (Fl)/(AY), (e_(1))/(e_(2)) = (Y_(2))/(Y_(1))`

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