Question

A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a a pulley and two weights W each are hung at the two ends, he elongation of he wire will be
(a) 0.5 m
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm

Answer 1

Correct option: (b) 1.0 mm

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{Young}\text{'}\mathrm{s}\mathrm{modulus}\mathrm{of}\mathrm{the}\mathrm{material}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{be}\mathrm{Y}. \\ \mathrm{Force}=\mathrm{Weight}=\mathrm{W}\left(\mathrm{given}\right) \\ \mathrm{Let}C.S.A.\hspace{0.17em}=\mathrm{A} \\ \mathrm{x}=1\mathrm{mm}=\mathrm{Elongation}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{case} \\ \mathrm{Length}=\mathrm{L} \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}}=\frac{\mathrm{WL}}{\mathrm{Ax}} \\ \mathrm{Let}\mathrm{y}\mathrm{be}\mathrm{the}\mathrm{elongation}\mathrm{on}\mathrm{one}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{when}\mathrm{put}\mathrm{in}\mathrm{a}\mathrm{pulley}. \\ \mathrm{When}\mathrm{put}\mathrm{in}\mathrm{a}\mathrm{pulley},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{on}\mathrm{each}\mathrm{side}=\frac{\mathrm{L}}{2} \\ \frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{\mathrm{L}}{2}}$}\right.}}=\mathrm{Y} \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{\mathrm{L}}{2}}$}\right.}}=\frac{\mathrm{WL}}{\mathrm{Ax}} \\ \Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2} \\ \mathrm{Total}\mathrm{elongation}\mathrm{in}\mathrm{the}\mathrm{wire}=2\mathrm{y}=2\left(\frac{\mathrm{x}}{2}\right)=\mathrm{x}=1\mathrm{mm} \end{gathered}$$