Question

A rod AD consisting of three segments AB, BC and CD are joined together is A. The length of the three segment are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly ${10}^{-4}{m}^2$.A weight of 10 kg is hung from D. Calculate the displacement of point D. If [$Y_{AB}=2.5\times {10}^{10}N/m^2$ , $Y_{BC}=4\times {10}^{10}N/m^2$ and $Y_{CD}=1\times {10}^{10}N/m^2$. Neglect the weight of the rod. Take g = 10 $m/s^2$]

• A. 4 $\times {10}^{-6}$ m
• B. 9 $\times {10}^{-6}$ m
• C. 24 $\times {10}^{-6}$ m
• D. 18 $\times {10}^{-6}$ m

Answer 1

The correct option is C 24 $\times {10}^{-6}$ m

$\begin{array}{c}\text{Eiven,}\\ \text{cross section of wire}\left(4\right)={10}^{-4}{m}^2\\ Y_{AB}\left(\text{young modulus}\right)=2.5\times {10}^{10}N/m^2\\ Y_{BC}=4\times {10}^{10}N/m^2\\ Y_{CD}=1\times {10}^{10}N/m^2\\ g=10m/s^2\end{array}$

$\begin{array}{cc}& Y=\frac{Fl}{A\Delta l}\\ & \Delta L=\frac{FL}{AY}\\ \Delta L_{AB}& =\frac{10\times 10}{{10}^{-4}}\times \frac{L_{AB}}{Y_{AB}}\\ & =\frac{{10}^6\times 0.1}{2.5\times {10}^{10}}\\ & =4\times {10}^{-6}m\end{array}$

$\begin{array}{cc}\Delta L_{BC}& ={10}^6\times \frac{L_{BC}}{Y_{BC}}\\ & ={10}^6\times \frac{0.2}{4\times {10}^{10}}\\ & =5\times {10}^{-6m}\\ \Delta L_{CD}& ={10}^6\times \frac{L_{CP}}{Y_{CD}}\\ & ={10}^6\times \frac{0.15}{{10}^{10}}\\ & =15\times {10}^{-6}m\end{array}$

$\therefore$displacement of $D=\Delta L_{AB}+\Delta L_{BC}+{\Delta }_{CD}$

$=(4\times {10}^{-6})+(5\times {10}^{-6})+(15\times {10}^{-6})$

$d=24\times {10}^{-6}$