Question

Two opposite forces $F_{1}=120 \, N$ and $F_{2}=80 \, N$ act on an elastic plank of modulus of elasticity $Y=2\times10^{11} N /m^{2} $ and length $l=1 m$ placed over a smooth horizontal surface. The cross-sectional area of the plank is $S=0.5$ $m^{2}$, The change in length of the plank is $x \times10^{-9} m$. The value of $x$ is

• A. $1.0$
• B. $1.5$
• C. $1.4$
• D. $1.1$

Answer 1

Answer: (A)

Consider an element of thickness $dx$.
Change in the length of the element is $dl$ $=\frac{T}{S}$ $\frac{dx}{Y}$ and
$T=F_{1}-\left(F_{1}-F_{2}\right)$ $\frac{x}{l}$
$\int\limits_{0}^{\Delta l} dl=\int\limits_{0}^{l}$ $\frac{F_{1}-\frac{\left(F_{1}-F_{2}\right)dx}{l} dx}{SY}$

$\Delta l$ $=\frac{\left(\left(F_{1}+F_{2}\right)\right)l}{2SY}$ $=\frac{200\times1}{2\times0.5\times2\times10^{11}}$ $=1\times10^{-9}m$
$\therefore $ $\quad $ $x=1$