Question

Two opposite forces $F_1=120N$ and $F_2=80N$ act on an elastic plank of modulus of elasticity $Y=2\times 10^{11}N/m^2$ and length $l=1m$ 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 $x\times 10^{-9}m$. 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
$T=F_1-\left(F_1-F_2\right)$ $\frac{x}{l}$
$\underset{0}{\overset{\Delta l}{\int }}dl=\underset{0}{\overset{l}{\int }}$ $\frac{F_1-\frac{\left(F_1-F_2\right)dx}{l}dx}{SY}$

$\Delta l$ $=\frac{\left(\left(F_1+F_2\right)\right)l}{2SY}$ $=\frac{200\times 1}{2\times 0.5\times 2\times 10^{11}}$ $=1\times 10^{-9}m$
$\therefore$ $x=1$