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 planck is $S=0.5m^2$, the change in length of the plank is $x\times {10}^{-9}m$:

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

Answer 1

Answer: (A)

$1.0$

Consider an element of thickness $dx$
Change in the length of the element is $dl=\frac{T}{S}\frac{dx}{Y}$ and
$T=F_1-(F_1-F_2)\frac{x}{l}$
${\int }_0^{\Delta l}dl={\int }_0^l\frac{F_1-(F_1-F_2)\frac{x}{l}}{SY}=\frac{F_1-\frac{F_1-F_2}{2}}{SY}$

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