Question

One end of a wire $2m$ long and $0.2cm$${}^2$ in cross section is fixed to a ceiling and a load of $4.8kg$ is attached to its free end. The elongation in the wire in mm is (if y $=$ 2.0 x 10${}^{11}$ N.m${}^{-2}$, g $=$10ms${}^{-2}$)

• A. $2.4$ x $10$${}^{-5}$
• B. $2.4$
• C. $0.024$
• D. $0.0024$

Answer 1

The correct option is C $0.024$

$Y=\frac{Fl}{A\Delta l}$
$\Delta l=\frac{Fl}{\Delta Y}$
$=\frac{45\times 2}{0.2\times {10}^{-4}\times 2\times {10}^{11}}$
$=24\times {10}^{-6}m$
$=24\times {10}^{-3}mm$
$=0.024mm$