Question

The elongation of a steel wire stretched by a force is '$e$'. If a wire of the same material of double the length and half the diameter is subjected to double the force, its elongation will be

• A. $16e$
• B. $4e$
• C. $\left(\frac{1}{4}\right)e$
• D. $\left(\frac{1}{16}\right)e$

Answer 1

The correct option is A $16e$

$Y=\frac{F/\pi (d/2{)}^2}{△l/l}$

$△l=\frac{Fl}{\pi (\frac{d}{2}{)}^{2}Y}$

Now, according to problem

$e=\frac{Fl}{\pi (\frac{d}{2}{)}^{2}Y}$

$Y=\frac{F/A}{△l_1/l}$

$△l_1=\frac{Fl}{YA}$

$=\frac{2F2l}{Y\pi (\frac{d}{2}{)}^2\frac{1}{4}}$

$=16\frac{Fl}{Y\pi (\frac{d}{2}{)}^2}$

$=16e$

So, new elongation is 16e