Question

A force of ${10}^3$ N stretches the length of a hanging wire by $1$ millimeter. The force required to stretch a wire of same material and length but having four times the diameter by $1$ millimeter is :

• A. $4\times {10}^3$ N
• B. $16\times {10}^3$ N
• C. $\frac{1}{4}\times {10}^3$ N
• D. $\frac{1}{16}\times {10}^3$ N

Answer 1

The correct option is B $16\times {10}^3$ N

$Y=\frac{F/A}{\Delta l/l}$ ; $Y$ & $\frac{\Delta l}{l}=constant$

$F\propto A\propto r^2$ $\frac{F_2}{F_1}={\left(\frac{r_2}{r_1}\right)}^2={\left(\frac{4r}{r}\right)}^2=16$

$F_2=16\times {10}^{3}N$