Question

A wire can be broken by $400kg.wt$. The load required to break the wire of double the thickness of the same material will be

• A. $800kg.wt$
• B. $1600kg.wt.$
• C. $3200kg.wt$
• D. $6400kg.wt$

Answer 1

Answer: (B)

$1600kg.wt.$

$F=stress\times area$
$F\alpha (thickness{)}^2$
$\frac{F_1}{F_2}={\left(\frac{thicknes{s}_1}{thicknes{s}_2}\right)}^2=\left(\frac{1}{2}\right)=\frac{1}{4}$
$F_2=4F_1$
$=4\left(400\right)$
$=1600kg.wt.$