Question

The breaking strength of a rod diameter $2cm$ is $2\times {10}^{-5}N$. Then that for rod of same material but diameter $1cm$ will be

• A. $2\times {10}^{5}N$
• B. $1\times {10}^{5}N$
• C. $8\times {10}^{5}N$
• D. $0.5\times {10}^{5}N$

Answer 1

The correct option is C $8\times {10}^{5}N$

Breaking force $\propto$ area of cross-section of wire.
$\Rightarrow$ $F\propto r^2$
$\Rightarrow$ $\frac{F_2}{F_1}={\left(\frac{r_2}{r_1}\right)}^2={\left(\frac{4}{2}\right)}^2=4$
$\therefore$ $F_2=4\times F_1=4\times 2\times {10}^5$
$\Rightarrow$ $F_2=8\times {10}^{5}N$