Question

A steel wire of diameter $2mm$ has a breaking strength of $4\times 10^{5}N$. What is the breaking force of similar, steel wire of diameter $1.5mm$ ?

• A. $2.3\times 10^{5}N$
• B. $2.6\times 10^{5}N$
• C. $3\times 10^{5}N$
• D. $1.5\times 10^{5}N$

Answer 1

Answer: (A)

We know
$\frac{\text{ Force }\times \text{ Length }}{\text{ Area }\times \text{ young’s modulus }}=$ elongation $\left\{\frac{FL}{Ay}=\Delta x\right\}$
$\Rightarrow F=\left(\frac{\Delta x\cdot y}{L}\right)A$
$F=\left(\frac{\Delta x\cdot y}{L}\cdot \frac{\pi }{4}\right)d^2$
We can say $F\propto d^2$
So we can use
$\frac{F_1}{F_2}=\frac{d_1^2}{d_2^2}$
$F_1=4\times 10^{5}N$
$d_1=2mm$
$F_2=?$
$d_2=1.5mm$
Substituting values
$\frac{4\times 10^5}{F_2}=\frac{(2{)}^2}{(1.5{)}^2}$
$F_2=2.3\times 10^{5}N$