Question

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

• 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 $\left\{\frac{F L}{A y}=\Delta x\right\}$
$\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}=2 \,mm $
$F_{2}=?$
$d_{2}=1.5\, mm$
Substituting values
$\frac{4 \times 10^{5}}{F_{2}}=\frac{(2)^{2}}{(1.5)^{2}}$
$F_{2}=2.3 \times 10^{5} N$