A steel wire of diameter $2\ mm$ has a breaking strength of $4 \times 10^5 N$.the breaking force $......... \times 10^5 N$ of similar steel wire of diameter $1.5\ mm ?$
Medium
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We know
$\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { young's modulus }}=$ elongation $\quad\left\{\frac{F L}{A y}=\Delta x\right\}$
$\Rightarrow F=\left(\frac{\Delta x \cdot y}{L}\right) A$
$F=\left(\frac{\Delta x \cdot y}{L} \cdot \frac{\pi}{4}\right) d^2$
We can say $F \propto d^2$
So we can use
$\frac{F_1}{F_2}=\frac{d_1^2}{d_2^2}$ $\left\{\begin{array}{l}F_1=4 \times 10^5 N \\ d_1=2 mm \\ F_2=? \\ d_2=1.5 mm \end{array}\right\}$
Substituting values
$\frac{4 \times 10^5}{F_2}=\frac{(2)^2}{(1.5)^2}$
$F_2=2.3 \times 10^5 \,N$
$\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { young's modulus }}=$ elongation $\quad\left\{\frac{F L}{A y}=\Delta x\right\}$
$\Rightarrow F=\left(\frac{\Delta x \cdot y}{L}\right) A$
$F=\left(\frac{\Delta x \cdot y}{L} \cdot \frac{\pi}{4}\right) d^2$
We can say $F \propto d^2$
So we can use
$\frac{F_1}{F_2}=\frac{d_1^2}{d_2^2}$ $\left\{\begin{array}{l}F_1=4 \times 10^5 N \\ d_1=2 mm \\ F_2=? \\ d_2=1.5 mm \end{array}\right\}$
Substituting values
$\frac{4 \times 10^5}{F_2}=\frac{(2)^2}{(1.5)^2}$
$F_2=2.3 \times 10^5 \,N$
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