A wire can be broken by applying a load of $20\, kg$ weight. The force required to break the wire of twice the diameter is .......... $kg\, wt$
Medium
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$3.$ Breaking stress $=\frac{\mathrm{F}}{\mathrm{A}}=$ constant $\frac{\mathrm{F}_{1}}{\mathrm{A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{A}_{2}} \Rightarrow \mathrm{F}_{2}=\frac{\mathrm{A}_{2}}{\mathrm{A}_{1}} \times \mathrm{F}_{1}=\frac{\pi \mathrm{r}_{2}^{2}}{\pi \mathrm{r}_{1}^{2}} \mathrm{F}_{1} \Rightarrow \mathrm{F}_{2}=\frac{(2)^{2}}{(1)^{2}} \mathrm{F}_{1}$
$\Rightarrow \mathrm{F}_{2}=4 \mathrm{F}_{1}=80 \mathrm{kg}-\mathrm{wt}$
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