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A force of $10 \mathrm{~N}$ is required to break a wire of radius $1 \mathrm{~mm}$. The force required to break the wire of same material, but radius $3 \mathrm{~mm}$ will be
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The correct answer is:
$90 \mathrm{~N}$
We know that, the required force depends upon the cross-sectional area. $F=$ Stress $\times$ Area
So, $\frac{F_{1}}{F_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}$
$\Rightarrow F_{2}=\frac{3^{2}}{1^{2}} \times 10=90 \mathrm{~N}$
So, $\frac{F_{1}}{F_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}$
$\Rightarrow F_{2}=\frac{3^{2}}{1^{2}} \times 10=90 \mathrm{~N}$
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