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Question: Answered & Verified by Expert
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
PhysicsMechanical Properties of SolidsMHT CETMHT CET 2020 (13 Oct Shift 2)
Options:
  • A $\frac{10}{9} \mathrm{~N}$
  • B $\frac{10}{3} \mathrm{~N}$
  • C $90 \mathrm{~N}$
  • D $30 \mathrm{~N}$
Solution:
1384 Upvotes Verified Answer
The correct answer is: $90 \mathrm{~N}$
We know that, the required force depends upon the cross-sectional area. $F=$ Stress $\times$ Area
So, $\frac{F_{1}}{F_{2}}=\frac{R_{1}^{2}}{R_{2}^{2}}$
$\Rightarrow F_{2}=\frac{3^{2}}{1^{2}} \times 10=90 \mathrm{~N}$

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