MHT CET · Physics · Mechanical Properties of Solids
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
- A \(\frac{10}{9} \mathrm{~N}\)
- B \(\frac{10}{3} \mathrm{~N}\)
- C \(90 \mathrm{~N}\)
- D \(30 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(90 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
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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