AP EAMCET · PHYSICS · Mechanical Properties of Solids
A metal wire with circular cross section and length one metre is pulled with tensile force of \(1000 \mathrm{~N}\) on each side. For the wire to be stretched not more than \(0.25 \mathrm{~cm}\), the minimum diameter of the wire required is (Young's modulus of the metal \(=10^{11} \mathrm{~Pa}\), take \(\sqrt{\pi}=1.77\) )
- A \(1.13 \mathrm{~mm}\)
- B \(2.26 \mathrm{~mm}\)
- C \(4.12 \mathrm{~mm}\)
- D \(3.1 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(2.26 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Tensile force, \(\mathrm{F}=1000 \mathrm{~N}\) Length, \(L=1 \mathrm{~m}\) Change in length, \(\Delta \mathrm{L}=0.25 \times 10^{-2} \mathrm{~m}\) Young modulus of the metal, \(Y=10^{11} \mathrm{pa}\)…
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