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JEE Mains · Physics · STD 11- Mechanical properties of solids
  • JEE Mains 2023
  • Medium
  • MCQ

A force is applied to a steel wire ' \(A\) ', rigidly clamped at one end. As a result elongation in the wire is \(0.2\,mm\). If same force is applied to another steel wire ' \(B\) ' of double the length and a diameter \(2.4\) times that of the wire ' \(A\) ', the elongation in the wire ' \(B\) ' will be \(............\times 10^{-2}\,mm\) (wires having uniform circular cross sections)

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Answer & Solution

Correct Answer

(D) \(6.9\)

Step-by-step Solution

Detailed explanation

\(Y =\frac{ F / A }{\frac{\Delta \ell}{\ell}}\)

\(\Rightarrow F =\frac{ YA }{\ell} \Delta \ell\)

\(\left(\frac{ A \Delta \ell}{\ell}\right)_1=\left(\frac{ A \Delta \ell}{\ell}\right)_2\) \(\Rightarrow \frac{\Delta \ell_2}{\Delta \ell_1}=\frac{ A _1}{ A _2} \times \frac{\ell_2}{\ell_1}\)

\(\Rightarrow \frac{\Delta \ell_2}{0.2}=\frac{1}{2.4 \times 2.4} \times \frac{2}{1}\) \(\Rightarrow \Delta \ell_2=6.9 \times 10^{-2}\,mm\)

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