WBJEE · Physics · Mechanical Properties of Solids
One end of a steel wire is fixed to the ceiling of an elevator moving up with an acceleration \(2 \mathrm{~m} / \mathrm{s}^2\) and a load of 10 kg hangs from the other end. If the cross section of the wire is \(2 \mathrm{~cm}^2\) then the longitudinal strain in the wire will be ( \(\mathrm{g}=\) \(10 \mathrm{~m} / \mathrm{s}^2\) and \(\left.\mathrm{Y}=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\right)\)
- A \(4 \times 10^{11}\)
- B \(3 \times 10^{-6}\)
- C \(8 \times 10^{-6}\)
- D \(2 \times 10^{-6}\)
Answer & Solution
Correct Answer
(B) \(3 \times 10^{-6}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\ & \mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a}) \\ & =10(10+2)=120 \mathrm{~N} \\ & \mathrm{y}=\frac{\mathrm{TL}}{\mathrm{A} \Delta \ell} \\ & \frac{\Delta \ell}{\mathrm{L}}=\frac{\mathrm{T}}{\mathrm{YA}}=\frac{\not 120…
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