JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2024
- Hard
- Numerical
Two blocks of mass \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is \(4.0 \times 10^{-5}\) \(\mathrm{m}\) and Young's modulus of the metal is \(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\). The longitudinal strain developed in the wire is \(\frac{1}{\alpha \pi}\). The value of \(\alpha\) is _______. [Use \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N}\)
\(\mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2\)
\(\text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}}\)
\(=\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi}\)
\(\alpha=12\)
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