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JEE Mains · Physics · STD 11- 8. mechanical properties of solids
Two blocks of masses \(m\) and \(M\) are connected by means of a metal wire of cross-sectional area \(A\) passing over a frictionless fixed pulley as shown in the figure. The system is then released. If \(M = 2\, m\), then the stress produced in the wire is
- A \(\frac{{2mg}}{{3A}}\)
- B \(\frac{{4mg}}{{3A}}\)
- C \(\frac{{mg}}{{A}}\)
- D \(\frac{{3mg}}{{4A}}\)
Answer & Solution
Correct Answer
(B) \(\frac{{4mg}}{{3A}}\)
Step-by-step Solution
Detailed explanation
Tension in the wire, \(T = \left( {\frac{{2mM}}{{m + M}}} \right)g.\) \(Stress = \frac{{Force/Tension}}{{Area}} = \frac{{2mM}}{{A\left( {m + M} \right)}}g\) \( = \frac{{2\left( {m \times 2m} \right)g}}{{A\left( {m + 2m} \right)}}\left( {M = 2m\,given} \right)\)…
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