JEE Mains · Physics · STD 11- 8. mechanical properties of solids
Two blocks of masses \(3 \,{kg}\) and \(5\, {kg}\) are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is \(\frac{24}{\pi} \times 10^{2}\, {Nm}^{-2}\). What is the minimum radius of the wire? (Take \(\left.g=10\, {ms}^{-2}\right)\) (in \(cm\))

- A \(125\)
- B \(1250\)
- C \(12.5\)
- D \(1.25\)
Answer & Solution
Correct Answer
(C) \(12.5\)
Step-by-step Solution
Detailed explanation
\({T}=\frac{2 {m}_{1} {m}_{2} {g}}{{m}_{1}+{m}_{2}}=\frac{2 \times 3 \times 5 \times 10}{8}\) \(=\frac{75}{2}\) Stress \(=\frac{{T}}{{A}}\) \(\frac{24}{\pi} \times 10^{2}=\frac{75}{2 \times \pi {R}^{2}}\) \({R}^{2}=\frac{75}{2 \times 24 \times 100}=\frac{3}{8 \times 24}\)…
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