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JEE Mains · Physics · STD 11- 8. mechanical properties of solids
A body of mass \(\mathrm{m}=10\; \mathrm{kg}\) is attached to one end of a wire of length \(0.3\; \mathrm{m} .\) The maximum angular speed (in \(rad \;s^{-1}\) ) with which it can be rotated about its other end in space station is (Breaking stress of wire \(=4.8 \times 10^{7} \;\mathrm{Nm}^{-2}\) and area of cross-section of the wire \(=10^{-2}\; \mathrm{cm}^{2}\) ) is
- A \(8\)
- B \(7\)
- C \(11\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=\mathrm{m\omega}^{2} \ell\) Breaking stress \(=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{mo}^{2} \ell}{\mathrm{A}}\) \(\Rightarrow \omega^{2}=\frac{4.8 \times 10^{7} \times\left(10^{-2} \times 10^{-4}\right)}{10 \times 0.3}=16\) \(\Rightarrow \omega=4\)
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