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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Advanced · Physics · Mechanical Properties of Solids

A horizontally oriented thin copper rod of length / is rotated...

A horizontally oriented thin copper rod of length / is rotated about vertical axis passing thorugh its middle. The breaking stress of copper is σ\sigma and density is ρ\rho. The frequency of rotation at which this rod just ruptures is

Options

  1. Aσρπ2l2rpm\sqrt{\frac{\sigma}{\rho \pi^2 l^2}} \mathrm{rpm}
  2. Bσ2π2ρl2rps\sqrt{\frac{\sigma}{2 \pi^2 \rho l^2}} \mathrm{rps}
  3. C2σπ2ρl2rps\sqrt{\frac{2 \sigma}{\pi^2 \rho l^2}} \mathrm{rps}
  4. D3σ2π2ρl2rpm\sqrt{\frac{3 \sigma}{2 \pi^2 \rho l^2}} \mathrm{rpm}

Correct answer

C. 2σπ2ρl2rps\sqrt{\frac{2 \sigma}{\pi^2 \rho l^2}} \mathrm{rps}

Step-by-step solution

The correct answer is option (C) Tma=mω2ℓ8=σ×Amω2ℓ8=σAω=8σρℓ2=2πnn=2σπ2ρℓ2rps\begin{aligned} & T_{m a}=\frac{m \omega^2 \ell}{8}=\sigma \times A \\ & \frac{m \omega^2 \ell}{8}=\sigma A \\ & \omega=\sqrt{\frac{8 \sigma}{\rho \ell^2}}=2 \pi n \\ & n=\sqrt{\frac{2 \sigma}{\pi^2 \rho \ell^2}} r p s\end{aligned}
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