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JEE Main 2020 (9 Jan, Shift 1) · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2020 (9 Jan, Shift 1)

Q12JEE Main 2020 · 9 JAN · SHIFT 1Medium

Thermal Stress & Breaking Stress

A body of mass m = 10 kg is attached to one end of a wire of length 0.3 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 × 107 Nm–2 and

area of cross-section of the wire = 10–2 cm2) is:

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. In a space station there is no weight; tension only provides the centripetal force: T=mω2LT = m\omega^2 L.
  2. Maximum tension T=4.8×107×10−6=48 NT = 4.8 \times 10^7 \times 10^{-6} = 48\text{ N} (as 10−2 cm2=10−6 m210^{-2}\text{ cm}^2 = 10^{-6}\text{ m}^2).
  3. ω2=4810×0.3=16⇒ω=4 rad s−1\omega^2 = \dfrac{48}{10 \times 0.3} = 16 \Rightarrow \omega = 4\text{ rad s}^{-1}
  4. So the answer is 4 rad/s.
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