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JEE Main 2022 (29 Jul, Shift 2) · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2022 (29 Jul, Shift 2)

Q8JEE Main 2022 · 29 JUL · SHIFT 2Medium

Thermal Stress & Breaking Stress

A metal wire of length 0.5 m0.5\text{ m} and cross-sectional area 10−4 m210^{-4}\text{ m}^2 has breaking stress 5×108 Nm−25 \times 10^{8}\text{ Nm}^{-2}. A block of 10 kg10\text{ kg} is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _________ ms−1\text{ms}^{-1}.

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

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Answer: 50

  1. Maximum tension T=SA=5×108×10−4=5×104 NT = S A = 5 \times 10^8 \times 10^{-4} = 5 \times 10^4\text{ N}.
  2. Tension gives the centripetal force: mv2r=T⇒v2=Trm=5×104×0.510=2500\dfrac{mv^2}{r} = T \Rightarrow v^2 = \dfrac{Tr}{m} = \dfrac{5 \times 10^4 \times 0.5}{10} = 2500
  3. v=50 m s−1v = 50\text{ m s}^{-1}
  4. So the answer is 50 m/s.
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