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JEE Main 2019 (10 Apr, Shift 2) · Thermal Stress & Breaking Stress

Thermal Stress & Breaking Stress question from JEE Main 2019 (10 Apr, Shift 2)

Q13JEE Main 2019 · 10 APR · SHIFT 2Easy

Thermal Stress & Breaking Stress

The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

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Answer: (a)

  1. A=Fσ=400379×106=1.055×10−6 m2A = \dfrac{F}{\sigma} = \dfrac{400}{379 \times 10^6} = 1.055 \times 10^{-6}\text{ m}^2
  2. d=4Aπ=1.344×10−6≈1.16×10−3 md = \sqrt{\dfrac{4A}{\pi}} = \sqrt{1.344 \times 10^{-6}} \approx 1.16 \times 10^{-3}\text{ m}
  3. So the answer is (a) 1.16 mm.
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