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JEE Main 2014 (12 Apr, Shift 1) · Shear Modulus & Poisson's Ratio

Shear Modulus & Poisson's Ratio question from JEE Main 2014 (12 Apr, Shift 1)

Q11JEE Main 2014 · 12 APR · SHIFT 1Medium

Shear Modulus & Poisson's Ratio

Steel ruptures when a shear of 3.5 ×\times 108 N m−2 is applied. The force needed to punch a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly:

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Answer: (c)

  1. Sheared area is the curved surface of the punched disc: A=πdtA = \pi d t.
  2. A=π×0.01×0.003=9.42×10−5 m2A = \pi\times0.01\times0.003 = 9.42\times10^{-5}\ \text{m}^2
  3. F=3.5×108×9.42×10−5≈3.3×104 NF = 3.5\times10^{8}\times9.42\times10^{-5} \approx 3.3\times10^{4}\ \text{N}
  4. So the answer is (c).
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