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Spring Systems question from JEE Main 2018

Q83JEE Main 2018Medium

Spring Systems

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)

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Answer: (c)

  1. For SHM k=mω2=4π2f2mk = m\omega^2 = 4\pi^2 f^2 m, with mm the mass of one atom.
  2. m=108×10−36.02×1023=1.79×10−25m = \frac{108\times10^{-3}}{6.02\times10^{23}} = 1.79\times10^{-25} kg
  3. k=4π2(1012)2(1.79×10−25)≈39.5×0.179≈7.1k = 4\pi^2(10^{12})^2(1.79\times10^{-25}) \approx 39.5\times0.179 \approx 7.1 N/m
  4. So the answer is (c) 7.17.1 N/m.
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