2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of\[{{10}^{12}}/\sec \]. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number\[\text{=6}\text{.02}\times \text{1}{{\text{0}}^{\text{23}}}\text{ gm mol}{{\text{e}}^{\text{-1}}}\]).                  [JEE Main Online 08-04-2018]

    A)  \[2.2\,\,N/m\]              

    B)  \[5.5\,\,N/m\]

    C)  \[6.4\,\,N/m\]              

    D)  \[7.1\,\,N/m\]

    Correct Answer: D

    Solution :

      \[kx=m{{\omega }^{2}}X\] \[\Rightarrow k=m{{\omega }^{2}}\] \[=\frac{108}{6.02\times {{10}^{23}}}\times {{(2\pi \times {{10}^{12}})}^{2}}\times {{10}^{-3}}\] \[k\approx 7.1N/m\]


You need to login to perform this action.
You will be redirected in 3 sec spinner