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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3. The time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (In 5 =1.601, In 2 = 0.693). [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 231 s                     

    B) 208 s

    C) 161 s                     

    D) 142 s

    Correct Answer: D

    Solution :

                    As we know,\[x={{x}_{0}}{{e}^{-bt/2m}}\]From question, \[8=10{{e}^{\frac{40b}{2m}}}\]                                  ....(i) Similarly,\[5=10{{e}^{-\frac{bt}{2m}}}\]                 ?..(ii) Solving eqns (i) and (ii) we get \[t\cong 142s\]


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