2. Solved Papers
  3. AIIMS

AIIMS AIIMS Solved Paper-2010

  • question_answer
    Two simple pendulums first of bob mass \[\therefore \] and length\[g=g\frac{{{R}^{2}}}{{{(R+h)}^{2}}}=\frac{9.8\times {{(6400)}^{2}}}{{{(6400+480)}^{2}}}\], second of bob mass \[=8.4m/{{s}^{2}}\] and length \[\left( P+\frac{{{\mu }^{2}}a}{{{V}^{2}}} \right)(V-\mu b)=\mu RT\] and\[P=\left( \frac{\mu RT}{V-\mu b} \right)-\frac{{{\mu }^{2}}a}{{{V}^{2}}}\]. If the vibrational energies of both are same. Then which is correct?

    A)  Amplitude of B greater than A

    B)  Amplitude of B smaller than A

    C)  Amplitude will be same

    D)  None of the above

    Correct Answer: B

    Solution :

    Frequency, \[K=\frac{233.33\times 4\times {{10}^{-3}}}{5\times {{10}^{-4}}\times 32}\] or         \[=58.33W/m{{\text{-}}^{o}}C\] \[v=\sqrt{2\left( \frac{g}{8} \right)h}\]   \[v=\frac{\sqrt{gh}}{2}\] \[t=\frac{v}{g}\left[ 1+\sqrt{1+\frac{2gh}{{{v}^{2}}}} \right]\]   \[=\frac{2\sqrt{gh}}{g}=2\sqrt{\frac{h}{g}}\] \[=\frac{\sqrt{gh}/2}{g}\left[ 1+\sqrt{1+\frac{2gh}{gh/4}} \right]\]\[=\frac{M{{R}^{2}}}{2}\] Energy,    \[=M\left( \frac{{{L}^{2}}}{I2}+\frac{{{L}^{2}}}{4} \right)\] \[\frac{M{{L}^{2}}}{12}+\frac{M{{R}^{2}}}{4}=\frac{M{{R}^{2}}}{2}\] and      \[L=\sqrt{3}R\]           (\[g\propto \frac{1}{{{R}^{2}}}\] E is same) \[g\propto \frac{1}{{{(R+h)}^{2}}}\]   \[\therefore \] Given, \[g=g\frac{{{R}^{2}}}{{{(R+h)}^{2}}}=\frac{9.8\times {{(6400)}^{2}}}{{{(6400+480)}^{2}}}\] and \[=8.4m/{{s}^{2}}\]\[\left( P+\frac{{{\mu }^{2}}a}{{{V}^{2}}} \right)(V-\mu b)=\mu RT\]\[P=\left( \frac{\mu RT}{V-\mu b} \right)-\frac{{{\mu }^{2}}a}{{{V}^{2}}}\]. So, amplitude of B smaller than A.


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