JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is\[{{t}_{0}}\]in air. Neglecting frictional force of water and given that the density of the bob is \[(4/3)\times 1000kg/{{m}^{3}}\].   What   relationship between\[t\]and\[{{t}_{0}}\]is true?

    A)  \[t={{t}_{0}}\]                                  

    B)  \[t={{t}_{0}}/2\]

    C)  \[t=2{{t}_{0}}\]                

    D)  \[t=4{{t}_{0}}\]

    Correct Answer: C

    Solution :

                    The time period of simple pendulum in air \[T={{t}_{.0}}=2\pi \sqrt{\left( \frac{l}{g} \right)}\]           ?. (i) \[l\],being the length of simple pendulum. In water, effective weight of bob \[w=\]weight of bob in air - upthrust \[\Rightarrow \]\[\rho V{{g}_{eff}}=mg-mg\]                 \[=\rho Vg-\rho Vg=(\rho -\rho )Vg\] where   \[\rho =\]density of bob, \[\rho =\]density of water \[\therefore \]  \[{{g}_{eff}}=\left( \frac{\rho -\rho }{\rho } \right)g=\left( 1-\frac{\rho }{\rho } \right)g\] \[\therefore \]  \[t=2\pi \sqrt{\left[ \frac{l}{\left( 1-\frac{\rho }{\rho } \right)g} \right]}\]                               ? (ii) Thus,     \[\frac{t}{{{t}_{0}}}=\sqrt{\left( \frac{1}{\left( 1-\frac{\rho }{\rho } \right)} \right)}\]                 \[=\sqrt{\left( \frac{1}{1-\frac{1000}{(4/3)\times 1000}} \right)}=\sqrt{\left( \frac{4}{4-3} \right)}\]                 \[=2\Rightarrow t=2{{t}_{0}}\]


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