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  • Stuck here, help me understand: - Oscillations and Waves - JEE Main-2

In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii r1 and r2.  The two spherical bobs have uniform mass distribution.  If the relative difference in the periods, is found to be 5×10−4 s, the difference in radii, \left | r1-r2 \right |   is best given by :

  • Option 1)

     1 cm

  • Option 2)

    0.1 cm

  • Option 3)

    0.5 cm

  • Option 4)

     0.01 cm

     

 

Answers (1)

best_answer

As we learnt in

Time period of oscillation of simple pendulum -

T=2\pi \sqrt{\frac{l}{g}}

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

 As we know that  T=2\pi\sqrt{\frac{l}{g}}\ \Rightarrow\ \; T \alpha \sqrt{l}

Differentiate both side

    \frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}

Change in length \Delta l=r_{1}-r_{2}

5\times 10^{-4}=\frac{1}{2}r_{1}-r_{2}\ \; \Rightarrow\ \; r_{1}-r_{2}=10\times 10^{-4} m

r_{1}-r_{2}=10^{-3} m\ \; \Rightarrow\ \; r_{1}-r_{2}=10^{-1}cm

\therefore\ \;r_{1}-r_{2}=0.1 cm

Correct option is 2.

 


Option 1)

 1 cm

This is an incorrect option.

Option 2)

0.1 cm

This is the correct option.

Option 3)

0.5 cm

This is an incorrect option.

Option 4)

 0.01 cm

 

This is an incorrect option.

Posted by

Plabita

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