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A bob of mass ‘m’ suspended by a thread of length I undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 1 4  times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be:-

Option 1 -

3 4 T

Option 2 -

3 2 T

Option 3 -

T

Option 4 -

4 3 T

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    Time period T = 2π l g e f f

    ⇒ T i = T = 2 π l g    

    ⇒ T f = T ' = 2 π 4 l / 3 g ( 1 − ρ σ ) = 2 π 1 6 l 9 g

    ⇒ T ' = 4 3 T

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