Two pendulums of length 121cm and 100cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

Option 1 -

11

Option 2 -

9

Option 3 -

10

Option 4 -

8

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

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago
    Correct Option - 1


    Detailed Solution:

    (n) ( 1.1 ) = ( n + 1 )

    0.1 ( n ) = 1

    n = 10

    Required number of oscillation = n + 1 = 10 + 1 = 11

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