Two pendulums of lengths 121 cm and 100 cm start vibrating. At some instant, the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum the two will be in phase in the mean position?

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answer is B.

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Detailed Solution

$T_{1} = 2 π \sqrt{\frac{121}{g}}, T_{2} = 2 π \sqrt{\frac{100}{g}}$
Let shorter pendulum makes n vibrations; then the longer will make one less than n to come in phase again.
${nT}_{2} = (n - 1) T_{1}$
OR
$n \times 2 π \sqrt{\frac{100}{g}} = (n - 1) 2 π \sqrt{\frac{121}{g}}$
Solving, we get; $n = 11$

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