Question: 36: Two pendulums of length 121 cm and 100 cm 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:
(1) 10
(2) 8
(3) 11
(4) 9
Answer: Option (3)
Explanation:
The time period of a simple pendulum is given by
T = 2\pi \sqrt{\frac{l}{g}},
where l is the length of the pendulum.
For two pendulums, the ratio of their time periods depends on the square root of their lengths.
Thus, \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}}.
Here, the lengths are l_1 = 121 \ \text{cm} and l_2 = 100 \ \text{cm}.
Therefore, \frac{T_1}{T_2} = \sqrt{\frac{121}{100}} = \frac{11}{10}.
This means the time periods are in the ratio 11 : 10.
Frequencies are inversely proportional to time period,
so the ratio of frequencies is 10 : 11.
For the pendulums to be again in phase at the mean position, each must complete an integer number of vibrations in the same time.
The minimum such time corresponds to 11 vibrations of the shorter pendulum and 10 vibrations of the longer pendulum.
Hence, the minimum number of vibrations of the shorter pendulum required is 11.
Therefore, the correct answer is Option (3).