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The length of a simple pendulum executing simple harmonic motion is increased by `21%`. The percentage increase in the time period of the pendulum of increased lingth is.A. `10%`B. `21%`C. `30%`D. `50%` |
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Answer» Correct Answer - A If initial length `l_(1) = 100` then `t_(2) = 121` By using `T = 2pi sqrt((1)/(g)) rrArr (T_(1))/(T_(2)) = sqrt((I_(1))/(I_(2)))` Hence `(T_(1))/(T_(2)) = sqrt((100)/(121)) rArr T_(2) = 1.1T_(1)` `%` increase `= (T_(2) - T_(1))/(T_(1)) xx 100= 10%` Alternative Time period of simple pendulum `T = 2pi sqrt((1)/(g))` `:. (Delta T)/(T) = (1)/(2) (DeltaI)/(I)` since `(DeltaI)/(I) = 21%` `:.Delta T)/(T) = (1)/(2) xx 21% = 10%` |
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