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If the length of a simple pendulum is decreased by 2%, find the percentage decrease in its period T, where `T = 2pi sqrt((l)/(g))` |
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Answer» Correct Answer - `1%` `log T = log ((2pi)/(sqrtg)) + (1)/(2) log l rArr (1)/(T).(dT)/(dl) = (1)/(2l)` `:. delta T = (T)/(2l).deltal rArr ((deltaT)/(T) xx 100) = ((1)/(2).(deltal)/(l) xx 100) = {(1)/(2) xx ((-2)/(100)) xx 100} = -1` |
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