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Compare the time periods of two simple pendulums of length 1 m and 16 m at at place. |
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Answer» Solution :Given:`l_1 = 1 m and l_2 = 16 m` Since `T PROP sqrtl THEREFORE (T_1)/(T_2) = sqrt((l_1)/(l_2))` `therefore (T_1)/(T_2) = sqrt(1/(16)) = 1/4 `i.e. `T_1: T_2= 1:4` |
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