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Time period of simple pendulum of length l at a place, where acceleration due to gravity is g and period T. Then period of simple pendulum of the same length at a place where the acceleration due to gravity 1.02 g will beA. TB. 1.02 TC. 0.99 TD. 1.01 T |
Answer» Correct Answer - C `T_(1)=2pisqrt((l)/(g))andT_(2)=2pisqrt((l)/((1.02)g))` `therefore (T_(2))/(T_(1))=sqrt((1)/(1.02))` `T_(2)=Tsqrt((100)/(102)) therefore T_(2)=0.99T` |
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