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The period of a simple pendulum is found to be increased by `50%` when the length of the pendulum is increased by `0.6m`. The initial length isA. 0.48 mB. 0.5 mC. 0.994 mD. 0.3 m |
Answer» Correct Answer - A `T_(2)=(1.5)T_(1)andl_(2)=l_(1)+0.6` `(T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt((l_(1)+0.6)/(l_(1)))therefore (1.5)^(2)=(l_(1)+0.6)/(l_(1))` `therefore 2.25=(l_(1)+0.6)/(l_(1)) therefore 1.25l_(1)=0.6` `l_(1)=(0.6)/(1.25)=(0.6xx4)/(5)=(2.4)/(5)` `=0.48m` |
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