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The bob of a simple pendulum performs SHM with period T in air and with period `T_(1)` in water. Relation between T and `T_1` is (neglect friction due to water, density of the material of the bob is = `9/8xx 10^3 (kg)/m^3`, density of water = `10^3(kg)/m^3`)A. `T_(1)=3T`B. `T_(1)=2T`C. `T_(1)=T`D. `T_(1)=(T)/(2)` |
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Answer» Correct Answer - A `T_(1)=(T)/(sqrt(1-(p)/(p_(m))))=(T)/(1-(10^(3))/((9)/(8)xx10^(-3)))` `=(T)/(sqrt(1-(8)/(9)))=(T)/(sqrt((9-8)/(9)))=(T)/(sqrt((1)/(9)))=(T)/((1)/(3))=3T` |
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