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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 kgm^3`, density of water = `1gcc^-1`)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 Time period of simple pendulum in water is given by `T=2pisqrt(l/g_(eff))` [Symbols have their usual meanings and `g_(eff)` = acceleration due to gravity in water] As we khow, `g_(eff)=g((sigma-rho)/sigma)` [Where,`sigma` = density of bob, p = density of water] `9.8((9/8xx10^3-10^3)/(9/8xx10^3))` `9.8((9/8-1)/(9/8))=9.8(1/8xx8/9)` `(9.8)/9` So, `T_1=2pisqrt((lxx9)/9.8) Rightarrow T_1=3T` [therefore Time period of simple pendulum in air, `T_1=2pisqrt((l)/9.8)` ] |
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