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The bob of simple pendulum executes SHM ...

The bob of simple pendulum executes SHM in water with a period T, while the period of oscillation of the bob is `T_(0)` in air. Neglecting frictional force of water and given that the density of the bob is `(4000)/(3)kgm^(-3)` , find the ration between T and `T_(0)`.

Text Solution

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`rho=(4000)/(3)kgm^(-3) , sigma=1000 kgm^(-3)`
`g'=(1-(sigma)/(rho))g=(1-(1000xx3)/(4000))g=(g)/(4)`
`t_(0)=2pisqrt((L)/(g)),t=2pisqrt((Lxx4)/(g))=2xx2pisqrt((L)/(g))`
or `t=2t_(0)`
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