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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)`.

A

`t=t_(0)`

B

`t=4t_(0)`

C

`t=2t_(0)`

D

`t=t_(0)//2`

Text Solution

Verified by Experts

The correct Answer is:
C
In air, the time period of the bob is `t_(0)=2pisqrt((l)/(g))`
In water apparent weight
`mg '=(4)/(3)pir^(3) p_(s)`,`g-(4)/(3)pir^(3)p_(w)g`
mg =`4/3pir^(3) Ps g(1-(pv)/Ps)=mg(1-(3)/(4))=mg/4 implies g'=g//4`
In water, the time period of the bob `t=2pisqrt(1/g)`,=`2pisqrt(41/g)=2t_(0)`
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