A simple pendulum with a brass bob hasa time period `2sqrt(2)` second. The bob is now immersed in a non viscous liquid and oscillated. If the density of liquid is `(1//9)` that of brass, find the time period in second of the same pendulum.

Answer

3

Explanation

Here, `T=2sqrt(2)s=2pisqrt((l)/(g))`. Let `V` be the volume and `rho` be the density of the brass bob. Mass of the bob, `m=Vrho`. Weight of the bob`=Vrhog`. Buoyancy force of liquid on bob`=V(rho//9)g=Vrhog//9.`
The effective weight of bob in liquid
`=Vrhog-Vrhog//9=(8)/(9)Vrhog`
`:.` Acceleration, `f^(')=(8Vrhog//9)/(m)`
`=(8Vrhog//9)/(Vrho)=(8g)/(9)`
Time period of the bob,
`T^(')=2pisqrt((l)/(g^(')))=wpisqrt((l)/(8g//9))=(3)/(2sqrt(2))T`
`=(3)/(2sqrt(2))xx2sqrt(2)s=3s`