Question

The bob of a simple pendulum executes SHM in water with time period T, while its period of oscillation is T' in air. Given that density of the bob is 4000/3 kg m^(-3) . Find the relation between T andT' by equating torque.

Answer 1

Let   V be the volume of the bob.Thus mass of the bob       M=ρbV=43×1000VkgAcceleration due to gravity in air is   g.Effective weight of the bob in water       Mg′=Mg−B       where    buoyant force   B=(Vρwater)g43×1000Vg′=43×1000Vg−V×1000g                       (as ρwater=1000kg/m3)
43×1000Vg′=13×1000Vg  ⟹g′=g4   
Now  time period of the bob in air           to=2π√lgTime period of the bob in water          t=2π√lg′  =2π√4lg    =2×2π√lg 
⟹t=2to

Answer 2

Answer:

Let   V be the volume of the bob.Thus mass of the bob       M=ρbV=43×1000VkgAcceleration due to gravity in air is   g.Effective weight of the bob in water       Mg′=Mg−B       where    buoyant force   B=(Vρwater)g43×1000Vg′=43×1000Vg−V×1000g                       (as ρwater=1000kg/m3)

43×1000Vg′=13×1000Vg  ⟹g′=g4    

Now  time period of the bob in air           to=2π√lgTime period of the bob in water          t=2π√lg′  =2π√4lg    =2×2π√lg  

⟹t=2to

Explanation: