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A simple pendulum bob oscillates with period t while immersed in water. If the same bob oscillates in air with period t0, and the viscous resistance of water is neglected, then for a bob whose density is (4/3) × 1000 kg/m³, the relation between t and t0 is
- t = 2t0
- t = t0/2
- t = t0
- t = 4t0
Correct answer: t = 2t0
Solution
Effective gravity in water g' = g(1 - rho_water/rho_bob) = g(1 - 1000/1333.3) = g/4. Since T is proportional to 1/sqrt(g), t/t0 = sqrt(g/g') = sqrt(4) = 2, so t = 2*t0.
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