# he bob of a simple pendulum executes simple harmonic motion in water with a period t , while the period of oscillation of the bob is t0  in air. Neglecting frictional force of water and given that the density of the bob is (4/3) x 1000 kg/m3. What relationship between t and t0  is true ? Option 1) t = t0 Option 2) t = t0/2 Option 3) t = 2t0 Option 4) t = 4t0

As we learnt in

Time period of oscillation of simple pendulum -

$T=2\pi \sqrt{\frac{l}{g}}$

- wherein

l = length of pendulum

g = acceleration due to gravity.

$\dpi{100} t_{0}=2\pi \sqrt{l/g}.........(i)$

Due to upthrust of water on the top, its apparent weight decreases

upthrust = weight of liquid displaced

$\dpi{100} \therefore \; \; \; Effective\; weight=mg-(V\sigma g)=V\rho g-V\sigma g$

$\dpi{100} V\rho g'=Vg(\rho -\sigma ),where\; \sigma \; is \; density \; of\; water\;$

$\dpi{100} or\; \; g'=g\left ( \frac{\rho -\sigma }{\rho } \right )$

$\dpi{100} \therefore \; \; t=2\pi \sqrt{l/g'}=2\pi \sqrt{\frac{l\rho }{g(\rho -\sigma )}}..........(ii)$

$\dpi{100} \therefore \; \; \; \frac{t}{t_{0}}=\sqrt{\frac{l\rho }{g(\rho -\sigma )}\times \frac{g}{l}}=\sqrt{\frac{\rho }{\rho -\sigma }}=\sqrt{\frac{4\times 1000/3}{(\frac{4000}{3}-1000)}}=2$

$\dpi{100} or\; \; \; t=t_{0}\times 2=2t_{0}$

Correct option is 3.

Option 1)

t = t0

This is an incorrect option.

Option 2)

t = t0/2

This is an incorrect option.

Option 3)

t = 2t0

This is the correct option.

Option 4)

t = 4t0

This is an incorrect option.

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