The bob of a simple pendulum executes simple | vedclass

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(C) The period of a simple pendulum in air is given by ${t_0} = 2\pi \sqrt{\frac{l}{g}}$.When the bob is submerged in water,it experiences an upward b

Vedclass · Accepted Answer

$t = 2{t_0}$

Vedclass · Answer

(C) The period of a simple pendulum in air is given by ${t_0} = 2\pi \sqrt{\frac{l}{g}}$.When the bob is submerged in water,it experiences an upward buoyant force. The effective acceleration due to gravity ${g_{eff}}$ is given by:${g_{eff}} = g \left(1 - \frac{ho_{water}}{ho_{bob}}ight)$.Given $ho_{bob} = \frac{4}{3} 	imes 10^3 \ kg/m^3$ and $ho_{water} = 10^3 \ kg/m^3$,we have:$\frac{ho_{water}}{ho_{bob}} = \frac{10^3}{(4/3) 	imes 10^3} = \frac{3}{4}$.Thus,${g_{eff}} = g \left(1 - \frac{3}{4}ight) = \frac{g}{4}$.The period in water is $t = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{g/4}} = 2 	imes 2\pi \sqrt{\frac{l}{g}}$.Substituting ${t_0}$,we get $t = 2{t_0}$.

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