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(B) The restoring force $F$ when the bottle is pushed down by a distance $x$ is given by the weight of the displaced water: $F = A \rho g x$,where $A

Vedclass · Accepted Answer

$7.9$

Vedclass · Answer

(B) The restoring force $F$ when the bottle is pushed down by a distance $x$ is given by the weight of the displaced water: $F = A ho g x$,where $A = \pi r^2$ is the cross-sectional area of the bottle.For simple harmonic motion,$F = m \omega^2 x$,where $m$ is the mass of the water inside the bottle.Equating the two: $\pi r^2 ho g x = m \omega^2 x$.Thus,$\omega = \sqrt{\frac{\pi r^2 ho g}{m}}$.Given $m = ho V$,where $V = 310\, ml = 310 	imes 10^{-6}\, m^3$ and $r = 2.5 	imes 10^{-2}\, m$.Substituting the values: $\omega = \sqrt{\frac{\pi r^2 ho g}{ho V}} = r \sqrt{\frac{\pi g}{V}}$.$\omega = (2.5 	imes 10^{-2}) \sqrt{\frac{3.14 	imes 10}{310 	imes 10^{-6}}} = (2.5 	imes 10^{-2}) \sqrt{\frac{31.4}{310 	imes 10^{-6}}} \approx (2.5 	imes 10^{-2}) \sqrt{10^5} \approx 2.5 	imes 316 	imes 10^{-2} \approx 7.9\, rad\, s^{-1}$.

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