A cylindrical plastic bottel of negligible mass is filled with 310 ml of water and left floating in a pound with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $ω$ . If the radius of the bottle is 2.5 cm, then $ω$ close to (density of water = $10^{3} k g / m^{3})$

5.00 r a d s^{- 1}

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1.25 r a d s^{- 1}

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3.75 r a d s^{- 1}

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7.95 r a d s^{- 1}

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Solution

The correct option is A $7.95 r a d s^{- 1}$
Extra Boyant force = $δ A x g$
$B_{o} + B \times m g = m a$
B = ma
$a = (\frac{δ A g}{m})$

$w^{2} = \frac{δ A g}{m}$

$w = \sqrt{\frac{10^{3} \times π (2.5)^{2} \times 10^{- 4} \times 10}{310 \times 10^{-} 6 \times 10^{3}}} = \sqrt{63.30} = 7.95 r a d s^{- 1}$

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The correct option is A 7.95 rads−1Extra Boyant force = δAxgBo+B×mg=ma B = maa=(δAgm) w2=δAgmw=√103×π(2.5)2×10−4×10310×10−6×103=√63.30=7.95 rads−1