A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond 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 = 103 kg / m3)

Question

A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond 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 = 103 kg / m3) (A) 5.00 rad s-1 (B) 1.25 rad s-1 (C) 3.75 rad s-1 (D) 2.5√10 rad s-1

Tardigrade Admin · Accepted Answer

Extra Boyant force = δ Axg B0 + B × mg = ma B= ma a =((δ Ag/m))x w2 =(δ Ag/m) w = √(103 ×π(2.5)2 ×10-4 ×10/310 ×10-6 ×103) = √63.30 = 7.95

$5.00 r a d s^{- 1}$

$1.25 r a d s^{- 1}$

$3.75 r a d s^{- 1}$

$2.5 10 r a d s^{- 1}$

Extra Boyant force = $δ$ Axg
$B_{0} + B \times m g = ma$
$B = ma$
$a = (\frac{δ A g}{m})^{x}$
$w^{2} = \frac{δ A g}{m}$
$w = \frac{1 0 ^{3} \times π ( 2.5 ) ^{2} \times 1 0 ^{- 4} \times 10}{310 \times 1 0 ^{- 6} \times 1 0 ^{3}}$
$= 63.30 = 7.95$

5.00rads−1

1.25rads−1

3.75rads−1

2.510​rads−1

Extra Boyant force = δ Axg B0​+B×mg=ma B=ma a=(mδAg​)x w2=mδAg​ w=310×10−6×103103×π(2.5)2×10−4×10​​ =63.30​=7.95

$5.00 r a d s^{- 1}$

$1.25 r a d s^{- 1}$

$3.75 r a d s^{- 1}$

$2.5 10 r a d s^{- 1}$

Extra Boyant force = $δ$ Axg
$B_{0} + B \times m g = ma$
$B = ma$
$a = (\frac{δ A g}{m})^{x}$
$w^{2} = \frac{δ A g}{m}$
$w = \frac{1 0 ^{3} \times π ( 2.5 ) ^{2} \times 1 0 ^{- 4} \times 10}{310 \times 1 0 ^{- 6} \times 1 0 ^{3}}$
$= 63.30 = 7.95$