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 in starts performing simple harmonic motion at angular frequency `omega`. If the radius of the bottle is 2.5 cm then `omega` is close

Options

A. `2.50 rad"/"s`
B. `3.75 rad"/"s`
C. `5.00 rad"/"s`
D. `1.25 rad"/"s`

Answer

C

Explanation

According to law of floatation,
(Weight of floating object)= (Weight of liquid displaced by floating object)
`therefore mg = m_(0)g`
`therefore m= m_(0) implies V. p_(0)= Alp_(0)`
`implies l= (m)/(A p_(0))= (m)/(pi R^(2) p_(0))`
`therefore l= (310g)/((3.14)(2.5 cm)^(2)(1-(g)/(cm^3))) approx 16 cm approx 0.16 m`
Now angular frequency,
`omega = sqrt((g)/(l)) = sqrt((9.8)/(0.16))= sqrt(61.25)= 7.826 (rad)/(s)`
The nearest value in given options for above answer is `5 rad"/"s`.
`therefore omega approx 5.0 (rad)/(s)`.