An open cylindrical vessel filled with water is rotated together with water at angular speed 1 rad s- 1 . At this angular speed, the centre of the bottom is just exposed. If the angular speed is increased to √(3/2) rad s- 1 , how much area (in cm2 ) of the bottom of the vessel is now exposed to surrounding? [Area of the cross-section of cylinder is 27 cm2 ]

Question

An open cylindrical vessel filled with water is rotated together with water at angular speed 1 rad s- 1 . At this angular speed, the centre of the bottom is just exposed. If the angular speed is increased to √(3/2) rad s- 1 , how much area (in cm2 ) of the bottom of the vessel is now exposed to surrounding? [Area of the cross-section of cylinder is 27 cm2 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let ω 1 be initial angular velocity Solution

(ω 12 R2/2 g)=h For angular velocity ω 2 let radius of exposed surface be x Solution

(ω 22/2 g)[R2 - x2]=h Putting the value of ω 1 and ω 2 π x2=9 Since π R2=27

Let ω1​ be initial angular velocity 2gω12​R2​=h For angular velocity ω2​ let radius of exposed surface be x 2gω22​​[R2−x2]=h Putting the value of ω1​ and ω2​ πx2=9 Since πR2=27

Let $ω_{1}$ be initial angular velocity

$\frac{ω _{1}^{2} R ^{2}}{2 g} = h$
For angular velocity $ω_{2}$ let radius of exposed surface be x

$\frac{ω _{2}^{2}}{2 g} [R^{2} - x^{2}] = h$
Putting the value of $ω_{1}$ and $ω_{2}$
$π x^{2} = 9$
Since $π R^{2} = 27$