A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is ω rad s-1. The difference in the height, h ( in cm ) of liquid at the centre of vessel and at the side will be:

Question

A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is ω rad s-1. The difference in the height, h ( in cm ) of liquid at the centre of vessel and at the side will be: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/303fbd7afe3457daff9f11e456aaa120-.png" /> (A) (25 ω2/2 g ) (B) (2 ω2/5 g) (C) (5 ω2/2 g) (D) (2 ω2/25 g )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a92fd70d94c8b95be2927c05e0ca2594-.png" /> Applying pressure equation from A to B P0+ρ ⋅ (R ω2/2) ⋅ R-ρ g h=P0 (ρ R2 ω2/2)=ρ g h h =( R 2 ω2/2 g )=(5)2 (ω2/2 g )=(25/2) (ω2/ g )

$\frac{25 ω ^{2}}{2 g}$

$\frac{2 ω ^{2}}{5 g}$

$\frac{5 ω ^{2}}{2 g}$

$\frac{2 ω ^{2}}{25 g}$

Applying pressure equation from A to B
$P_{0} + ρ \cdot \frac{R ω ^{2}}{2} \cdot R - ρ g h = P_{0}$
$\frac{ρ R ^{2} ω ^{2}}{2} = ρ g h$
$h = \frac{R ^{2} ω ^{2}}{2 g} = (5)^{2} \frac{ω ^{2}}{2 g} = \frac{25}{2} \frac{ω ^{2}}{g}$

2g25ω2​

5g2ω2​

2g5ω2​

25g2ω2​

Applying pressure equation from A to B P0​+ρ⋅2Rω2​⋅R−ρgh=P0​ 2ρR2ω2​=ρgh h=2gR2ω2​=(5)22gω2​=225​gω2​

$\frac{25 ω ^{2}}{2 g}$

$\frac{2 ω ^{2}}{5 g}$

$\frac{5 ω ^{2}}{2 g}$

$\frac{2 ω ^{2}}{25 g}$

Applying pressure equation from A to B
$P_{0} + ρ \cdot \frac{R ω ^{2}}{2} \cdot R - ρ g h = P_{0}$
$\frac{ρ R ^{2} ω ^{2}}{2} = ρ g h$
$h = \frac{R ^{2} ω ^{2}}{2 g} = (5)^{2} \frac{ω ^{2}}{2 g} = \frac{25}{2} \frac{ω ^{2}}{g}$