Question

A liquid is kept in a cylindrical vessel which is. being rotated about its axis .The liquid rises at the sides Of the radius of the vessel is 0.05m and the speed of rotation is 2 rotations per second, find the difference in the height of the liquid at the centre of the vessel and at its sides. Take g=9.8ms

Answer 1

Answer:

Explanation

Given R=5cm,ω=2π(4) rad/s

We apply the Bernoulli's equation between the center point and the point at the circumference in the water we get:

p1​−p2​=21​ρv2

hρg=21​ρ(rω)2

h=2gw2r2​=2g4×π2×42×0.052​=0.08m=8cm

Answer 2

Answer:

The correct answer is 2 cm.

Explanation:

According to the Bernoulli's theorem, the sum of the energies possessed via a flowing ideal liquid at a point is steady provided that the liquid is incompressible and non-viscous and go with the flow in streamline.

If the fluid flows horizontally so that no exchange in gravitational potential energy takes place, then a decrease in fluid pressure is related to an increase in fluid velocity.

ρ = $\frac{1}{2}$ρ$v^{2}$ = constant

Close to the ends, the velocity of liquid is greater so that pressure is lesser as a end result the liquid rise at the edges to make amends for this drop of pressure

i.e., ρgh = $\frac{1}{2}$ρ$v^{2}$ = $\frac{1}{2}$ρ$r^{2}w^{2}$

Therefore,

h=$\frac{r^{2}w^{2}}{2g}$

=$\frac{r^{2}(2\pi v)^{2}}{2g}$

=$\frac{r^{2}2\pi ^{2} v^{2}}{g}$

=$\frac{(0.05)^{2}2\pi ^{2} (2)^{2}}{9.8}$

= 0.02 m

= 2 cm

Therefore, the correct answer is 2 cm.

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