An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

where;

height h which is the height for the free surface in a rotating tank is expressed as:

at the bottom surface of the tank;

r = 0, h = 0

∴

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

replacing the value of h in equation (2); we have:

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then

Replacing equation (1) and (3)

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

c, 1a hope this

it is the second option.