Question

An hourglass consists of two sets of congruent composite figures on either end. Each composite figure is made up of a cone and a cylinder.Each cone of the hourglass has a height of 12 millimeters. The total height of the sand within the top portion of the hourglass is 47 millimeters. The radius of both the cylinder and cone is 4 millimeters. Sand drips from the top of the hourglass to the bottom at a rate of 10π cubic millimeters per second. How many seconds will it take until all of the sand has dripped to the bottom of the hourglass?

Answer 1

@Nurali can you help me ?

Answer 2

A)6.4
B)62.4
C)8.5
D)56.0

Answer 3

The total volume of sand consists of the sand in the cone portion plus whatever sand is in the cylindrical portion of the hourglass. We'll call the total volume V, the conical volume Vcn, and the cylindrical volume Vcy.

V = Vcn + Vcy

The cone is 12mm high; and since the total amount of sand is 47mm high, that means the cone is entirely filled.

Vcn = (1/3)(pi)(r^2)(h)
Vcn = (1/3)(pi)(4^2)(12)
Vcn = 64pi

Now... if the sand is 47mm high but the cone is only 12mm high, that means the height of the sand in the cylinder is 47 - 12 = 35mm. So...

Vcy = (pi)(r^2)(h)
Vcy = (pi)(4^2)(35)
Vcy = 560pi

Now substitute back into your original equation.

V = Vcn + Vcy
V = 64pi + 560pi
V = 624pi

The sand drips at a rate of 10pi cu mm / sec.

624pi / 10pi = 62.4

Answer 4

thank you !!!

Answer 5

Anytime.