**Situation**

A closed conical vessel has a base radius of 2 m and is 6 m high. When in upright position, the depth of water in the vessel is 3 m.

Part 1: What is the volume of water?

A. 22 m^{3}

B. 25 m^{3}

C. 28 m^{3}

D. 32 m^{3}

Part 2: If the vessel is held in inverted position, how deep is the water?

A. 4.53 m

B. 5.74 m

C. 4 m

D. 5 m

Part 3: What is the weight of water in quintals. Unit weight of water is 9,800 N/m^{3}.

A. 263.4

B. 195.4

C. 219.7

D. 247.2

**Solution**

$V_{cone} = 8\pi \, \text{ m}^3$

In upright position,

$\dfrac{V_{air}}{V_{cone}} = \dfrac{3^3}{6^3}$

$\dfrac{V_{air}}{8\pi} = \dfrac{27}{216}$

$V_{air} = \pi \, \text{ m}^3$

$V_{water} = V_{cone} - V_{air}$

$V_{water} = 8\pi - \pi$

$V_{water} = 7\pi \, \text{ m}^3$

$V_{water} = 21.991 \, \text{ m}^3$ Part 1: [ A ]

In inverted position:

$\dfrac{V_{water}}{V_{cone}} = \dfrac{h^3}{6^3}$

$\dfrac{7\pi}{8\pi} = \dfrac{h^3}{216}$

$h = 5.7388 \, \text{ m}$ Part 2: [ B ]

$\text{Weight of water} = \gamma_w V_{water} = 9800(21.991) = 215\,511.8 \, \text{ N}$

$\text{Weight of water} = 215\,511.8 \, \text{ N}\left( \dfrac{1 \, \text{ kg}}{9.81 \text{ N}} \right)\left( \dfrac{1 \, \text{ quintal}}{100 \text{ kg}} \right)$

$\text{Weight of water} = 219.686 \, \text{ quintals}$ Answer: [ C ]

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