A sphere of diameter 12cm, is dropped in the right circular cylinder vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 359 cm . Find the diameter of the cylindrical vessel.

A
20 cm
B
18 cm
C
22.5 cm
D
16 cm

Solution:

Concept- Consider the volume of water level raised in terms of cylindrical vessel’s volume of height equals to water level raised. And then equate it with the volume of the sphere to calculate the diameter of the cylindrical vessel.
Assume,
Height of the cylinder = h cm.
Diameter of cylindrical vessel =d cm.
Given that,
Diameter of sphere =12 cm.
Radius of sphere

= \frac{12}{2} = 6 cm .

\Rightarrow

Volume of water raised = Volume of sphere submerged into the cylindrical vessel

Volume of cylinder with height, h_{waterlevel} = 3 \frac{5}{9} cm

Radius of cylinder = r_{c} cm .

Volume of sphere of radius (r_{s}) cm = π {(r_{s})}^{2} h_{waterlevel} = \frac{4}{3} π {(r_{s})}^{3}

Substituting the values in the above equation, we get

\Rightarrow π {r_{c}}^{2} h_{waterlevel} = \frac{4}{3} π {r_{s}}^{3}

\Rightarrow π {r_{c}}^{2} (3 \frac{5}{9}) = \frac{4}{3} π {(6)}^{3}

On further simplifying, we get

{\Rightarrow r}_{c}^{2} (\frac{32}{9}) = \frac{4}{3} (216)

\Rightarrow {r_{c}}^{2} (\frac{32}{9}) = 4 \times 72

\Rightarrow {r_{c}}^{2} = \frac{288 \times 9}{32}

\Rightarrow {r_{c}}^{2} = 81

\Rightarrow r_{c} = \sqrt{81}

\Rightarrow r_{c} = 9 cm

Diameter of the cylindrical vessel

d_{c} = 2 \times r_{c} = 2 \times 9 = 18 cm .

So, the diameter of the cylindrical vessel is 18 cm.
Hence, the correct option is 2.

A sphere of diameter 12cm, is dropped in the right circular cylinder vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $3 \frac{5}{9}$ cm . Find the diameter of the cylindrical vessel.

Solution:

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