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A sphere of diameter 6 cm is dropped into a right circular cylinder vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel ?

Mensuration

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Answer

Radius of sphere (r) = 62\dfrac{6}{2} = 3 cm.

Radius of cylinder (R) = 122\dfrac{12}{2} = 6 cm.

Let height of water raised be h cm.

Volume of sphere = Volume of water rise in cylinder

43πr3=πR2h43×33=62×h4×32=62×hh=3636=1 cm.\dfrac{4}{3}πr^3 = πR^2h \\[1em] \dfrac{4}{3} \times 3^3 = 6^2 \times h \\[1em] 4 \times 3^2 = 6^2 \times h \\[1em] h = \dfrac{36}{36} = 1 \text{ cm}.

Hence, the height of water raised is 1 cm.

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