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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink]
Volume of water increase = volume of sphere

So, pie* 6^2 *1 = 4/3 *pie *r^3 or, r^3 = 27 or r = 3

So, It is D. :)
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink]
Change in the height of the cone's volume = Volume of the cone

\(r_{cy} = 6\); \(h_{cy}=1\)

\(πr_{cy}^2h=\frac{4}{3}πr_{sp}^3\)

\(r_{sp}^3=\frac{6*6*3}{4}=3^3\)

\(r_{sp}=3\)

Ans D
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink]
Volume of the sphere = Volume of the of the water rose

\(\frac{4}{3}\)*pi*\(r^3\) = pi*\(R^2\)*h

\(\frac{4}{3}\)*pi*\(r^3\) = pi*36*1

\(r^3\) = 27
r=3
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink]
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Math Expert
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