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On completely immersing a solid metal sphere into a cylindrical bucket

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Bunuel
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On completely immersing a solid metal sphere into a cylindrical bucket [#permalink] New post   13 Jan 2021, 09:00
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On completely immersing a solid metal sphere into a cylindrical bucket partially filled with water, the level of water rose 1cm without overflow. If the internal radius of the bucket is 6cm, what is the radius of the sphere?

A. 0.5
B. 1
C. 2
D. 3
E. 4
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D
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink] New post   13 Jan 2021, 09:29
As the water level rose by 1, so we assume the volume of cylinder with height 1 and radius 6 (radius of bucket) is equal to the volume of the sphere.

radius of bucket (r) =6
Height (h) = 1

Substituting the formula, gives radius of sphere (R) as,

πr^2h=4/3πR^3

We get R as 3, option (D)

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TarunKumar1234
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink] New post   13 Jan 2021, 09:38
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] New post   13 Jan 2021, 22:41
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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stne
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On completely immersing a solid metal sphere into a cylindrical bucket [#permalink] New post   18 Jan 2021, 05:56
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Bunuel wrote:
On completely immersing a solid metal sphere into a cylindrical bucket partially filled with water, the level of water rose 1cm without overflow. If the internal radius of the bucket is 6cm, what is the radius of the sphere?

A. 0.5
B. 1
C. 2
D. 3
E. 4


Volume of the sphere = Volume of water displaced

\(\frac{4}{3}\pi(r^3)= \pi(6)^2*1\)
\(r=3\)

Where \(r= \)radius of the sphere
\(6=\) radius of the cylinder or bucket
\(1=\) rise in height of the cylinder or bucket
Volume of water displaced = \(\pi(6)^2*1\)
Volume of Sphere= \(\frac{4}{3}\pi(r^3)\)

Ans-D

Hope it helps.
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Re: On completely immersing a solid metal sphere into a cylindrical bucket [#permalink] New post   18 Jan 2021, 08:44
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]
18 Jan 2021, 08:44
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