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A solid metal sphere rests in a rectangular tray. The radius of the metal ball is 2cm. The width and the length of the base of the tray are 7cm and 12cm respectively. The depth of water in the tray is 2cm.
(a)Find the volume of the water in the tray
(b)If the height of the tray is 55cm, find the volume of water needed to be poured into the tray so that the metal sphere is just submerged.

 Feb 28, 2020
 #1
avatar+1131 
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A solid metal sphere rests in a rectangular tray. The radius of the metal ball is 2cm. The width and the length of the base of the tray are 7cm and 12cm respectively. The depth of water in the tray is 2cm.
(a)Find the volume of the water in the tray
(b)If the height of the tray is 55cm, find the volume of water needed to be poured into the tray so that the metal sphere is just submerged.

 

Archimedes' principle

Sphere volume   V = 4/3 * pi * r³  = 33.51 cm³

a)     7cm * 12cm * 2cm = 168cm³             168cm³ - (33.51cm³/2) = 151.245cm³

 

b)     ( 2 *168cm³) - 33.51cm³ = 302.49 cm³   wink

 

( It's been corrected )  laugh

 Feb 28, 2020
edited by Dragan  Feb 28, 2020
 #2
avatar+111433 
+1

(b)If the height of the tray is 55cm, find the volume of water needed to be poured into the tray so that the metal sphere is just submerged.

 

If the radius  of  the sphere  is   2cm.......it will be submereged  when the height  of the water  = 4cm

 

So....if  there  were no sphere, the  volume of  water  would be  4 * 7 * 12  =    336 cm^3

 

And  the  volume  of the sphere  = (4/3)pi (2)^3  =  (32/3) pi  cm^3

 

So....once  the sphere  is submereged, we only need

 

[336  - (32/3)pi  ]  ≈ 302.49   cm^3   of water to submerge  the sphere

 

 

cool cool cool

 Feb 28, 2020

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