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.

Guest Feb 28, 2020

#1**0 **

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 **

**( It's been corrected ) **

Dragan Feb 28, 2020

#2**+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

CPhill Feb 28, 2020