Question 1153132
let the volume of the water in the tray be {{{V[w]}}}

let the volume of the tray be {{{V[t]}}}

let the volume of the sphere be {{{V[s]}}}

given:

The radius of the metal ball is {{{r=2cm}}}.
 
The depth of water in the tray is {{{h[1]=2cm}}}

The width and the length of the base of the tray are {{{7cm}}} and {{{12cm}}} respectively. .
The height of the tray is {{{h=55cm}}}


(a)Find the volume of the water in the tray

{{{V[w]}}} will be less then should be because of a ball

since given that the depth of water in the tray is {{{h[1]=2cm}}}, the length of tray is {{{7cm}}} and the width is {{{12cm}}}

{{{V[w]=2*7*12}}} (assuming there is no ball in)

since the depth of water in the tray is {{{h[1]=2cm}}} and radius of the metal ball is {{{r=2cm}}}, means half of the metal sphere is in the water

so, we need to deduct {{{1/2}}}  volume of the metal sphere from {{{V[w]=2*7*12}}}


{{{V[w]=2*7*12-V[s]/2}}}

{{{1/2}}}  volume of the ball is: 

{{{V[s]/2=((4/3)*pi*2^3)/2=((4/3)*pi*2^2)=(16/3)*pi}}}.....{{{pi=3.14}}}, round it to whole number

{{{V[w]=168-(16/3)*3}}}

{{{V[w]=168-16}}}

{{{V[w]=152}}}


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


the volume  the volume of water needed to be poured into the tray so that the metal sphere is just submerged is same as in (a)

{{{V[w]=152}}}-> the volume of water that covers another half of the metal sphere