# How to calculate the height of water displaced by a metal sphere droppped in a cylinder of water?I would like to see the mathematical working

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To determine the change in height of the water in a cylinder, we need to use the formula for the volume of sphere and cylinder.

Volume of sphere is:

`V_s= 4/3piR^3` where R is the radius

Initila volume of cylinder is:

`V_c=Bh_o` where B is the area of the base

`h_o` is the height of the water

When the sphere is dropped into the cylinder, the volume of the cylinder increases. The total volume is:

`V_T= V_s + V_c`

`V_T = 4/3piR^3 + Bh_o`

Note that the total volume is equal to `Bh_f` where `h_f` is the new height of the water. And B, area of the base, remains the same.

`Bh_f = 4/3piR^3 + Bh_o`

`Bh_f - Bh_o = 4/3piR^3`

`B(h_f-h_o) = 4/3 piR^3`

`h_f-h_o = (4/3piR^3)/B`

`Deltah = (4piR^3)/(3B)`

**Hence, the height of water displaced by a metal sphere dropped into the cylindepr is `Delta h =(4piR^3)/(3B)` .**