A cube of $9\ cm$ edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are $15\ cm$ and $12\ cm$, find the rise in water level in the vessel.
Given:
A cube of $9\ cm$ edge is immersed completely in a rectangular vessel containing water.
The dimensions of the base are $15\ cm$ and $12\ cm$.
To do:
We have to find the rise in the water level in the vessel.
Solution:
Edge of the cube $= 9\ cm$
This implies,
Volume of the cube $= (9)^3$
$= 729\ cm^3$
Length of the vessel $(l) = 15\ cm$
Breadth of the vessel $(b) = 12\ cm$
Therefore,
Height of the water $(h)=\frac{\text { volume of water }}{l \times b}$
$=\frac{729}{15 \times 12}$
$=\frac{729}{180}$
$=\frac{81}{20}$
$=4.05 \mathrm{~cm}$
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