Question
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.
✓
Answer
Volume of the cube $= S^3 = 9^3 = 729\ cm^3$
Area of the base $= l \times b = 15 \times 12 = 180\ cm^2$
Rise in water level $=\frac{\text{Volume of the cube }}{\text{Area of base of rectangular vessel}}$
$=\frac{729}{180}$
$= 4.05\text{ cm}.$
Area of the base $= l \times b = 15 \times 12 = 180\ cm^2$
Rise in water level $=\frac{\text{Volume of the cube }}{\text{Area of base of rectangular vessel}}$
$=\frac{729}{180}$
$= 4.05\text{ cm}.$
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