A cube of 9 cm edge is immersed completely in a rectangular vessel containing water.
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.

Solution:

Volume of the cube $=\mathrm{s}^{3}=9^{3}=729 \mathrm{~cm}^{3}$

Area of the base $=\mathrm{l}^{*} \mathrm{~b}=15^{*} 12=180 \mathrm{~cm}^{2}$

Rise in water level $=\frac{\text { Volume of the cube }}{\text { Area of base of rectangular vessel }}$

= 729/180

= 4.05 cm

 

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