A cylindrical container with diameter of base $$56\ cm$$ contains sufficient water to submerge a rectangular solid of iron with dimensions $$(32\ cm\times 22\ cm\times 14\ cm)$$. Find the rise in the level of water when the solid is completely submerged
Consider $$h\ cm$$ as the rise in level of water
We know that
Volume of cylinder of height $$h$$ and base radius $$28\ cm=$$ volume of rectangle iron solid
By substituting the values
$$\dfrac{22}{7}\times 28^2 \times h=32\times 22\times 14$$
On further calculation
$$22\times 28\times 4\times h=32\times 22\times 14$$
So we get
$$h=\dfrac{(32\times 22\times 14)}{ (22\times 28\times 4)}$$
By division
$$h=4cm$$
Therefore, the rise in the level of water when the solid is completely submerged is $$4 cm$$.