Math Problem Statement

. A cuboid of base area P sq units is filled with water upto a height of Q units. A sphere of volume R cu units is dropped into the cuboid such that it is completely submerged. A representation of the submerged sphere is shown below.

Solution

To understand this problem clearly, we need to analyze the situation step by step.

Given:

  • Cuboid with a base area of PP square units.
  • Water filled up to a height of QQ units.
  • Sphere of volume RR cubic units is completely submerged in the water inside the cuboid.

Objective:

We need to determine the effect of the sphere being submerged into the cuboid. Typically, this results in water displacement and a rise in the water level within the cuboid.

Approach:

  1. Water initially in the cuboid: The volume of water already in the cuboid before the sphere is submerged can be found by: Vwater=P×Q(cubic units)V_{\text{water}} = P \times Q \quad \text{(cubic units)} where PP is the base area, and QQ is the initial height of water.

  2. Volume of the sphere: The volume of the sphere is given by RR cubic units.

  3. New height of water after submersion: After the sphere is submerged, the water level will rise due to the displacement of water. The displaced water volume equals the volume of the sphere, RR. Let the new height of water in the cuboid be HH.

    The new volume of water in the cuboid is: Vnew=P×HV_{\text{new}} = P \times H

    Since the additional volume of water is due to the sphere, the total volume of water after submersion is the initial water volume plus the sphere's volume: P×H=P×Q+RP \times H = P \times Q + R

  4. Solving for the new height: Rearranging the equation to solve for HH: H=Q+RPH = Q + \frac{R}{P}

Result:

After the sphere is submerged, the new height of the water in the cuboid will be: H=Q+RPH = Q + \frac{R}{P}

This equation shows that the new water height increases by RP\frac{R}{P}, which is the volume of the sphere divided by the base area of the cuboid.

Would you like further details or clarifications?

Here are 5 related questions to consider:

  1. How would the result change if the cuboid had a slanted side?
  2. What happens if the sphere isn't completely submerged?
  3. How would the water height change if multiple spheres are submerged?
  4. How can we calculate the volume of a sphere given its radius?
  5. How can this problem be generalized to irregular shapes?

Tip: When working with volumes and areas, always ensure the units are consistent across the problem.

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Math Problem Analysis

Mathematical Concepts

Volume
Water Displacement
Geometry
Algebra

Formulas

V = P × Q
H = Q + R / P

Theorems

Principle of Archimedes (displacement of water)

Suitable Grade Level

Grades 9-11