D AY 152 – A PPLICATION OF VOLUME OF A SPHERE
I NTRODUCTION We encounter spherical objects in real life such as tennis balls, spherical water tanks, round- bottomed flasks, and beads. Anything that is contained in a spherical object has a volume equivalent to the volume of that object or a fraction of it. In this lesson, we will discuss the application of volume of a sphere.
V OCABULARY Sphere This is a set of points equidistant from a single point in three dimensional space.
To discuss, the real life applications of the volume a sphere, we will start by reminding ourselves the formula of calculating the volume of the sphere. Consider the sphere below centered at O. 3 𝑗𝑜 O 4 The volume of the sphere is given by the formula, 3 𝜌𝑠 3 . 4 3 × 3.142 × 3 3 = 113.1 𝑗𝑜 3 𝑊 =
The volume of the sphere has very many applications We will discuss some of the applications of the volume of the sphere. Spherical tanks 1. When manufacturing spherical tanks of a specific volume, the manufacturers must ensure that the radius of the tank gives the required volume. For instance, if a manufacturer wants a spherical tank that can hold a volume of 66 cubic yards, then he will first calculate the radius of the tank as follows. 4 3 𝜌𝑠 3 ⟹ 𝑠 3 = 3𝑊 3×66 V = 4𝜌 = 4×3.142 = 15.75 3 15.75 = 2.5 yards 𝑠 =
Thus the manufacturer will ensure that the tank has a radius of 2.5 yards. At times a tank might not be full of the liquid that is contained inside. In this case the volume of the liquid is a fraction of the volume of the whole tank. Example 1 A spherical water tank is half way full of water. If the tank has a radius of the tank is 5 𝑔𝑢, calculate the volume of the water contained in the tank.
Solution Since the water is half way full, its volume is given by 1 4 3 𝜌𝑠 3 𝑊 = 2 × 1 4 3 × 3.142 × 5 3 𝑊 = 2 × = 261.8 cubic feet
2. Estimation of volume of gas molecules and their radii. In the estimation of the volume of a gas molecule, the molecule is assumed to be spherical. The gas molecules are pumped at a certain rate, say 1000,000 molecules per second, into a container whose volume is known. Then the volume of the container is divided by the number of molecules inside to get the volume of a single molecule. Example 2 In the estimation of the diameter of the oxygen molecule, oxygen gas was pumped into a container of volume, 10 cubic inches at a rate of 100,000,000 molecules per second. If it took 3 𝑡𝑓𝑑𝑝𝑜𝑒𝑡 to fill the container. Calculate the diameter of the oxygen molecule.
Solution Number of molecules in the container = 100000000 × 3 = 300000000 10 300000000 = 3.333 × 10 −8 𝑗𝑜 3 Volume of one molecule 3×3.333×10 −8 = 7.957 × 10 −9 𝑗𝑜 Radius of the molecule = 4×3.142 Diameter of the molecule = 2 × 7.957 × 10 −9 𝑗𝑜 = 1.591 × 10 −8 𝑗𝑜
3. Manufacture of spherical objects Objects like ball bearings, beads and snooker balls are manufactured from molten substances such as plastic and steel. Their size determines the amount of raw material to be prepared or purchased to manufacture a number of these objects.
Example 3 Calculate the amount of molten steel needed to make 1500 ball bearings each with a radius of 0.2 𝑗𝑜. Solution 4 3 × 3.142 × 0.2 3 = 0.0335 𝑗𝑜 3 Volume of one ball bearing = Volume of 1500 ball bearings = 0.0335 × 1500 = 50.25 𝑗𝑜 3
HOMEWORK A student wanted to find out the amount of plastic contained in a snooker ball. He measured the diameter of the ball with a caliper and found that the diameter of the ball was 1.82 𝑗𝑜 . Calculate the volume of the plastic contained in snooker ball.
A NSWERS TO HOMEWORK 3.157 𝑗𝑜 3
THE END
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