Question

3. The diameter of a spherical balloon shrinks to one-half of its original size.How does this affect the volume?Hint: Test two scenarios and compare the volumes! Show your work!!A. The volume is cut in halfB. The volume doublesC. The volume is 1/8 the original volumeD. The volume is 1/4 the original volume

Answer 1

C. The volume is 1/8 the original volume

Step-by-step explanation:

Step 1. To find how the volume changes if the diameter is reduced to one-half of the original size, we will test two scenarios:

• In the first scenario, the diameter will be 2, and in the second scenario, the diameter will be one-half of 2 which is 1.

We will find the volume for these two cases and see how it changes.

Step 2. For a diameter of d=2.

If the diameter is 2, the radius is:

`\begin{gathered} r=d/2 \\ r=2/2 \\ r=1 \end{gathered}`

Using the volume formula for a sphere:

`V=(4\pi)/(3)r^3`

In this case:

`V=(4\pi)/(3)(1)^3`

We will call this volume, V1:

`V_1=(4\pi)/(3)`

Step 3. Now we will find the volume for the second case in which the diameter is d=1.

The radius is:

`\begin{gathered} r=d/2 \\ r=1/2 \\ \end{gathered}`

Using the same formula for the volume:

`V=(4\pi)/(3)((1)/(2))^3`

Solving the operations:

`V=(4\pi)/(3)((1)/(8))^`

We will call this V2:

`V_2=(4\pi)/(3)((1)/(8))^`

Step 4. As you can see, the first part of the previous expression is V1:

`\begin{gathered} V_(1)=(4\pi)/(3) \\ V_2=(4\pi)/(3)((1)/(8)) \end{gathered}`

Therefore:

`V_2=V_1((1)/(8))`

The second volume is the first volume multiplyed by 1/8 ⇒ it is 1/8 of the original volume.

Answer:

C. The volume is 1/8 the original volume