To solve this, we use the concept of similarity for the two jars. If the jars are mathematically similar, their corresponding dimensions are proportional, and the volumes relate to the cube of the ratio of their corresponding heights.

Given:

• Volume of the larger jar ($V_1$) = 1.6 L
• Height of the larger jar ($h_1$) = 30 cm
• Height of the smaller jar ($h_2$) = 24 cm
• Volume of the smaller jar ($V_2$) = ?

Step 1: Ratio of heights

$\text{Height ratio} = \frac{h_2}{h_1} = \frac{24}{30} = \frac{4}{5}$

Step 2: Ratio of volumes

Since the volumes are proportional to the cube of the height ratio: $\frac{V_2}{V_1} = \left(\frac{h_2}{h_1}\right)^3 = \left(\frac{4}{5}\right)^3 = \frac{64}{125}$

Step 3: Solve for $V_2$

$V_2 = V_1 \times \frac{64}{125}$ Substitute $V_1 = 1.6 \, \text{L}$: $V_2 = 1.6 \times \frac{64}{125} = \frac{102.4}{125} = 0.8192 \, \text{L}$

Final Answer:

The volume of the smaller jar is approximately 0.82 L.

Would you like further details or a breakdown of the calculations?

Related Questions:

1. How does the cube of the ratio affect the volumes of similar objects?
2. Can this method apply to shapes other than jars?
3. What if the smaller jar's height were halved? How would its volume change?
4. How do we verify if two objects are mathematically similar?
5. How would the surface area ratio compare for these jars?

Tip:

When working with similar shapes, remember that linear dimensions scale linearly, areas scale by the square of the ratio, and volumes scale by the cube of the ratio!