Question

If a ball bearing is dropped into a beaker of water 3 cm deep, by how much does the water rise? (1) The radius of the ball bearing is \(7 \mathrm{~mm}\) (2) The radius of the beaker is \(5 \mathrm{~cm}\) A. 1 alone, not 2 alone B. 2 alone, not 1 alone C. 1 and 2 together (need both) D. 1 alone or 2 alone E. 1 and 2 together are not sufficient

Short answer

C. 1 and 2 together (need both)

Step-by-step solution

Understand the problem

The exercise asks for how much the water level will rise when a ball bearing is dropped into a beaker of water. We are given the radius of the ball bearing and the radius of the beaker. This involves calculating the volume of the ball bearing and comparing it to the cross-sectional area of the beaker.

Calculate the volume of the ball bearing

The ball bearing is a sphere, so use the formula for the volume of a sphere: \[V_{sphere} = \frac{4}{3} \pi r^3\]Given that the radius of the ball bearing is 7 mm, convert this to cm first: \[7 \text{ mm} = 0.7 \text{ cm}\]Substitute the radius into the volume formula: \[V_{sphere} = \frac{4}{3} \pi (0.7)^3\]Calculate the volume.

Calculate the cross-sectional area of the beaker

The beaker is cylindrical. Given the radius of the beaker is 5 cm, find the cross-sectional area using the formula for the area of a circle: \[A_{circle} = \pi r^2\]Substitute the radius of the beaker into the formula: \[A_{circle} = \pi (5)^2 = 25\pi\ \text{cm}^2\]

Determine the rise in water level

To find the rise in water level, divide the volume of the ball bearing by the cross-sectional area of the beaker:\[\text{Water rise} = \frac{V_{sphere}}{A_{circle}}\]Substitute the calculated volume and area into the formula to find the rise in the water level.

Evaluate the provided options

Since both the radius of the ball bearing and the radius of the beaker are needed to solve for the rise in water level, analyze the provided options in the original exercise. Based on our calculations, both values are necessary.

Key concepts

volume calculation

Volume calculation is a fundamental concept in geometry. When solving problems involving three-dimensional objects, finding the volume is often essential. In the given exercise, the volume of the ball bearing, which is a sphere, needs to be calculated. The formula for the volume of a sphere is: \(V_{sphere} = \frac{4}{3} \pi r^3\). First, you need to convert the radius from millimeters to centimeters (since consistent units are crucial): 7 mm = 0.7 cm. Then, plug this value into the volume formula: \(V_{sphere} = \frac{4}{3} \pi (0.7)^3\). It’s important to remember that knowing the volume helps in understanding how much space the object being submerged can displace. This displacement is what causes the rise in water level in the beaker.

geometry

Geometry involves the study of shapes and their properties. In solving the problem, understanding the geometry of both the ball bearing (a sphere) and the beaker (a cylinder) is essential. The sphere's volume formula \(V_{sphere} = \frac{4}{3} \pi r^3\) gives the amount of space it occupies. For the beaker, the important geometric property is its cross-sectional area, which is calculated using the formula for the area of a circle, \(A_{circle} = \pi r^2\). Given the radius of the beaker as 5 cm, the area is: \(A_{circle} = \pi (5)^2 = 25\pi \ \text{cm}^2\). Together, these geometric properties are used to solve for the rise in water level when the ball bearing is dropped into the beaker.

problem-solving steps

Solving geometry problems involves clearly defined steps. Here’s the approach used in this exercise:

• Understand the problem: Identify what is being asked and the given information.
• Calculate the volume of the ball bearing using \(V_{sphere} = \frac{4}{3} \pi r^3\).
• Convert units if necessary (from mm to cm in this case).
• Calculate the cross-sectional area of the beaker using \(A_{circle} = \pi r^2\).
• Use the geometric properties to determine the rise in water level by dividing the volume of the ball by the beaker’s cross-sectional area.

Carefully following each step ensures the problem is tackled methodically and accurately.

radius conversion

Radius conversion is crucial for consistency in units, especially when dealing with measurements in different units such as mm and cm. In this exercise, the radius of the ball bearing is given in millimeters (7 mm), but it needs to be converted to centimeters. The conversion factor is that 1 cm = 10 mm. Thus, 7 mm is equal to 0.7 cm. This consistent unit helps when plugging into the formula: \(V_{sphere} = \frac{4}{3} \pi (0.7)^3\). Keeping consistent units throughout your calculations avoids errors and makes the problem easier to solve.