# Diameter 0.8 m height 1.2 m floats in water

• darkmagic
In summary, the problem involves a cone with a diameter of 0.8 m and a height of 1.2 m, which is floating in water. The question is how far will the cone sink when placed in a liquid with a density of 1360 kg/m^3. The solution involves using the buoyant force equation and writing the diameter as a function of the length. The attempt at a solution involves using a triangle to determine the submerged volume, but this approach is incorrect. Instead, the correct equation is 0.4/1.2 = x/L.

## Homework Statement

Given:
diameter 0.8 m
height 1.2 m
floats in water
how far will it sink in a liquid (density =1360 kg/m^3)

Buoyant force

## The Attempt at a Solution

In the attachment
I'm stuck on how to find the volume of the submerged cone in the liquid

#### Attachments

• pr 9.jpg
6 KB · Views: 341
• solution.jpg
12.6 KB · Views: 351
You are doing well, so far. Hint 1: Write d (or r, or x) as a function of L.

This what I think: since the liquid is more dense than water, then the cone will float a little. From triangle, 0.8 becomes 0.8+y and 0.4 becomes 0.4-y. The ratio and proportion becomes 0.4/1.2 = x/(0.8 + y). Am I correct? or the cone will sink a little?

No. Hint 2: 0.4/1.2 = x/L.

To find the volume of the submerged cone, you can use the formula V = (1/3)πr²h, where V is the volume, r is the radius of the base (diameter divided by 2), and h is the height of the cone. In this case, the radius would be 0.4 m and the height would be the distance the cone sinks in the liquid, which we will call x.

Next, you can use the formula for the buoyant force, Fb = ρVg, where Fb is the buoyant force, ρ is the density of the liquid, V is the volume of the submerged object, and g is the acceleration due to gravity (9.8 m/s²). We can set this equal to the weight of the cone, which is equal to its mass multiplied by the acceleration due to gravity, mg.

So, we have ρVg = mg. Substituting in the values we know, we get (1360 kg/m³)(π(0.4m)²x)(9.8 m/s²) = (m)(9.8 m/s²). We can solve for x by dividing both sides by (1360 kg/m³)(π(0.4m)²)(9.8 m/s²), giving us x = m/π.

Since we know the height of the cone is 1.2 m, we can set x equal to 1.2 m and solve for m. This will give us the mass of the cone, which we can then use to find the volume of the submerged cone. Once we have the volume, we can use the formula for the volume of a cone to find the radius of the base.

I hope this helps you find the solution to your problem. If you have any further questions, please let me know.

## 1. What is the volume of a diameter 0.8 m height 1.2 m float in water?

The volume of the float can be calculated using the formula V = πr²h, where r is the radius (half the diameter) and h is the height. In this case, the radius is 0.4 m and the height is 1.2 m, so the volume of the float is approximately 0.48 cubic meters.

## 2. How much weight can a diameter 0.8 m height 1.2 m float hold in water?

The weight that a float can hold in water depends on its density, which is determined by the material it is made of. If we assume the float has a density similar to that of water (1 g/cm³), then it can hold approximately 480 kg of weight (0.48 cubic meters x 1000 g/cm³).

## 3. Will the diameter 0.8 m height 1.2 m float sink or float in water?

The float will most likely float in water, as its volume is larger than the weight of the object itself. However, if the float is made of a material with a higher density than water, it may sink.

## 4. How does the diameter and height of the float affect its buoyancy in water?

The diameter and height of the float do not directly affect its buoyancy. The buoyancy of an object is determined by its volume and the density of the liquid it is submerged in. However, a larger float may displace more water and have a greater buoyant force.

## 5. What is the purpose of a diameter 0.8 m height 1.2 m float in water?

A float of this size could be used for various purposes, such as marking a swimming area, supporting a dock or platform, or providing buoyancy for a watercraft. The specific purpose would depend on the design and use of the float.