Question

Maher is trying to find the density of a plastic toy. The toy’s mass is 160 g.

Maher placed the toy in a graduated cylinder that has 70 ml of water, the water level increased
till 150 ml.
a) Find the toy’s volume.

Answer 1

The volume of the toy is 80 ml.

To find the volume of the plastic toy, Maher can use the principle of buoyancy. When an object is placed in a fluid, it will displace an amount of fluid equal to its own volume. The volume of the displaced fluid can be measured and used to calculate the volume of the object.

In this case, Maher has placed the toy in a graduated cylinder filled with water, and he has observed that the water level increased from 70 ml to 150 ml. This means that the toy displaced 150 - 70 = 80 ml of water.

The volume of the toy is equal to the volume of the displaced water, so the toy's volume is 80 ml. This is the volume of the toy when it is completely submerged in water.

It's important to note that the volume of an object can change depending on its temperature, pressure, and other factors. To get an accurate measurement of the volume of the toy, Maher should make sure to measure the volume of the displaced water carefully and under controlled conditions.

Answer 2

Volume of the toy:
`80\; {\rm mL}`.

Average density of the toy:
`2\; {\rm g\cdot mL^(-1)}` (or equivalently,
`2\; {\rm g \cdot cm^(-3)}`.)

Step-by-step explanation:

The graduated cylinder initially measures the volume of water in this cylinder:

`V(\text{water}) = 70\; {\rm mL}`.

Assume that the toy is submerged in the water. The graduated cylinder would then measure the volume of the water and the toy, combined:

`V(\text{water}) + V(\text{toy}) = 150\; {\rm mL}`.

Rearrange to find the volume of the toy:

`\begin{aligned}V(\text{toy}) &= 150\; {\rm mL} - V(\text{water}) \\ &= 150\; {\rm mL} - 70\; {\rm mL} \\ &= 80\; {\rm mL}\end{aligned}`.

To find the average density of this toy, divide mass by volume:

`\begin{aligned}(\text{average density}) &= \frac{(\text{mass})}{(\text{volume})} \\ &= \frac{160\; {\rm g}}{80\; {\rm mL}} \\ &= 2\; {\rm g\cdot mL^(-1)}\end{aligned}`.