Answer
Verified
396.9k+ views
Hint: Density is defined as the mass of a substance per unit volume. Thus, if we know the mass and volume of a specific substance, we can find out its density by dividing the mass value by the volume. It is expressed generally in the units of grams per liter.
\[\text{Density }\left( \text{d} \right)=\dfrac{\text{Mass }\left( \text{g} \right)}{\text{Volume }\left( \text{L} \right)}\]
Complete answer:
Any substance that occupies space has a mass and volume. The term which is determined as the mass of a substance per unit volume is known as the density of the substance. It is a characteristic property of any matter.
Usually, solids are the densest, liquids are intermediate and gases are the least dense state of matter.
Let us now take a brief look at how to calculate the density of a rubber eraser.
1 – First we have to find the mass of the eraser by weighing it on the weighing machine. From here, we get the value of mass in grams.
2 – Now finding out the volume of the eraser is a bit difficult. So, to make our work easier, a cylindrical vessel full of water is taken. The initial height of the water is marked $\left( {{\text{h}}_{1}} \right)$ and noted. Then, solid rubber is immersed in the water. Due to this rubber will displace some amount of water and the level of water rises in the vessel. The new height of the water level is noted $\left( {{\text{h}}_{2}} \right)$.
So, the volume of the rubber eraser will be equal to the difference between the volume of the cylindrical vessel filled with water before and after the immersion of rubber in water.
\[\begin{align}
& \because \text{Volume of cylindrical vessel}=\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h} \\
& \therefore \text{Volume of rubber eraser}=\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\Delta \text{h} \\
& \text{Where }\Delta \text{h}={{\text{h}}_{\text{2}}}-{{\text{h}}_{\text{1}}} \\
& \text{And r}=\dfrac{\text{Diameter of vessel}}{2}=\dfrac{\text{d}}{2} \\
& \Rightarrow \\
\end{align}\]
3 – As density is equal to mass divided by volume. So, we can write –
\[\text{Density of rubber eraser (d)}=\dfrac{\text{m}}{\text{ }\!\!\pi\!\!\text{ }{{\left( \dfrac{\text{d}}{2} \right)}^{\text{2}}}\left( {{\text{h}}_{\text{2}}}-{{\text{h}}_{\text{1}}} \right)}\]
Hence, in this way, we can determine the density of a rubber eraser.
Note:
There can be however some error in case the rubber has pores and it absorbs some of the water. This in turn will effectively displace the air inside the rubber and increase its real density. So, all the measurements need to be done very carefully.
\[\text{Density }\left( \text{d} \right)=\dfrac{\text{Mass }\left( \text{g} \right)}{\text{Volume }\left( \text{L} \right)}\]
Complete answer:
Any substance that occupies space has a mass and volume. The term which is determined as the mass of a substance per unit volume is known as the density of the substance. It is a characteristic property of any matter.
Usually, solids are the densest, liquids are intermediate and gases are the least dense state of matter.
Let us now take a brief look at how to calculate the density of a rubber eraser.
1 – First we have to find the mass of the eraser by weighing it on the weighing machine. From here, we get the value of mass in grams.
2 – Now finding out the volume of the eraser is a bit difficult. So, to make our work easier, a cylindrical vessel full of water is taken. The initial height of the water is marked $\left( {{\text{h}}_{1}} \right)$ and noted. Then, solid rubber is immersed in the water. Due to this rubber will displace some amount of water and the level of water rises in the vessel. The new height of the water level is noted $\left( {{\text{h}}_{2}} \right)$.
So, the volume of the rubber eraser will be equal to the difference between the volume of the cylindrical vessel filled with water before and after the immersion of rubber in water.
\[\begin{align}
& \because \text{Volume of cylindrical vessel}=\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h} \\
& \therefore \text{Volume of rubber eraser}=\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\Delta \text{h} \\
& \text{Where }\Delta \text{h}={{\text{h}}_{\text{2}}}-{{\text{h}}_{\text{1}}} \\
& \text{And r}=\dfrac{\text{Diameter of vessel}}{2}=\dfrac{\text{d}}{2} \\
& \Rightarrow \\
\end{align}\]
3 – As density is equal to mass divided by volume. So, we can write –
\[\text{Density of rubber eraser (d)}=\dfrac{\text{m}}{\text{ }\!\!\pi\!\!\text{ }{{\left( \dfrac{\text{d}}{2} \right)}^{\text{2}}}\left( {{\text{h}}_{\text{2}}}-{{\text{h}}_{\text{1}}} \right)}\]
Hence, in this way, we can determine the density of a rubber eraser.
Note:
There can be however some error in case the rubber has pores and it absorbs some of the water. This in turn will effectively displace the air inside the rubber and increase its real density. So, all the measurements need to be done very carefully.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE