Question

The density of material in CGS system of units is \[4\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}\] . In a system of units in which unit of length is \[10\,cm\] and unit of mass is \[100\,{\text{g}}\] , the value of density of material will be:
(A) \[0.4\]
(B) \[40\]
(C) \[400\]
(D) \[0.04\]

Answer 1

Hint:First of all, we will convert the mass of \[4\,{\text{g}}\] into the scale of \[100\,{\text{g}}\] and the length of \[1\,{\text{cm}}\] into the scale of \[10\,{\text{cm}}\] to find the number of units equivalent. Then we will substitute the required values in the equation of density to find the density in the new system of units.

Complete step by step solution:
In the given question, we are supplied the following data:
The density of material in the CGS system of units is given as \[4\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}\].
We are asked to find the value of density of the material if the unit of length is \[10\,{\text{cm}}\] and the unit of mass is \[100\,{\text{g}}\].
To begin with, we will first need to convert the given units into the new system of units as mentioned in the question.
The second case says the unit of mass is \[100\,{\text{g}}\] and the unit of length is \[10\,{\text{cm}}\] .
We know,
\[\rho = \dfrac{m}{v}\]
Where,
\[\rho \] indicates the density of a material.
\[m\] indicates the mass of the material.
\[v\] indicates the volume of the material.
\[\therefore \rho = \dfrac{{4\,{\text{g}}}}{{1\,{\text{c}}{{\text{m}}^3}}}\] …… (1)
When the mass is \[100\,{\text{g}}\] , then mass of \[4\,{\text{g}}\] in new system of units is equivalent to \[\dfrac{4}{{100}}\] units.
Again, when unit of length is \[10\,{\text{cm}}\] , then \[1\,{\text{cm}}\] in new system of units is equivalent to \[\dfrac{1}{{10}}\] units.
Now, in the equation (1), we can modify the values as well, which is shown below:
$\rho = \dfrac{{4 \times 1\,{\text{g}}}}{{{{\left( {1\,{\text{cm}}} \right)}^3}}} \\
\Rightarrow \rho = \dfrac{{4 \times \dfrac{1}{{100}}}}{{{{\left( {\dfrac{1}{{10}}} \right)}^3}}} \\
\Rightarrow \rho = \dfrac{4}{{100}} \times {10^3} \\
\Rightarrow \rho = 40\,{\text{units}} \\$

Hence, the value of density of material is \[40\,{\text{units}}\].The correct option is (b).

Additional information:
Density is a per-volume indicator of mass. An object's average density is proportional to its total mass, which is divided by its total volume. There would be less volume for an object made of a comparatively dense material (such as iron) than for an object of equivalent mass made of a less dense material (such as water).

Note:While solving the problem, it is important to remember that the density of the material is already given. All we need to find is the density of the same material in the new system of units, where mass of \[100\,{\text{g}}\] forms one unit and length of \[10\,{\text{cm}}\] forms one unit. Most of the students tend to make mistakes by just dividing \[100\,{\text{g}}\] by the cube of \[10\,{\text{cm}}\] which is completely wrong.