Solveeit Logo

Question

Question: The density of material in CGS system of units is \[4\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}\] ....

The density of material in CGS system of units is 4gcm34\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}} . In a system of units in which unit of length is 10cm10\,cm and unit of mass is 100g100\,{\text{g}} , the value of density of material will be:
(A) 0.40.4
(B) 4040
(C) 400400
(D) 0.040.04

Explanation

Solution

First of all, we will convert the mass of 4g4\,{\text{g}} into the scale of 100g100\,{\text{g}} and the length of 1cm1\,{\text{cm}} into the scale of 10cm10\,{\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 4gcm34\,{\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 10cm10\,{\text{cm}} and the unit of mass is 100g100\,{\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 100g100\,{\text{g}} and the unit of length is 10cm10\,{\text{cm}} .
We know,
ρ=mv\rho = \dfrac{m}{v}
Where,
ρ\rho indicates the density of a material.
mm indicates the mass of the material.
vv indicates the volume of the material.
ρ=4g1cm3\therefore \rho = \dfrac{{4\,{\text{g}}}}{{1\,{\text{c}}{{\text{m}}^3}}} …… (1)
When the mass is 100g100\,{\text{g}} , then mass of 4g4\,{\text{g}} in new system of units is equivalent to 4100\dfrac{4}{{100}} units.
Again, when unit of length is 10cm10\,{\text{cm}} , then 1cm1\,{\text{cm}} in new system of units is equivalent to 110\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 40units40\,{\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 100g100\,{\text{g}} forms one unit and length of 10cm10\,{\text{cm}} forms one unit. Most of the students tend to make mistakes by just dividing 100g100\,{\text{g}} by the cube of 10cm10\,{\text{cm}} which is completely wrong.