
The density of material in CGS system of units is $3 g / \mathrm{cm}^{3} .$ In a system of units in which unit of length is $10 \mathrm{~cm}$ and unit of mass is $100 \mathrm{~g},$ the value of density of material will be
A.$0.4$
B.30
C.400
D.0.04
Answer
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Hint: First of all we will convert the $3 \mathrm{~g}$ mass into the $100 \mathrm{~g}$ scale and the $1 \mathrm{~cm}$ length into the $10 \mathrm{~cm}$ scale to find the equivalent number of units. In order to find the density in the new system of units, we will then substitute the values required in the density equation.
Formula used:
$\rho=\dfrac{m}{v}$
Complete answer:
In the question given, the following data is provided:
Material density is given as $3 \mathrm{~g} \mathrm{~cm}^{-3}$ in the CGS system of units.
If the unit of length is $10 \mathrm{~cm}$ and the unit of mass is $100 \mathrm{~g}$, we are asked to find the density value of the material. To begin with as mentioned in the question, we will first need to convert the units given into the new system of units. The second case says $100 \mathrm{~g}$ is the unit of mass and $10 \mathrm{~cm}$ is the unit of length.
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.
$\therefore \rho=\dfrac{3 \mathrm{~g}}{1 \mathrm{~cm}^{3}} \ldots \ldots(1)$
When the mass is $100 \mathrm{~g}$, then the mass of $4 \mathrm{~g}$ in a new system of units is equivalent to $\dfrac{3}{100}$ units. Again, when the unit of length is $10 \mathrm{~cm},$ then $1 \mathrm{~cm}$ in a new system of units is equivalent to $\dfrac{1}{10}$ units.
Now we can also change the values in the equation (1), which is shown below:
$\rho=\dfrac{3 \times 1 \mathrm{~g}}{(1 \mathrm{~cm})^{3}}$
$\Rightarrow \rho=\dfrac{3 \times \dfrac{1}{100}}{\left(\dfrac{1}{10}\right)^{3}}$
$\Rightarrow \rho=\dfrac{3}{100} \times 10^{3}$
$\Rightarrow \rho=30 \mathrm{units}$
The value of density of material will be $\Rightarrow \rho=30 \mathrm{units}$
The correct option (B)
Note:
In the question given, the following data is provided: Material density is given as $4 \mathrm{~g} \mathrm{~cm}^{-3}$ in the CGS system of units. If the unit of length is $10 \mathrm{~cm}$ and the unit of mass is $100 \mathrm{~g}$, we are asked to find the density value of the material.
Formula used:
$\rho=\dfrac{m}{v}$
Complete answer:
In the question given, the following data is provided:
Material density is given as $3 \mathrm{~g} \mathrm{~cm}^{-3}$ in the CGS system of units.
If the unit of length is $10 \mathrm{~cm}$ and the unit of mass is $100 \mathrm{~g}$, we are asked to find the density value of the material. To begin with as mentioned in the question, we will first need to convert the units given into the new system of units. The second case says $100 \mathrm{~g}$ is the unit of mass and $10 \mathrm{~cm}$ is the unit of length.
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.
$\therefore \rho=\dfrac{3 \mathrm{~g}}{1 \mathrm{~cm}^{3}} \ldots \ldots(1)$
When the mass is $100 \mathrm{~g}$, then the mass of $4 \mathrm{~g}$ in a new system of units is equivalent to $\dfrac{3}{100}$ units. Again, when the unit of length is $10 \mathrm{~cm},$ then $1 \mathrm{~cm}$ in a new system of units is equivalent to $\dfrac{1}{10}$ units.
Now we can also change the values in the equation (1), which is shown below:
$\rho=\dfrac{3 \times 1 \mathrm{~g}}{(1 \mathrm{~cm})^{3}}$
$\Rightarrow \rho=\dfrac{3 \times \dfrac{1}{100}}{\left(\dfrac{1}{10}\right)^{3}}$
$\Rightarrow \rho=\dfrac{3}{100} \times 10^{3}$
$\Rightarrow \rho=30 \mathrm{units}$
The value of density of material will be $\Rightarrow \rho=30 \mathrm{units}$
The correct option (B)
Note:
In the question given, the following data is provided: Material density is given as $4 \mathrm{~g} \mathrm{~cm}^{-3}$ in the CGS system of units. If the unit of length is $10 \mathrm{~cm}$ and the unit of mass is $100 \mathrm{~g}$, we are asked to find the density value of the material.
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