Question

The density of a substance is 4 g/cc. In a system in which the unit of length is 5 cm and the unit of mass is 20 g, the density of the substance is:

• A. 16 units
• B. 40 units
• C. 25 units
• D. 50 units

Hint:

When converting the numerical value of a physical quantity between two systems of units, use the relationship: $n_1 [U_1] = n_2 [U_2]$, where $n$ is the numerical value and $[U]$ represents the units in each system. For density ($\rho$), the unit is proportional to $\frac{\text{Mass}}{(\text{Length})^3}$.

Answer 1

The Correct Option is C

Step 1: Identify the density in the CGS system.
The density of the substance is given as 4 g/cc, which is in the CGS system of units (grams for mass and centimeters for length). $\rho_{CGS} = 4 \frac{\text{g}}{\text{cm}^3}$
Step 2: Define the units in the new system.
In the new system:
Unit of length ($L_1$) = 5 cm
Unit of mass ($M_1$) = 20 g

Step 3: Express the density in terms of the new units.
Let the numerical value of the density in the new system be $n$. The dimensions of density are [Mass]/[Length]$^3$.
$\rho_{new} = n \frac{\text{Unit of mass}}{(\text{Unit of length})^3} = n \frac{20 \text{ g}}{(5 \text{ cm})^3} = n \frac{20}{125} \frac{\text{g}}{\text{cm}^3} = n \frac{4}{25} \frac{\text{g}}{\text{cm}^3}$
Step 4: Equate the density in both systems.
Since the density of the substance remains the same, we can equate the expressions from Step 1 and Step 3:
$4 \frac{\text{g}}{\text{cm}^3} = n \frac{4}{25} \frac{\text{g}}{\text{cm}^3}$
Step 5: Solve for the numerical value in the new system ($n$).
Divide both sides by $\frac{4}{25} \frac{\text{g}}{\text{cm}^3}$:
$n = \frac{4}{\frac{4}{25}} = 4 \times \frac{25}{4} = 25$ Thus, the density of the substance in the new system is 25 units.