The density of a material in CGS system is 2 g / cm^3. In a system of units in which the unit of length is 2 cm and the unit of mass is 4 g, what is the numerical value of the density of the material? A) 4 B) 2 C) 1 D) 0.5
Question
The density of a material in CGS system is $2 g / \mathrm{cm}^{3}$. In a system of units in which the unit of length is $2 \mathrm{~cm}$ and the unit of mass is $4 \mathrm{~g}$, what is the numerical value of the density of the material?
A) 4
B) 2
C) 1
D) 0.5
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Answer
To determine the numerical value of the density of the material in a different system of units, let's break down and simplify the given information step-by-step.
The density of the material in the CGS system ($g/cm^3$) is $2 , \text{g/cm}^3$.
New Units:
- Unit of Length ($L$): $2 , \text{cm}$
- Unit of Mass ($M$): $4 , \text{g}$
We need to find the density in the new system of units. Recall that density ($\rho$) is mass divided by volume ($M/L^3$).
We use the relation between the old and new systems of units to find the new density. Let’s denote:
- $\rho_1$ as the density in the CGS system.
- $\rho_2$ as the density in the new system.
The formula to convert density between units is: $$ \rho_2 = \rho_1 \left(\frac{M_1}{M_2}\right) \left(\frac{L_1}{L_2}\right)^{-3} $$
Substituting the given values:
- $\rho_1 = 2 , \text{g/cm}^3$
- $M_1 = 1 , \text{g}$ (standard unit of mass in CGS is 1 g)
- $M_2 = 4 , \text{g}$
- $L_1 = 1 , \text{cm}$ (standard unit of length in CGS is 1 cm)
- $L_2 = 2 , \text{cm}$
Now, we apply the values to the formula: $$ \rho_2 = 2 \cdot \left(\frac{1}{4}\right) \cdot \left(\frac{1}{2}\right)^{-3} $$
Simplify the terms: $$ \rho_2 = 2 \cdot \frac{1}{4} \cdot \left(\frac{1}{8}\right)^{-1} $$ $$ \rho_2 = 2 \cdot \frac{1}{4} \cdot 8 $$ $$ \rho_2 = 2 \cdot 2 $$ $$ \rho_2 = 4 $$
Therefore, the numerical value of the density of the material in the new system of units is:
Answer: (A) 4
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