Question

A cube of edge 4 cm has mass 256 g. The density of the material in SI unit is:

• A. 4 kg/m$^3$
• B. 1600 kg/m$^3$
• C. 4000 kg/m$^3$
• D. 1000 kg/m$^3$

Hint:

Always be careful with units. You can convert to SI units at the very beginning (mass = 0.256 kg, edge = 0.04 m) or at the end. Often converting at the end is easier to avoid working with many leading zeros.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to find the density of a cubic object given its mass and edge length, and then express it in the International System of Units (SI), which is kilograms per cubic meter (kg/m$^3$).

Step 2: Key Formula or Approach:
1. Volume of a cube: \( V = l^3 \), where \( l \) is the edge length.
2. Density formula: \( \rho = \frac{\text{mass}}{\text{volume}} \).
3. Conversion: To convert g/cm$^3$ to kg/m$^3$, multiply by 1000.

Step 3: Detailed Explanation:
Given data:
- Edge length (\( l \)) = 4 cm
- Mass (\( m \)) = 256 g
Calculation:
1. Calculate the volume in cm$^3$:
\[ V = (4 \text{ cm})^3 = 64 \text{ cm}^3 \] 2. Calculate the density in g/cm$^3$:
\[ \rho = \frac{256 \text{ g}}{64 \text{ cm}^3} = 4 \text{ g/cm}^3 \] 3. Convert to SI units (kg/m$^3$):
Since \( 1 \text{ g} = 10^{-3} \text{ kg} \) and \( 1 \text{ cm}^3 = 10^{-6} \text{ m}^3 \):
\[ 1 \text{ g/cm}^3 = \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 1000 \text{ kg/m}^3 \] Therefore, \( \rho = 4 \times 1000 \text{ kg/m}^3 = 4000 \text{ kg/m}^3 \).

Step 4: Final Answer:
The density of the material is 4000 kg/m$^3$.