Mass and length of a metal cube are `10 kg +- 0.1 kg` and `1 m +- 0.02 m`. Its density with percentage error is

Options

A. `10 kg/m +-7%`
B. `10 kg/m +-3%`
C. `10//3 kg/m +- 7%`
D. `10//3 kg/m +- 3%`

Answer

To find the density of the metal cube along with its percentage error, we can follow these steps:

### Step 1: Understand the formula for density
Density (D) is defined as mass (M) divided by volume (V). For a cube, the volume can be expressed in terms of its length (L):
\[ D = \frac{M}{V} = \frac{M}{L^3} \]

### Step 2: Identify the given values
From the problem statement, we have:
- Mass, \( M = 10 \, \text{kg} \pm 0.1 \, \text{kg} \)
- Length, \( L = 1 \, \text{m} \pm 0.02 \, \text{m} \)

### Step 3: Calculate the density without error
Using the values provided:
\[ D = \frac{M}{L^3} = \frac{10 \, \text{kg}}{(1 \, \text{m})^3} = 10 \, \text{kg/m}^3 \]

### Step 4: Calculate the errors in density
To find the percentage error in density, we need to consider the errors in mass and length. The formula for the percentage error in density is given by:
\[ \frac{\Delta D}{D} = \frac{\Delta M}{M} + 3 \frac{\Delta L}{L} \]

Where:
- \( \Delta M = 0.1 \, \text{kg} \)
- \( \Delta L = 0.02 \, \text{m} \)

### Step 5: Substitute the values into the error formula
Now substituting the values into the error formula:
\[ \frac{\Delta D}{D} = \frac{0.1 \, \text{kg}}{10 \, \text{kg}} + 3 \cdot \frac{0.02 \, \text{m}}{1 \, \text{m}} \]

### Step 6: Calculate each term
Calculating the first term:
\[ \frac{0.1}{10} = 0.01 \]

Calculating the second term:
\[ 3 \cdot \frac{0.02}{1} = 0.06 \]

### Step 7: Combine the terms
Now, combine both terms:
\[ \frac{\Delta D}{D} = 0.01 + 0.06 = 0.07 \]

### Step 8: Convert to percentage
To convert this to a percentage, multiply by 100:
\[ \Delta D \text{ (percentage)} = 0.07 \times 100 = 7\% \]

### Final Result
Thus, the density of the metal cube is \( 10 \, \text{kg/m}^3 \) with a percentage error of \( 7\% \).

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