Question

Density and volume of a body are given as \( (20\pm 4)\text{ gm/cm^3 \) and \( (10\pm 1)\text{ cm}^3 \) respectively. The absolute error in measurement of mass is}

• A. 20 gm
• B. 30 gm
• C. 45 gm
• D. 60 gm

Hint:

Always remember that in multiplication or division, relative errors add up, whereas in addition or subtraction, absolute errors add up. Converting relative error to absolute error at the end is a standard two-step process: find the fractional error, then multiply by the nominal product value.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The density \( \rho \) and volume \( V \) of a body are given with their respective absolute errors. We need to find the absolute error in the calculated mass \( m \).

Step 2: Key Formula or Approach:
1. Relation between mass, density, and volume:
\[ m = \rho \cdot V \]
2. Error Propagation for Multiplication: For a product \( z = x \cdot y \), the relative errors are summed to find the relative error of the result:
\[ \frac{\Delta z}{z} = \frac{\Delta x}{x} + \frac{\Delta y}{y} \]
Therefore, for mass:
\[ \frac{\Delta m}{m} = \frac{\Delta \rho}{\rho} + \frac{\Delta V}{V} \]

Step 3: Detailed Explanation:
From the problem, we have:
- Density, \( \rho = 20 \text{ gm/cm}^3 \) and \( \Delta \rho = 4 \text{ gm/cm}^3 \)
- Volume, \( V = 10 \text{ cm}^3 \) and \( \Delta V = 1 \text{ cm}^3 \)
Calculate the nominal value of mass \( m \):
\[ m = \rho \cdot V = 20 \times 10 = 200 \text{ gm} \]
Now, calculate the relative error in mass:
\[ \frac{\Delta m}{200} = \frac{4}{20} + \frac{1}{10} \]
\[ \frac{\Delta m}{200} = \frac{1}{5} + \frac{1}{10} = \frac{2 + 1}{10} = \frac{3}{10} \]
\[ \frac{\Delta m}{200} = 0.3 \]
Calculate the absolute error \( \Delta m \):
\[ \Delta m = 0.3 \times 200 = 60 \text{ gm} \]

Step 4: Final Answer:
The absolute error in the measurement of mass is 60 gm.