Question

A certain body weighs $22.42\,g$ and has a measured volume of $4.7\,cc$. The possible error in the measurement of mass and volume are $0.01\,gm$ and $0.1\,cc$. Then maximum error in the density will be

• A. $22\%$
• B. $2\%$
• C. $0.2\%$
• D. $0.02\%$

Answer 1

The Correct Option is B

To find the maximum error in the density of the given body, let's first understand the problem and the formula involved.

The density (\rho) of a body is calculated using the formula:

\rho = \frac{m}{V}

where:

• m is the mass of the body.
• V is the volume of the body.

According to error propagation in division, the relative error in the calculated quantity is given by the sum of the relative errors of the values involved in the division. The formula to calculate the percentage error in density is:

\text{Percentage Error in }\rho = \left( \frac{\Delta m}{m} + \frac{\Delta V}{V} \right) \times 100\%

Where:

• \Delta m is the possible error in mass.
• \Delta V is the possible error in volume.

Given data:

• Weight of the body, m = 22.42 \, \text{g}
• Volume of the body, V = 4.7 \, \text{cc}
• Possible error in mass, \Delta m = 0.01 \, \text{g}
• Possible error in volume, \Delta V = 0.1 \, \text{cc}

Substitute these values into the percentage error formula:

\begin{align*} \text{Percentage Error in }\rho & = \left( \frac{0.01}{22.42} + \frac{0.1}{4.7} \right) \times 100\% \\ & = \left( 0.000446 + 0.021277 \right) \times 100\% \\ & = 0.021723 \times 100\% \\ & \approx 2.17\% \end{align*}

As we need the maximum possible percentage error, this rounds to approximately 2\%.

Therefore, the maximum error in the density is 2\%.

The correct answer is hence:

2%