Question

A wire has mass \((0.3\pm 0.003)\) gram, radius \((0.5\pm 0.005)\) cm and length \((6\pm 0.06)\) cm. The maximum percentage error in the measurement of density is

• A. \(2\,\%\)
• B. \(3\,\%\)
• C. \(4\,\%\)
• D. \(5\,\%\)

Hint:

Density is m / (pi r^2 l), so add the mass error, twice the radius error and the length error.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The wire is a cylinder, so its density is \(\rho = \dfrac{m}{\pi r^2 l}\). In a product or quotient the maximum fractional errors add, and a power multiplies the fractional error by that power.

Step 2: Key Formula or Approach:
\[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \]

Step 3: Detailed Explanation:
\(\dfrac{\Delta m}{m} = \dfrac{0.003}{0.3} = 1\%\).
\(\dfrac{\Delta r}{r} = \dfrac{0.005}{0.5} = 1\%\), so the radius contributes \(2\times1 = 2\%\).
\(\dfrac{\Delta l}{l} = \dfrac{0.06}{6} = 1\%\).
\[ \frac{\Delta\rho}{\rho}\times100 = 1 + 2 + 1 = 4\% \]
Option (A) 2% and (B) 3% come from forgetting the square on the radius or adding only some terms. Option (D) 5% adds an extra term.

Step 4: Final Answer:
The maximum percentage error in density is 4%.

Final Answer:
Density error is 1 + 2 + 1 = 4 percent. \[ \boxed{\text{(C) }4\%} \]