Question

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

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

Hint:

Density = m / (pi r^2 l), so the maximum fractional error is dm/m + 2 dr/r + dl/l.

Answer 1

The Correct Option is A

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

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 \(2\dfrac{\Delta r}{r} = 2\%\)
\(\dfrac{\Delta l}{l} = \dfrac{0.06}{6} = 1\%\)
\[ \frac{\Delta\rho}{\rho} = 1 + 2 + 1 = 4\% \]
Option (B), 3%, results from forgetting that radius appears squared.

Final Answer:
The maximum error in density is 4%, option (A). \[ \boxed{4\%} \]