Question

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

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

Hint:

Density is mass divided by volume of the cylinder; the radius error counts twice.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Density \(\rho=\dfrac{m}{\pi r^{2}l}\).
For such a product, the maximum fractional error is \(\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\).

Step 2: Individual percentage errors
\(\dfrac{\Delta m}{m}=\dfrac{0.003}{0.3}=1\%\)
\(\dfrac{\Delta r}{r}=\dfrac{0.005}{0.5}=1\%\)
\(\dfrac{\Delta l}{l}=\dfrac{0.06}{6}=1\%\)

Step 3: Calculation
Total \(=1+2(1)+1=4\%\).

Step 4: Compare the options
The answer is option (C).

Final Answer:
The maximum percentage error in the density is 4%. \[ \boxed{\text{(C) }4\%} \]