Question

A wire has mass \((0.4\pm 0.002)\) g, radius \((0.6\pm 0.006)\) mm and length \((8\pm 0.04)\) cm. The percentage error in the measurement of its density is

• A. \(2\%\)
• B. \(2.55\%\)
• C. \(3\%\)
• D. \(3.55\%\)

Hint:

Write density as mass divided by volume of a cylinder and add the fractional errors, doubling the one for the radius.

Answer 1

The Correct Option is C

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

Step 2: Formula for the error
\(\frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+2\frac{\Delta r}{r}+\frac{\Delta l}{l}\).

Step 3: Calculation
\(\frac{\Delta m}{m}=\frac{0.002}{0.4}=0.5\%\), \(\frac{\Delta r}{r}=\frac{0.006}{0.6}=1\%\), \(\frac{\Delta l}{l}=\frac{0.04}{8}=0.5\%\).

Step 4: Add the errors
\(\frac{\Delta\rho}{\rho}\times 100=0.5+2(1)+0.5=3\%\), which is option (C).

Final Answer:
The percentage error in the density is 3 percent. \[ \boxed{\text{(C) }3\%} \]