Question

A metal wire has mass (0.4±0.002)g, radius (0.3±0.001)mm and length (5±0.02)cm. The maximum possible percentage error in the measurement of density will nearly be:

• A. 1.4%
• B. 1.2%
• C. 1.3%
• D. 1.6%

Answer 1

The Correct Option is D

To find the maximum possible percentage error in the measurement of density, we use the formula for density, which is given by:

Density (\( \rho \)) = \( \frac{\text{mass}}{\text{volume}} \)

The volume \( V \) of the wire, which is a cylinder, is calculated by: 

\( V = \pi r^2 h \)

where \( r \) is the radius and \( h \) is the length of the wire.

The percentage error in density, using error propagation, can be calculated by:

\(\frac{\Delta \rho}{\rho} \times 100 = \left(\frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta h}{h}\right) \times 100\)

Given:

Mass, \( m = 0.4 \) g, \( \Delta m = 0.002 \) g

Radius, \( r = 0.3 \) mm = 0.03 cm, \( \Delta r = 0.001 \) mm = 0.0001 cm

Length, \( h = 5 \) cm, \( \Delta h = 0.02 \) cm

Calculate percentage errors:

\(\frac{\Delta m}{m} = \frac{0.002}{0.4} = 0.005\)

\(\frac{\Delta r}{r} = \frac{0.0001}{0.03} = 0.0033\)

\(\frac{\Delta h}{h} = \frac{0.02}{5} = 0.004\)

Percentage error in density:

\((0.005 + 2 \times 0.0033 + 0.004) \times 100 = 0.0156 \times 100 = 1.56\%\)

Rounding to nearest significant digits, the maximum possible percentage error is about 1.6%.

Answer 2

We know that,
\(\rho =  \frac{M}{V}\)

\(\rho= \frac{M}{\pi r^2l}\)

\(\frac{\Delta\rho}{\rho}= \frac{\Delta M}{M}+ \frac{2\Delta r}{r}+ \frac{\Delta l}{l}\)

\(\frac{\Delta\rho}{\rho}\)% = [\(\frac{0.002}{0.4}+ \frac{2(0.001)}{0.3}+ \frac{0.02}{5}\)]\(\times 100\)

\(\frac{\Delta\rho}{\rho}\)% \(= \frac{1}{2}\)%+\(\frac{2}{3}\)%+\(\frac{2}{5}\)%

\(\frac{\Delta\rho}{\rho}\)% = 1.6%

So, the correct option is (D): 1.60%