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:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q21 – NEET-UG 2023

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Home Physics Units & MeasurementsQuestion #568

A metal wire has mass

(0.4 \pm 0.002)

g, radius

(0.3 \pm 0.001)

mm and length

(5 \pm 0.02)

cm. The maximum possible percentage error in the measurement of density will nearly be:

Clue 1 of 3: Density Formula

Recall that density ( $ρ$ ) is defined as mass ( $M$ ) per unit volume ( $V$ ). For a cylindrical wire, the volume is given by $V = π r^{2} L$ , where $r$ is the radius and $L$ is the length.

🧠 Full Solution Path

Step 1: Formulate Density and Error Equation✦ Active

The density of a wire is given by $ρ = \frac{Mass}{Volume}$ . For a cylindrical wire, the volume is $V = π r^{2} L$ . Therefore, the density formula is $ρ = \frac{M}{π r^{2} L}$ . The maximum possible fractional error in density is given by the sum of the fractional errors of the measured quantities, considering their powers:

\frac{Δ ρ}{ρ} = \frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ L}{L}

💡 Teacher's Secret Hint

Remember that constants like $π$ do not contribute to the error.

Step 2: Calculate Individual Fractional Errors○ Expand

Given values and their uncertainties are: Mass: $M = 0.4$ g, $Δ M = 0.002$ g Radius: $r = 0.3$ mm, $Δ r = 0.001$ mm Length: $L = 5$ cm, $Δ L = 0.02$ cm Calculate the fractional errors for each quantity:

\frac{Δ M}{M} = \frac{0.002}{0.4} = 0.005

\frac{Δ r}{r} = \frac{0.001}{0.3} = \frac{1}{300} \approx 0.003333

\frac{Δ L}{L} = \frac{0.02}{5} = 0.004

💡 Teacher's Secret Hint

Ensure units are consistent for each fractional error calculation, though for fractional error, units cancel out.

Step 3: Calculate Total Percentage Error○ Expand

Substitute the individual fractional errors into the total fractional error formula:

\frac{Δ ρ}{ρ} = 0.005 + 2 \times (\frac{1}{300}) + 0.004

\frac{Δ ρ}{ρ} = 0.005 + 0.006666 . . . + 0.004

\frac{Δ ρ}{ρ} = 0.015666 . . .

To find the percentage error, multiply by $100 %$ :

Percentage error = 0.015666 . . . \times 100 % \approx 1.566 % \approx 1.6 %

💡 Teacher's Secret Hint

Round the final answer to the nearest given option.

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Physics Question 21 – NEET-UG 2023

Clue 1 of 3: Density Formula

🧠 Full Solution Path