Question

The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m, l$ and $d$ are $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is

• A. 0.63%
• B. 0.82%
• C. 0.72%
• D. 0.25%

Hint:

Density of a cylinder is $\rho = \frac{4m}{\pi d^2 l}$. Apply the formula for fractional error propagation for multiplication and division.

Answer 1

The Correct Option is B

The problem asks for the percentage error in the density of a uniform cylinder based on measured dimensions. We utilize the principle that for a function $z = x^a y^b w^c$, the relative error is $\frac{\Delta z}{z} = |a|\frac{\Delta x}{x} + |b|\frac{\Delta y}{y} + |c|\frac{\Delta w}{w}$.
Step 1: Express density $\rho$ in terms of measured variables $m, d, l$.
$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{\pi (d/2)^2 l} = \frac{4m}{\pi d^2 l}$$
Step 2: Derive the percentage error expression.
Percentage error in $\rho$ is given by:
$$\left(\frac{\Delta \rho}{\rho} \times 100\right)\% = \left(\frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}\right) \times 100\%$$
Step 3: Substitute the given values.
$m = 97.42, \Delta m = 0.02$
$l = 8.35, \Delta l = 0.05$
$d = 20.20, \Delta d = 0.02$
$$\text{Error} = \left( \frac{0.02}{97.42} + \frac{2 \times 0.02}{20.20} + \frac{0.05}{8.35} \right) \times 100$$
Step 4: Compute the numeric values.
$\frac{0.02}{97.42} \times 100 \approx 0.0205\%$
$\frac{0.04}{20.20} \times 100 \approx 0.1980\%$
$\frac{0.05}{8.35} \times 100 \approx 0.5988\%$
Step 5: Total percentage error.
$$\text{Total Error} = 0.0205\% + 0.1980\% + 0.5988\% = 0.8173\%$$
Rounding this to the closest option gives $0.82\%$.