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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 $97.42 \pm 0.02 \text{ g}$, $8.35 \pm 0.05 \text{ mm}$ and $20.20 \pm 0.02 \text{ mm}$, respectively. Calculated percentage fractional error in $\rho$ is _________.

The correct answer is
0.82%

Calculating Cylinder Density Fractional Error

The density ($\rho$) of a uniform cylinder is given by the formula:

$ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{V} $

The volume ($V$) of a cylinder is calculated using its length ($l$) and radius ($r$):

$ V = \pi r^2 l $

Since the diameter ($d$) is given, the radius is $r = \frac{d}{2}$. Substituting this into the volume formula:

$ V = \pi \left(\frac{d}{2}\right)^2 l = \frac{\pi d^2 l}{4} $

Now, substitute the volume formula back into the density formula:

$ \rho = \frac{m}{\frac{\pi d^2 l}{4}} = \frac{4m}{\pi d^2 l} $

Determining Fractional Error

To find the fractional error in density ($\frac{\Delta \rho}{\rho}$), we use the error propagation rules for multiplication and division. Constants like $4$ and $\pi$ do not contribute to the error. The formula becomes:

$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l} $

We are given the following measurements:

  • Mass ($m$): $97.42 \pm 0.02$ g
  • Length ($l$): $8.35 \pm 0.05$ mm
  • Diameter ($d$): $20.20 \pm 0.02$ mm

Calculating Individual Fractional Errors

  • Fractional error in mass ($\frac{\Delta m}{m}$):

    $ \frac{\Delta m}{m} = \frac{0.02}{97.42} \approx 0.000205 $

  • Fractional error in diameter ($\frac{\Delta d}{d}$):

    $ \frac{\Delta d}{d} = \frac{0.02}{20.20} \approx 0.000990 $

  • Fractional error in length ($\frac{\Delta l}{l}$):

    $ \frac{\Delta l}{l} = \frac{0.05}{8.35} \approx 0.005988 $

Calculating Total Fractional Error and Percentage Error

Now, substitute these values into the total fractional error formula:

$ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l} $

$ \frac{\Delta \rho}{\rho} \approx 0.000205 + 2 \times (0.000990) + 0.005988 $

$ \frac{\Delta \rho}{\rho} \approx 0.000205 + 0.001980 + 0.005988 $

$ \frac{\Delta \rho}{\rho} \approx 0.008173 $

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

$ \text{Percentage Error} = \frac{\Delta \rho}{\rho} \times 100\% \approx 0.008173 \times 100\% \approx 0.8173\% $

Rounding to two decimal places gives 0.82%.

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