Estimating Percentage Error in Sphere Mass

The mass ($M$) of a sphere is related to its volume ($V$) and density ($\rho$). The formula is $M = \rho V$. Assuming the density $\rho$ is constant, the mass is directly proportional to the volume.

The volume ($V$) of a sphere is given by the formula:

\(V = \frac{4}{3}\pi r^3\)

where $r$ is the radius of the sphere. Since $M \propto V$ and $V \propto r^3$ (with $\frac{4}{3}\pi\rho$ as constants), we can write the relationship between mass and radius as:

\(M = k r^3\)

where $k = \frac{4}{3}\pi\rho$ is a constant.

Error Propagation Formula

When a physical quantity $y$ depends on another quantity $x$ raised to a power, i.e., $y = k x^n$, the fractional error in $y$ is related to the fractional error in $x$ by the formula:

\(\frac{\Delta y}{y} = |n| \frac{\Delta x}{x}\)

In terms of percentage error, this is:

\(\frac{\Delta y}{y} \times 100\% = |n| \left(\frac{\Delta x}{x} \times 100\%\right)\)

In our case, mass $M$ depends on radius $r$ as $M \propto r^3$. So, $y = M$, $x = r$, and $n = 3$.

The percentage error in mass is therefore:

\(\frac{\Delta M}{M} \times 100\% = 3 \left(\frac{\Delta r}{r} \times 100\%\right)\)

Calculating Percentage Error in Radius

The given radius is $r = 40.0$ mm and the estimated error in radius is $\Delta r = 0.5$ mm.

The fractional error in radius is:

\(\frac{\Delta r}{r} = \frac{0.5}{40.0}\)

The percentage error in radius is:

\(\frac{\Delta r}{r} \times 100\% = \frac{0.5}{40.0} \times 100\%\)

\(= \frac{5}{400} \times 100\%\)

\(= \frac{1}{80} \times 100\%\)

\(= \frac{100}{80}\%\)

\(= \frac{10}{8}\%\)

\(= \frac{5}{4}\%\)

\(= 1.25\%\)

So, the percentage error in the radius is $1.25\%$.

Calculating Percentage Error in Mass

Now, we use the error propagation formula for mass:

\(\frac{\Delta M}{M} \times 100\% = 3 \left(\frac{\Delta r}{r} \times 100\%\right)\)

Substitute the calculated percentage error in radius:

\(\frac{\Delta M}{M} \times 100\% = 3 \times 1.25\%\)

\(= 3.75\%\)

The estimated percentage error in the mass of the sphere is $\pm 3.75\%$. The $\pm$ sign indicates the error can be positive or negative, meaning the true value could be $3.75\%$ higher or lower than the calculated mass based on the measured radius.

Summary of Calculation Steps:

• Identify the relationship between the quantity whose error is given (radius) and the quantity whose error is requested (mass): $M \propto r^3$.
• Apply the error propagation rule for powers: $\frac{\Delta M}{M} = 3 \frac{\Delta r}{r}$.
• Calculate the percentage error in the given quantity (radius): $\frac{\Delta r}{r} \times 100\% = \frac{0.5}{40.0} \times 100\% = 1.25\%$.
• Calculate the percentage error in the requested quantity (mass): $\frac{\Delta M}{M} \times 100\% = 3 \times (\text{percentage error in radius}) = 3 \times 1.25\% = 3.75\%$.

The estimated error in its mass is $\pm 3.75\%$.