Question

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left[\frac{x}{100}\right]\%$. If the relative errors in measuring the mass and the diameter are $6.0\%$ and $1.5\%$ respectively, the value is$x$is_______.

Answer 1

Step 1: Given data :

The maximum error in the density of the sphere is $\frac{d\rho }{\rho }\times 100$ $=\left[\frac{x}{100}\right]\%$

The relative errors in measuring the mass $\frac{dm}{m}\times 100=6.0\%$

The relative errors in measuring the diameter are$\frac{dD}{D}\times 100=1.5\%$

Step 2: Draw the diagram and formula used :

Density can be given using the formula-

$density\left(d\right)=\frac{mass\left(m\right)}{volume\left(v\right)}$

The volume of a solid sphere $=\frac{4}{3}\pi R^3$

Where, $R$ is radius of solid sphere.

Step 3: Calculating the value of $x$

Using the density formula,

$\begin{array}{l}\rho =\frac{m}{\left[\frac{4}{3}\pi R^3\right]}\\ \rho =\frac{m}{\left[\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right]}\end{array}$$\left[diameter\left(D\right)=2R(twiceradius)\right]$

Now calculate the density and take the log, let $k=\frac{1}{\left({\displaystyle \frac{4}{3}}\pi \left({\displaystyle \frac{1}{8}}\right)\right)}$

$\begin{array}{l}\rho =k\left(\frac{m}{D^3}\right)\\ \log \rho =\log k+\log m-3\log D\end{array}$

Differentiating,

$\begin{array}{l}\frac{d\rho }{\rho }\times 100=\left(\frac{dm}{m}\times 100\right)+3\left(\frac{dD}{D}\times 100\right)\end{array}$

As we have maximum error density

Simplify further,

$\begin{array}{c}\frac{d\rho }{\rho }=6.0+3\times 1.5\\ \frac{d\rho }{\rho }=10.5\%\\ \frac{x}{100}\%=10.5\%\\ x=1050\end{array}$

Therefore the value of $x$is $1050$.