Question

The density of a cube is measured by measuring the mass and length of its side. If the maximum errors in the measurement of its mass and length are $3\%$ and $2\%$ respectively, the maximum error in the measurement of density would be:

Answer 1

Step 1: Given parameters

Measurement of mass, $\mathrm{m}=3\%$

Measurement of length, $\mathrm{l}=2\%$

Step 2: To find

We have to find the maximum error in the measurement of density.

Step 3: Calculation

We know the formula,

$$\begin{gathered} \mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}} \\ =\frac{\mathrm{m}}{{\mathrm{l}}^3} \end{gathered}$$

Here, ${\mathrm{l}}^3$ is the volume of a cube.

Now,

${\left[\left(\frac{∆\mathrm{\rho }}{\mathrm{\rho }}\right)\times 100\right]}_{\max }=\left(\left(\frac{\mathrm{\Delta m}}{\mathrm{m}}\right)\times 100\right)+3\left(\left(\frac{∆\mathrm{l}}{\mathrm{l}}\right)\times 100\right)$

Here,

$\left(\frac{\mathrm{\Delta m}}{\mathrm{m}}\right)\times 100$ is the maximum error measurement of the mass.

$\left(\frac{\mathrm{\Delta l}}{\mathrm{l}}\right)\times 100$ is the maximum error in the measurement of the length.

Now,

The maximum error in the measurement of density $$\begin{gathered} =3\%+3(2\%) \\ =3\%+6\% \\ =9\% \end{gathered}$$

Therefore, the maximum error is $9\%$.