Question

A sphere has a mass of $12.2\pm 0.1$ kg and radius $10\pm 0.1$ cm , find the maximum $\%$ error in density.
A. $10\%$
B. $2.4\%$
C. $3.83\%$
D. $4.2\%$

Answer 1

Hint: First we will define what we mean by density and then we will write a definition of the sphere. After that we will apply the formula for the error of density that is $\dfrac{\Delta \rho }{\rho }=\dfrac{\Delta m}{m}+3\dfrac{\Delta r}{r}$ , we will put the values given in the question and hence get the answer.

Complete step by step answer:
The density of material shows the denseness of that material in a specific given area. A material’s density is defined as its mass per unit volume. Density is essentially a measurement of how tightly matter is packed together. It is a unique physical property for a particular object. The formula for density is as follows:
$\text{Density}=\dfrac{Mass}{Volume}\Rightarrow \rho =\dfrac{m}{V}$ , where
$\rho $ is the density, $m$ is the mass , $V$ is the volume.
Though SI unit of density is $\dfrac{kg}{{{m}^{3}}}$ , for convenience we use $\dfrac{g}{c{{m}^{3}}}$ for solids, $\dfrac{g}{ml}$ for liquids, and $\dfrac{g}{L}$ for gases.
Let’s define what a sphere is. So a sphere is defined as the three-dimensional round solid figure in which every point on its surface is equidistant from its centre. The fixed distance is called the radius of the sphere and the fixed point is called the centre of the sphere. When the circle is rotated, we will observe the change of shape and thus sphere will be generated.
It is given that the mass of the sphere is: $m\pm \Delta m=\left( 12.2\pm 0.1 \right)$ kg and we have been given the radius of sphere as: $r+\Delta r=10\pm 0.1$ cm :
Now, we know that the volume of sphere is: $\dfrac{4}{3}\pi {{r}^{3}}$ , where $r$ is the radius of sphere.
$\rho =\dfrac{m}{V}=\dfrac{m}{\left( \dfrac{4}{3} \right)\pi {{r}^{3}}}$
Now, error in density: $\dfrac{\Delta \rho }{\rho }=\dfrac{\Delta m}{m}+3\dfrac{\Delta r}{r}$ , now we will put the values from above: $\begin{align}
  & \Rightarrow \dfrac{\Delta \rho }{\rho }=\dfrac{\Delta m}{m}+3\dfrac{\Delta r}{r}\Rightarrow \dfrac{\Delta \rho }{\rho }=\dfrac{0.1}{12.2}+3\dfrac{0.1}{10} \\
 & \Rightarrow \dfrac{\Delta \rho }{\rho }=\dfrac{1}{122}+\dfrac{0.3}{10}=0.0383 \\
\end{align}$
Now, the maximum percentage error will be $0.0383\times 100=3.83\%$.

Hence, the correct option is C.

Note:
Always remember to multiply the obtained error into $100$ to convert it into the percentage. Since there are various decimals involved student can make mistakes during calculations so one needs to be careful while doing that. Also, don’t forget to put the $\%$ sign at the end.