The measured mass and volume of a body are 2.00 g and 5.0 cm³ resrectively. With possible errors of 0.01 g and 0.1 cm³ , what would be the percent error in density?

The correct option is 2.5%.

Step-by-Step Explanation:

Density ($d$) of a body is defined as the ratio of its mass ($M$) to its volume ($V$):
$d=\frac{M}{V}$

When two quantities are divided, the maximum fractional error in the resulting value is the sum of the fractional errors in the individual quantities. Therefore, the relative error in density ($\frac{\Delta d}{d}$) is given by:
$\frac{\Delta d}{d}=\frac{\Delta M}{M}+\frac{\Delta V}{V}$

To find the percentage error in density, we multiply the relative error by 100:
$\text{Percentage error in density}=\left(\frac{\Delta M}{M}+\frac{\Delta V}{V}\right)\times 100\%$

Given values in the problem:
Mass, $M=2.00\text{g}$
Error in mass, $\Delta M=0.01\text{g}$
Volume, $V=5.0{\text{cm}}^3$
Error in volume, $\Delta V=0.1{\text{cm}}^3$

Let us calculate the individual fractional errors:
Fractional error in mass:
$\frac{\Delta M}{M}=\frac{0.01}{2.00}=0.005$

Fractional error in volume:
$\frac{\Delta V}{V}=\frac{0.1}{5.0}=0.02$

Adding these fractional errors to find the total relative error in density:
$\frac{\Delta d}{d}=0.005+0.02=0.025$

Finally, we calculate the percentage error:
$\text{Percentage error}=0.025\times 100\%=2.5\%$