A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible errors in the measurement of mass and volume are 0.01 g and 0.1 cc. Then maximum error in the density will be

The correct option is 2%.

To find the maximum percentage error in the density of the body, we can use the formula for density:
$d=\frac{M}{V}$
where:
$d$ is the density,
$M$ is the mass, and
$V$ is the volume.

The relative error in density, $\frac{\Delta d}{d}$, is given by the sum of the relative errors of the individual measurements:
$\frac{\Delta d}{d}=\frac{\Delta M}{M}+\frac{\Delta V}{V}$

Let us identify the given values from the problem statement:
Measured mass, $M=22.42\text{g}$
Error in mass, $\Delta M=0.01\text{g}$
Measured volume, $V=4.7\text{cc}$
Error in volume, $\Delta V=0.1\text{cc}$

Now, we calculate the maximum percentage error in density:
$\text{Percentage Error in Density}=\left(\frac{\Delta d}{d}\right)\times 100=\left(\frac{\Delta M}{M}+\frac{\Delta V}{V}\right)\times 100$

Substitute the values into the equation:
$\text{Percentage Error in Density}=\left(\frac{0.01}{22.42}+\frac{0.1}{4.7}\right)\times 100$

Calculate each term:
$\frac{0.01}{22.42}\times 100\approx 0.0446\%$
$\frac{0.1}{4.7}\times 100\approx 2.1277\%$

Add the two values to get the total percentage error:
$\text{Total Percentage Error}\approx 0.0446\%+2.1277\%=2.1723\%$

Rounding to the nearest significant figure or matching with the closest option provided in the question, the maximum error in the density is approximately 2%.