The density of a sphere is measured by measuring the mass and diameter. If it is known that the maximum percentage errors in measurement of mass and diameter are `2%` and `3%` respectively then the maximum percentage error in the measurement of density is

Answer

To find the maximum percentage error in the measurement of density, we can follow these steps:

### Step 1: Understand the formula for density
The density \( \rho \) of a sphere is given by the formula:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass and \( V \) is the volume.

### Step 2: Find the volume of the sphere in terms of diameter
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Since the diameter \( d \) is twice the radius \( r \) (i.e., \( r = \frac{d}{2} \)), we can express the volume in terms of the diameter:
\[ V = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{6} \]

### Step 3: Substitute the volume back into the density formula
Now substituting the volume back into the density formula:
\[ \rho = \frac{m}{\frac{\pi d^3}{6}} = \frac{6m}{\pi d^3} \]

### Step 4: Determine the percentage errors
Given:
- The maximum percentage error in mass \( \left(\frac{\Delta m}{m}\right) \) is \( 2\% \).
- The maximum percentage error in diameter \( \left(\frac{\Delta d}{d}\right) \) is \( 3\% \).

### Step 5: Calculate the maximum percentage error in density
Using the formula for the maximum percentage error in a quotient:
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \cdot \frac{\Delta d}{d} \]
Here, the coefficient of \( \frac{\Delta d}{d} \) is \( 3 \) because density is inversely proportional to the cube of the diameter.

Substituting the values:
\[ \frac{\Delta \rho}{\rho} = 2\% + 3 \cdot 3\% \]
\[ \frac{\Delta \rho}{\rho} = 2\% + 9\% = 11\% \]

### Conclusion
Thus, the maximum percentage error in the measurement of density is:
\[ \boxed{11\%} \]