The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is `0.5mm` and there are `50` divisions on the circular scale. The reading on the main scale is `2.5mm` and that on circular scale is `20` divisions. if the measured mass of the ball has a relative error of `2%`, the relative percentage error in the density is

Options

A. `0.9%`
B. `2.4%`
C. `3.1%`
D. `4.2%`

Answer

To determine the relative percentage error in the density of a solid ball, we will follow these steps:

### Step 1: Calculate the Least Count of the Screw Gauge
The least count (LC) of the screw gauge is given by the formula:
\[ \text{Least Count} = \frac{\text{Pitch}}{\text{Total Number of Circular Scale Divisions}} \]
Given:
- Pitch \( P = 0.5 \, \text{mm} \)
- Total Circular Scale Divisions \( = 50 \)

Calculating the least count:
\[ \text{LC} = \frac{0.5}{50} = 0.01 \, \text{mm} \]

### Step 2: Calculate the Diameter of the Ball
The diameter \( D \) can be calculated using the formula:
\[ D = \text{Main Scale Reading (MSR)} + \left(\text{Circular Scale Reading (CSR)} \times \text{Least Count}\right) \]
Given:
- MSR \( = 2.5 \, \text{mm} \)
- CSR \( = 20 \)

Calculating the diameter:
\[ D = 2.5 + (20 \times 0.01) = 2.5 + 0.2 = 2.7 \, \text{mm} \]

### Step 3: Calculate the Uncertainty in Diameter
The uncertainty in the diameter \( \Delta D \) is equal to the least count:
\[ \Delta D = \text{LC} = 0.01 \, \text{mm} \]

### Step 4: Understand the Relation of Density with Mass and Diameter
The density \( \rho \) of the ball is given by:
\[ \rho = \frac{m}{V} \]
For a solid sphere, the volume \( V \) is:
\[ V = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 \]
Thus, density can be expressed as:
\[ \rho = \frac{m}{\frac{4}{3} \pi \left(\frac{D}{2}\right)^3} \]

### Step 5: Calculate the Relative Percentage Error in Density
The relative error in density can be calculated using the formula:
\[ \frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta m}{m} \times 100 + 3 \left(\frac{\Delta D}{D} \times 100\right) \]
Given:
- Relative error in mass \( \frac{\Delta m}{m} \times 100 = 2\% \)

Calculating the second term:
\[ \frac{\Delta D}{D} = \frac{0.01}{2.7} \]
Calculating \( \frac{\Delta D}{D} \):
\[ \frac{\Delta D}{D} \approx 0.0037 \]
Thus,
\[ \frac{\Delta D}{D} \times 100 \approx 0.37\% \]

Now substituting back into the relative error formula:
\[ \frac{\Delta \rho}{\rho} \times 100 = 2 + 3 \times 0.37 \approx 2 + 1.11 = 3.11\% \]

### Final Result
The relative percentage error in the density of the solid ball is approximately:
\[ \boxed{3.1\%} \]