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 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the 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

The correct answer is 3.1 %.

Here is the step-by-step educational explanation to determine the relative percentage error in the density of the solid ball:

Step 1: Calculate the least count (LC) of the screw gauge
The least count of a screw gauge is defined as the pitch divided by the total number of divisions on the circular scale:
$\text{Least Count (LC)}=\frac{\text{Pitch}}{\text{Number of circular scale divisions}}$
Substituting the given values:
$\text{LC}=\frac{0.5\text{mm}}{50}=0.01\text{mm}$

Step 2: Determine the measured diameter ($d$) and its absolute error ($\Delta d$)
The measured diameter is given by the formula:
$d=\text{Main Scale Reading (MSR)}+(\text{Circular Scale Reading (CSR)}\times \text{LC})$
Given that MSR = 2.5 mm and CSR = 20 divisions:
$d=2.5\text{mm}+(20\times 0.01\text{mm})=2.5\text{mm}+0.20\text{mm}=2.70\text{mm}$
The absolute error in the diameter measurement is equal to the least count of the screw gauge:
$\Delta d=0.01\text{mm}$

Step 3: Relate density to mass and diameter
The density ($\rho$) of a sphere is given by:
$\rho =\frac{\text{Mass}}{\text{Volume}}=\frac{M}{\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3}=\frac{6M}{\pi d^3}$

Step 4: Calculate the relative percentage error in density
Taking the relative error on both sides of the density equation, we get:
$\frac{\Delta \rho }{\rho }\times 100\%=(\frac{\Delta M}{M}\times 100\%)+3\times (\frac{\Delta d}{d}\times 100\%)$
Given that the relative error in mass is 2%:
$\frac{\Delta M}{M}\times 100\%=2\%$
Now substitute the values for diameter and its error into the equation:
$\frac{\Delta \rho }{\rho }\times 100\%=2\%+3\times (\frac{0.01}{2.70}\times 100\%)$
Simplify the error term:
$\frac{\Delta \rho }{\rho }\times 100\%=2\%+3\times \frac{1}{270}\times 100\%$
$\frac{\Delta \rho }{\rho }\times 100\%=2\%+\frac{10}{9}\%$
$\frac{\Delta \rho }{\rho }\times 100\%\approx 2\%+1.11\%=3.11\%\approx 3.1\%$

Thus, the relative percentage error in the density is approximately 3.1 %.