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 percentage error of $2 %,$ the relative percentage error in the density is

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2.4%

3.1%

4.2%

0.9%

answer is C.

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Detailed Solution

Least count of screw gauge = $\frac{0.5}{50} = 0.01 mm = Δ r$

Diameter, $r = 2.5 mm + 20 \times \frac{0.5}{50}$

$= 2.70 mm$

$\frac{Δ r}{r} = \frac{0.01}{2.70}$

$\frac{Δ r}{r} \times 100 = \frac{1}{2.7}$

Now, density $d = \frac{m}{V} = \frac{m}{\frac{4}{3} π {(\frac{r}{2})}^{3}}$

Here, r is the diameter

$\frac{Δ d}{d} \times 100 = \{\frac{Δ m}{m} + 3 (\frac{Δ r}{r})\} \times 100$

$= \frac{Δ m}{m} \times 100 + 3 \times (\frac{Δ r}{r}) \times 100$

$= 2 % + 3 \times \frac{1}{2.7}$

$= 3.1 %$

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