The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with…
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 $0.5 \mathrm{~mm}$ and there are 50 divisions on the circular scale. The reading on the main scale is $2.5 \mathrm{~mm}$ 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
$0.9 \%$
$2.4 \%$
$3.1 \%$
$4.2 \%$
Solution
Least count of screw gauge $=\frac{0.5}{50}$
$
=0.01 \mathrm{~mm}=\Delta r
$
Diameter, $r=2.5 \mathrm{~mm}+20 \times \frac{0.5}{50}$
$
=270 \mathrm{~mm}
$
$
\frac{\Delta r}{r}=\frac{0.01}{2.70}
$
or $\frac{\Delta r}{r} \times 100=\frac{1}{2.7}$
Now, density $d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi\left(\frac{r}{2}\right)^3}$
Here, $r$ is the diameter.
$
\begin{aligned}
\therefore \quad \frac{\Delta d}{d} & \times 100=\left\{\frac{\Delta m}{m}+3\left(\frac{\Delta r}{r}\right)\right\} \times 100 \\
= & \frac{\Delta m}{m} \times 100+3 \times\left(\frac{\Delta r}{r}\right) \times 100
\end{aligned}
$
$
\begin{aligned}
& =2 \%+3 \times \frac{1}{27} \\
& =3.11 \%
\end{aligned}
$
Analysis of Question
(i) Question is moderately difficult.
(ii) In practical part, questions in JEE are asked from vernier callipers and screw gauge.