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JEE Advanced · Physics · 1. Math in Physics

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

  1. A \(0.9 \%\)
  2. B \(2.4 \%\)
  3. C \(3.1 \%\)
  4. D \(4.2 \%\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3.1 \%\)

Step-by-step Solution

Detailed explanation

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.
\(\therefore \frac{\Delta d}{d} \times 100=\left\{\frac{\Delta m}{m}+3\left(\frac{\Delta r}{r}\right)\right\} \times 100\)
\(=\frac{\Delta m}{m} \times 100+3 \times\left(\frac{\Delta r}{r}\right) \times 100\)
\(
\begin{aligned}
& =2 \%+3 \times \frac{1}{27} \\
& =3.11 \%
\end{aligned}
\)
Analysis of Question
(i) Question is moderately difficult.
(ii) In practical part, questions in JEE are asked from vernier callipers and screw gauge.
Same subject
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