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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Physics
Chapter
Mathematics in Physics

JEE Advanced 2011 · Paper 2 · Physics · Mathematics in Physics

The density of a solid ball is to be determined in an experiment....

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 mm0.5 \mathrm{~mm} and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm2.5 \mathrm{~mm} and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%2 \%, the relative percentage error in the density is

Options

  1. A0.9%0.9 \%
  2. B2.4%2.4 \%
  3. C3.1%3.1 \%
  4. D4.2%4.2 \%

Correct answer

C. 3.1%3.1 \%

Step-by-step solution

Least count of screw gauge =0.550=\frac{0.5}{50} =0.01  mm=Δ\mathrm{~mm}=\Deltar Diameter, r=2.5 mm+20×0.550r=2.5 \mathrm{~mm}+20 \times \frac{0.5}{50} =270  mmΔrr=0.012.70\mathrm{~mm} \frac{\Delta r}{r}=\frac{0.01}{2.70}or Δrr×100=12.7\frac{\Delta r}{r} \times 100=\frac{1}{2.7} Now, density d=mV=m43π(r2)3d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi\left(\frac{r}{2}\right)^3}Here,Here,risthediameter.is the diameter.∴Δdd×100={Δmm+3(Δrr)}×100=Δmm×100+3×(Δrr)×100\begin{aligned} \therefore \quad\frac{\Delta d}{d}& \times100=\left\{\frac{\Delta m}{m}+3\left(\frac{\Delta r}{r}\right)\right\} \times100 \\ = & \frac{\Delta m}{m} \times100+3 \times\left(\frac{\Delta r}{r}\right) \times100 \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.
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