Q. 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
Explanation
Least count of screw gauge $= \frac{0.5}{50}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.01\ mm = \mathrm{\Delta}r$ Diameter $r = 2.5\ mm + 20 \times \frac{0.5}{50} = 2.70\ mm$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\mathrm{\Delta}r}{r} = \frac{0.01}{2.70}$ Or$\frac{\mathrm{\Delta}r}{r} \times 100 = \frac{1}{2.7}$ Now density $d = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi\left( \frac{r}{2} \right)^{3}}$ Here, $r$ is the diameter. $\therefore\ \ \ \ \ \frac{\mathrm{\Delta}d}{d} \times 100 = \left\{ \frac{\mathrm{\Delta}m}{m} + 3\ \left( \frac{\mathrm{\Delta}r}{r} \right) \right\}\ \times 100$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\mathrm{\Delta}m}{m} \times 100 + 3 \times \left( \frac{\mathrm{\Delta}r}{r} \right) \times 100\$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\% + 3 \times \frac{1}{2.7}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.11\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.