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
A) $0.9\%$
B) $2.4\%$
C) $3.1\%$
D) 4.2%
Explanation
Least count $= \frac{0.5}{50} = 0.01\ mm$ Diameter of ball $D = 2.5\ mm + (20)(0.01)$ $D = 2.7\ mm$ $\rho = \frac{M}{\upsilon ol} = \frac{M}{\frac{4}{3}\pi\left( \frac{D}{2} \right)^{3}} \Rightarrow \left( \frac{\mathrm{\Delta}\rho}{\rho} \right)_{\max} = \frac{\mathrm{\Delta}M}{M} + 3\frac{\mathrm{\Delta}D}{D}$ $\left( \frac{\mathrm{\Delta}\rho}{\rho} \right)_{\max} = 2\% + 3\left( \frac{0.01}{2.7} \right) \times 100\% \Rightarrow \frac{\mathrm{\Delta}\rho}{\rho} = 3.1\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.