Q. The maximum percentage error in the measurement of density of a wire is [Given, mass of wire $=(0.60\pm0.003)$ g, radius of wire $=(0.50\pm0.01)$ cm, length of wire $=(10.00\pm0.05)$ cm]
A) 8
B) 5
C) 4
D) 7
Explanation
[Solution by Dr. Abubacker] $\rho=\frac m{\pi r^2l}$, so $\frac{\Delta\rho}\rho=\frac{\Delta m}m+2\frac{\Delta r}r+\frac{\Delta l}l=0.5\%+2(2\%)+0.5\%=5\%$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.