Q. A body of mass $0.05\ kg$ is observed to fall with an acceleration of $9.5\ ms^{- 2}$. The opposing force of air on the body is $(g = 9.8\ ms^{- 2})$
A) $0.015\ N$
B) $0.15\ N$
C) $0.030\ N$
D) Zero
Explanation
(a) Here, mass of the body
$m = 0.05\ kg$
[image:5_image279.emf]
Acceleration $a = 9.5\ ms^{- 2},\ g = 9.8\ m/s^{- 2}$
$\therefore mg - f_{air} = ma\ \Rightarrow f_{air} = m(g - a)$
$= 0.05(9.8 - 9.5) = 0.015\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.