CET Karnataka Engineering
CET - Karnataka Engineering Solved Paper-2009
question_answer
A body of mass 0.05 kg is observed to fall with an acceleration of \[9.5\text{ }m{{s}^{-2}}\]. The opposing force of air on the body is \[(g=9.8\text{ }m{{s}^{-2}})\]
A) 0.015 N
B) 0.15 N
C) 0.030 N
D) zero
Correct Answer:
A
Solution :
From, Newton's laws of motion, \[mg-{{F}_{air}}=ma\] or \[{{F}_{air}}=m\,(g-a)\] \[=0.05\,\,(9.8-9.5)\] \[=0.015\,N\]