A) \[g,g\]
B) \[g-a.\text{ }g-a\]
C) \[g-a,g\]
D) \[a,g\]
Correct Answer: C
Solution :
Apparent weight of ball \[w=w-R\] \[R=ma\] acts upward\[=ma\] \[w=mg-ma=m(g-a)\] Hence, apparent acceleration in the lift is\[g-a\] Now if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is g.You need to login to perform this action.
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