A) Net force on X is F.
B) Force applied by X on Y is\[\frac{F}{3}\].
C) Force applied by Y on Z is equal to net force on X.
D) Net force on Y is equal to zero.
Correct Answer: C
Solution :
Let a be the common acceleration, then \[F=3ma\,\,or\,\,a=\frac{F}{3m}\] Net force on \[X=ma=m\times \frac{F}{3m}=\frac{F}{3}\] Force applied by \[X\] on Y=(m+m)a\[=2m\times \frac{F}{3m}=\frac{2F}{3}\] Force applied by Y on Z \[=ma=m\times \frac{F}{3m}=\frac{F}{3}\]
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