A) \[\frac{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})F}{{{m}_{3}}}\]
B) \[\frac{{{m}_{3}}F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\]
C) \[F-({{m}_{1}}+{{m}_{2}})g\]
D) F
Correct Answer: B
Solution :
[b]| \[F=({{m}_{1}}+{{m}_{2}}+{{m}_{3}})g\sin \theta \] | ...(i) |
| \[N={{m}_{3}}g\sin \theta \] | ..(ii) |
| Dividing Eq. (ii) by Eq. (i), we get | |
| \[N=\frac{{{m}_{3}}F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] | |
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