Three masses `M_(1), M_(2)` and `M_(3)` are lying on a frictionless table. The masses are connected by massless threads as shown. The mass `M_(3)` is pulled by a constant force F as shown. The tension in the thread between masses `M_(2)` and `M_(3)` is

Options

A. `((M_(1)+M_(2))/(M_(1)+M_(2)+M_(3)))F`
B. `((M_(2)+M_(3))/(M_(1)+M_(2)+M_(3)))F`
C. `((M_(1)+M_(3))/(M_(1)+M_(2)+M_(3)))F`
D. `((M_(1)+M_(2))/(M_(1)+M_(2)+M_(3)))F`

Answer

A

Explanation

Free body diagram of masses are
`therefore F-T_(1)=M_(3)a`... (i)
`T_(1)-T_(2)=m_(2)a`....(ii)
`T_(2)=M_(1)a` ....(iii)
Adding (i), (ii) and (iii) we get
`W=a(M_(1)+M_(2)+M_(3))`
`thereforea=(F)/((M_(1)+M_(2)+M_(3)))` .....(iv)
From (i), `F-T_(1)=M_(3)a`
`F-T_(1)=(M_(3)F)/((M_(1)+M_(2)+M_(3)))` using (iv)
`therefore T_(1)=F-(M_(3)F)/((M)_(1)+M_(2)+M_(3)))`
`=F[1-(M_(3))/(M_(1)+M_(2)+M_(3))]`
`T_(1)=(F(M_(1)+M_(2))/(M_(1)+M_(2)+M_(3))`