In fig the blocks A,B,and C of mass m each have acceleration `a_(1),a_(2), and a_(3)`, repectively. `F_(1) and F_(2)` are external force of magnitude `2mg and mg`, respectively. Then

Options

A. `a_(1)=a_(2)=a_(3)`
B. `a_(1)gta_(3)gta_(2)`
C. `a_(1)=a_(2),a_(2)gta_(3)`
D. `a_(1)gta_(2),a_(2)=a_(3)`

Answer

B

Explanation

`F_(1)-mg=ma_(1)`
`2mg-mg=ma_(1)`
`a_(1)=g`

`a_(2)=(2mg-mg)/(2m+m)=(g)/(3)`

`a_(3)=(2mg-mg)/(m+m)=(g)/(2)`
`a_(1)gta_(3)gta_(2)`