In fig the blocks A, B, and C of mass m each have acceleration `a_(1),a_(2), and a_(3)`, respectively. `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_(2)gta_(3)`
C. `a_(1)=a_(2)` , `a_(2)gta_(3)`
D. `a_(1)gta_(2)` , `a_(2)=a_(3)`

Answer

B

Explanation

Case I: `a_(1)=(2mg-mg)/(m)=g`
case II: `a_(2)=((2mg-mg))/((2m+m))g=(g)/(3)`
Case III: `a_(3)=((2mg+mg-mg))/(2m)=(g)/(2)` Hence `a_(1)gta_(3)gta_(2)`