The correct option is
B 13NLet us take the tension between A and B be T and tension between B and C be T1.
The masses are connected to a massless string moving over a frictionless pulley.
Now, if they are allowed to move freely, then they move with common magnitude of acceleration 'a'.
Mass of A, B and C = 2kg (m)
Now,
For C,
=> ma=mg−T1
=> T1=mg−ma. ....(1)
For B,
=> ma=mg+T1−T. ....(2)
For A,
=> ma=T−mg
=> T=ma+mg. ....(3)
Now, from (1) and (3) we will put value of The and T1 in (2), we get
=> ma=mg+(mg−ma)−(ma+mg)
=> ma=mg+mg−ma−ma−mg
=> 3ma=mg
=> a=g/3
Now, putting the value in (1) and (3), we will get the tensions..
Tension between B and C = m(g−a)
= m(g−g/3)= 2(2g/3)= 4g/3= 40/3 = 13.3N