Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. The wedge is fixed on horizontal surface. Friction is absent everwhere. When the system is released from rest.

Options

A. tension in string is `(mg)/(2)`
B. tension is strinig is `(mg)/(4)`
C. acceleration of A is `(g)/(2)`
D. acceleration of A is `(3)/(4)g`

Answer

B, D

Explanation

Let a be acceleration of system and T be tension in the string.
F.B.D of block A
`mgsin30^@+T=ma`
`(mg)/(2)+T=ma` ..(i)
F.B.D of block B
`mg-T=ma` ..(ii)
Adding equation (i) and (ii), we get
`2ma=(3mg)/(2)impliesa=(3)/(4)g`