In the situation shown in figure, both the pulleys and the strings are light and all the surfaces are frictionless. The acceleration of mass M, tension in the string PQ and force exerted by the clamp on the pulley, will respectively be -

Explanation

As pulley Q is not fixed, so if it moves a distance d the length of string between P and Q will change by 2d (d from above and d from below), i.e., M will move 2d This in turn implies that if `a(rarr 2d)` is the acceleration of M, the acceleration of Q and of 2M will be `(a//2) (rarr d)`
Now if we consider the motion of mass M, it is accelerated downward, so
`T=M(g-a)` ....(i)
and for the motion of Q,
`2T-T' = 0xx(a//2)=0, i.e. T' =2T` ...(ii)
and for the motion of mass 2M
`T' = 2M(a//2) i.e., T' =Ma` ...(iii)
From Eqns. (ii) and (iii) as `T=(1//2)Ma` so Eqn. (i) reduces to
`(1//2)Ma=M(g-a) or a=(2//3)g`
So, the acceleration of mass M is `(2//3)` g while tension in the string PQ from Eqn (i) will be,
`T=M[g-(2//3)g]=(1//3)Mg`
Now from (b) it is clear that force on pulley by the clamp will equal and opposite to the resultant of T and T at `90^(@)` to each other, i.e.,
`(R_(2))=sqrt(T^(2)+T^(2))=sqrt(2)T=(sqrt(2)//3)Mg`
Note: In this problem the acceleration of pulley Q or mass 2M will be `a//2 =(1//3)` g and tension in the string connecting 2M to the pulley Q, T' will be `2T=(2//3)Mg`