for question 32
let the accelaration of the mass 2M be a downwards
therefore by constraint relation acceleration of block of mass M will be 2a upwards along the inclined plane
now let the tension be T in the string connected to block of mass M
now by making FBD of pulley(connected to block of mass 2M) we get tension in the string connected to block of mass 2M as 2T (since net force on the pulley has to be zero as it is massless)
now equation for block of mass M
= T - Mgsin30 = M(2a)
=> T - Mg/2 = 2Ma ------------------------- 1
now equation for block of mass 2M
= 2M - 2T = 2Ma -------------------------------------------2
from equation 1 & 2
a =g/6
therefore acceleration of block of mass M = 2a i.e g/3 upwards along the inclined plane.
HOPE YOU UNDERSTOOD