Seven pulleys are connected with the help of three light strings as shown in figure. Consider P3, P4, P5 as light pulleys and pulleys P6 and P7 have masses m each. For this arrangement, mark the correct statement(s).

First of all, draw FBD of P3. Let tensions, in three strings be T1, T2 and T3, respectively 2T1−T1=0×a ⇒ T1=0Now draw FBD of P4 and P5.2T1−T2=0 ⇒ T2=02T2−T3=0 ⇒ T2=T3=0So forces acting on P6 and P7 will be that of gravity and they will be in free fall. Hence, acceleration of each of them will be g downwards.