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).

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Tension in the string connecting P1, P3, and P4 is zero.

Tension in the string connecting P1, P3, and P4 is mg/3.

Tensions in all the three strings are same and equal to zero.

Acceleration of P6 is g downwards and that of P7 is g upwards.

answer is A.

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Detailed Solution

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.

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