Seven pulleys are connected with the help of three light strings as shown in the figure below. Consider $P_{3}, P_{4}, P_{5}$ as light pulleys and pulleys $P_{6}$ and $P_{7}$ have masses m each. For this arrangement, mark the correct statement(s)

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Acceleration of $P_{6}$ is g downwards and that of $P_{7}$ is g upwards

Tension in the string connecting $P_{1}, P_{3}$ and $P_{4}$ is zero

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

Tension in the string connecting $P_{1}, P_{3}$ and $P_{4}$ is mg/3

answer is A, C.

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Detailed Solution

First of all draw FBD of $P_{3} .$ Let tensions, in three strings be $T_{1}, T_{2}$ and $T_{3},$ respectively
$2 T_{1} - T_{1} = 0 \times a \Rightarrow T_{1} = 0$
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Now draw FBD of $P_{4}$ and $P_{5}$
$2 T_{1} - T_{2} = 0 \Rightarrow T_{2} = 0$
$2 T_{2} - T_{3} = 0 \Rightarrow T_{2} = T_{3} = C$

So force acting on $P_{6}$ and $P_{7}$ will be that of gravity and they will be in free fall. Hence, acceleration of each of them will be g downwards