Question

Seven pulleys are connected with the help of three light strings as shown in figure. 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).

• A. Tension in the string connecting $P_{1}, P_{3}$, and $P_{4}$ is zero.
• B. Tension in the string connecting $P_{1}, P_{3}$ and $P_{4}$ is $m g / 3$.
• C. Tensions in all the three strings are same and equal to zero.
• D. Acceleration of $P_{6}$ is $g$ downwards and that of $P_{7}$ is $g$ upwards.

Answer 1

Answer: (A), (C)

First of all, draw FBD of $P_3$.
Let tensions, in three strings be $T_1, T_2$ and $T_3$, respectively.
$2 T_1-T_1=0 \times a$

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=0$
So forces 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.