Question

Seven pulleys are connected with the help of three light strings as shown in the figure given below. Consider $P_3,P_4$ and $P_5$ 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 $mg/3$ .
• C. Tension in all the 3 strings is same and equal to zero.
• D. Acceleration of $P_6$ is $g$ downward and that of $P_7$ is $g$ upward.

Answer 1

Answer: (A), (C)

Tension in all the 3 strings is same and equal to zero.

First of all draw FBD of $P_3$. Let the tension in three strings be $T_1,T_2$ and $T_3$ respectively.

$2T_1-T_1=0\times a\Rightarrow T_1=0$
Now draw FBD of $P_4$ and $P_5$
$2T_1-T_2=0\Rightarrow T_2=0$
$2T_2-T_3=0\Rightarrow T_2=T_3=0$


Similarly for the acceleration, draw the FBD of $P_6$ and $P_7$ and get the values of acceleration which will be in downward directions for both. Hence, option d is wrong.