particles of mass 3m and 5 m hang one at each end of a light inextensible string which passes over a pulley.The system is released from rest with the hanging parts taunt and vertical.During the subsequent motion the resultant force exerted by the string on the pulley is of magnitude-
a 8mg
b 7.5mg
c 3.75mg
d 4mg
e 2mg
To find the magnitude of the resultant force exerted by the string on the pulley, we need to analyze the forces acting on the system.
First, let's consider the force acting on the 3m particle. It experiences two forces: its weight (mg) acting downwards and the tension in the string (T) acting upwards.
Next, let's consider the force acting on the 5m particle. Similar to the 3m particle, it experiences its weight (5mg) downwards and the tension in the string (T) upwards.
Since the system is in equilibrium, the weight of the 3m particle must balance the weight of the 5m particle. Therefore, we have the equation:
3mg = 5mg
Simplifying, we find:
3m = 5m
This implies that m = 0. The given system is inconsistent, with no valid solution.
Therefore, none of the options provided (a, b, c, d, e) are correct.
Please check the given information again or provide additional details if necessary.