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

Well, let's see here. According to my calculations and my impeccable sense of humor, the correct answer is... (drumroll, please)... c) 3.75mg!

Why, you ask? Well, because when the two masses are hanging vertically and at rest, the tension in the string on one side is equal to the weight of that mass (3m) and on the other side it is equal to the weight of the other mass (5m). When the system is released, the heavier mass will start moving downwards, causing the pulley to rotate. This rotation will cause the tension in the string to change.

Now, if we take a closer look at the forces involved, we can observe that the tension in the string on the heavier side must be greater than the tension on the lighter side. This is because the heavier mass wants to accelerate downwards, so the tension must be strong enough to counteract that. As a result, the pulley will experience a net force in the direction of the lighter mass.

So, the magnitude of the resultant force exerted by the string on the pulley can be calculated by finding the difference between the tensions on each side. The tension on the heavier side is 5mg, and the tension on the lighter side is 3mg. Subtracting the two, we get 5mg - 3mg = 2mg.

But hold on, we're not done yet! Since the pulley is experiencing a net force in the direction of the lighter mass, we need to double the result to take into account both sides of the pulley. Therefore, the correct answer is 2mg * 2 = 4mg!

Oh wait, that's not right, is it? I apologize for the confusion, my dear friend. It seems I have stumbled upon a mathematical blunder. The correct answer is indeed e) 2mg. I'm sorry for the mix-up, but I hope my attempt at humor brought a smile to your face nonetheless!

To find the magnitude of the resultant force exerted by the string on the pulley, we need to consider the forces acting on each particle and the properties of the pulley.

Let's analyze the forces acting on each particle:

Particle with mass 3m:
- Weight: mg (downward)
- Tension in the string: T (upward)

Particle with mass 5m:
- Weight: 5mg (downward)
- Tension in the string: T (upward)

Now, let's consider the pulley. Since it is an inextensible light string passing over the pulley, the tension on both sides of the string will be the same. Therefore, the tension T in the string is equal to the force exerted by the string on the pulley.

Since the system is in equilibrium, the sum of the downward forces on the pulley must be equal to the sum of the upward forces:

Sum of downward forces = Sum of upward forces

The downward forces on the pulley are due to the weights of the two particles: mg + 5mg = 6mg.

The upward forces on the pulley are due to the tensions in the string: 2T.

Setting these forces equal, we have:

6mg = 2T

Simplifying the equation, we find T = 3mg.

Therefore, the magnitude of the resultant force exerted by the string on the pulley is 3mg.

Hence, the correct option is c) 3.75mg.

To determine the magnitude of the resultant force exerted by the string on the pulley, we need to analyze the forces acting on the system.

Let's label the mass of the particle at one end of the string as 3m and the mass of the particle at the other end as 5m. Since the system is released from rest, we can assume that it is in equilibrium at the beginning of the motion.

There are three forces acting on each particle: the weight (mg), the tension in the string (T), and the resultant force (R) exerted by the string on the particle. The tension in the string is the same throughout the system due to the inextensible nature of the string. The weight acts vertically downward, and the resultant force also acts vertically upward.

Now, let's consider the motion of the system. As the heavier particle (5m) falls, the lighter particle (3m) will rise. The pulley acts as a pivot point, and the string unwinds as the particles move vertically.

Since the system is released from rest and the pulley is light, we can assume the string remains taut during the motion. This means the magnitudes of the tensions in the string on either side of the pulley are equal.

As the 5m particle falls, the resultant force on it is the difference between its weight (5mg) and the tension in the string (T). Since the resultant force acts upward, its magnitude is given by:

R_5m = 5mg - T

As the 3m particle rises, the resultant force on it is the sum of its weight (3mg) and the tension in the string (T). Since the resultant force acts upward, its magnitude is given by:

R_3m = 3mg + T

Since the string is taut, the tensions in both sides are equal:

T = T

Now, we can apply Newton's second law for each particle:

For the 5m particle:
R_5m = 5mg - T = 5ma_5m
where a_5m is the acceleration of the 5m particle.

For the 3m particle:
R_3m = 3mg + T = 3ma_3m
where a_3m is the acceleration of the 3m particle.

Since the particles are connected by the string, they have the same magnitude of acceleration, but opposite in direction:

a_5m = -a_3m

By substituting T = T into the equations above, we can solve for the magnitudes of R_5m and R_3m:

5mg - T = 5ma_5m
3mg + T = 3ma_3m

Rearranging the equations gives:
T = 5mg - 5ma_5m
T = 3ma_3m - 3mg

Equating the expressions for T:
5mg - 5ma_5m = 3ma_3m - 3mg

Simplifying:
8mg = (5m + 3m)a
8mg = 8ma

We can cancel out 'm' from both sides:
8g = 8a

Dividing by 8 gives:
g = a

This tells us that the acceleration of the system is equal to the acceleration due to gravity, 'g'.

Now, we can substitute 'a' back into one of the equations for tension:
T = 5mg - 5ma_5m

Since g = a, we have:
T = 5mg - 5mg
T = 0

Therefore, the tension in the string is zero throughout the motion.

Finally, we can find the resultant force exerted by the string on the pulley. Since the tension is zero and the weight of the particles act vertically downwards, the resultant force exerted by the string on the pulley is equal to the weight of the particles:

Resultant force = 5mg + 3mg
Resultant force = 8mg

Hence, the magnitude of the resultant force exerted by the string on the pulley is 8mg. Therefore, the correct answer is option (a) 8mg.