Just imagine this system to be made of two point masses of 0.5 kg and 1 kg located at the middle of the original rod and the centre of the sphere respectively.

The distance between these point masses is 30 cm (half the length of rod + radius of the sphere).

The centre of mass of the system is at a distance ‘d’ from the 0.5 kg mass of the rod such that
0.5×d = (30–d)×1.
This gives d = 20 cm.

The distance ‘x’ of this centre of mass from the end of the rod is given by
x = 25+d = 25+20 = 45 cm.

Therefore, the knife edge is to be placed at 45 cm from the end.

Problem Analysis:

We have a metallic sphere and a thin rod, which are welded together. The system will balance horizontally on a knife edge. We need to find the distance 'x' at which the knife edge should be placed in order to balance the system.

Given:
- Mass of the sphere (m1) = 1 kg
- Radius of the sphere (r) = 5 cm = 0.05 m
- Mass of the rod (m2) = 0.5 kg
- Length of the rod (l) = 50 cm = 0.5 m

Approach:
To balance the system, the net torque about the knife edge should be zero. Torque is given by the product of force and perpendicular distance from the point of rotation.

1. Calculate the torque due to the sphere.
2. Calculate the torque due to the rod.
3. Set up the equation for torque balance and solve for 'x'.
4. Determine the distance 'x' at which the system will balance.

Solution:

1. Torque due to the sphere:
The weight of the sphere acts vertically downward through its center of mass. The torque due to the sphere can be calculated as follows:

Torque (τ1) = force (F1) × distance (d1)

The force acting on the sphere is its weight (mg1) and the distance is the perpendicular distance from the point of rotation (which is the radius of the sphere, r).

So, τ1 = mg1 × r

Substituting the given values, we get:
τ1 = (1 kg × 9.8 m/s²) × 0.05 m

2. Torque due to the rod:
The weight of the rod acts vertically downward through its center of mass. The torque due to the rod can be calculated as follows:

Torque (τ2) = force (F2) × distance (d2)

The force acting on the rod is its weight (mg2) and the distance is the perpendicular distance from the point of rotation (which is half the length of the rod, l/2).

So, τ2 = mg2 × (l/2)

Substituting the given values, we get:
τ2 = (0.5 kg × 9.8 m/s²) × (0.5 m/2)

3. Setting up the torque balance equation:
To balance the system, the net torque about the knife edge should be zero. Therefore, we have:

τ1 + τ2 = 0

Substituting the calculated values of τ1 and τ2, we get:
(1 kg × 9.8 m/s²) × 0.05 m + (0.5 kg × 9.8 m/s²) × (0.5 m/2) = 0

4. Solving for 'x':
Now, we need to find the distance 'x' at which the system will balance. The distance 'x' is the perpendicular distance from the point of rotation (knife edge) to the center of mass of the rod.

We can rearrange the torque balance equation to solve for 'x':

(1 kg × 9.8 m/s²) × 0.