Where (at what mark) should a third mass of 0.30 kg be hung to keep the stick in balance? A uniform meter stick is supported by a knife edge at the 50-cm mark and has masses of 0.40 kg and 0.60 kg hanging at the 20-cm and 80-cm marks, respectively.
1. 20 cm
2. 25 cm
3. 30 cm
4. 70 cm

The correct answer is 30 cm.

Solution:-

This question is based on torque.
Torque = (distance from the pivot point) x (mass)
For balancing, we need the torques to be equal on both sides of the ruler.
First weight for the torque is 0.40kg and at the 20 cm mark, which is 30 cm from the pivot point (50cm mark).
So, by the formula, torque is equal to 30(0.40) = 12

The second wight for the other torque is 0.60 kg at the 80 cm mark, also 30 cm from the pivot point
Calculated torque is 30 (0.60) = 18

Now, we need to balance the torque with a weight of 0.30kg
It will need to go on the side of the 20 cm mark because we need balance to it.
We know the weight but not the distance, so let the distance is 'x', then the torque will be x (0.30)
This gives the equation,
12 + x (.30) = 18
0.3x = 6
x = 20

It needs to be 20 cm from the pivot point towards the 20 cm end which puts it at the 30 cm mark (50-20).