Final answer:
Using the principle of torque balance and the given information, the mass of the meter stick is calculated to be 70g, which does not match any of the provided options.
Step-by-step explanation:
To find the mass of the meter stick, we can use the concept of the torque balance. Torque, also known as the moment of force, is the rotational equivalent of linear force and is calculated as the product of the force and the distance from the pivot point at which the force is applied. In this scenario, the meter stick is in equilibrium, meaning the torques around the pivot point are balanced.
Let's call the mass of the meter stick m. Since the weight of 10g is at a distance of 35 cm from the balance point (45 cm - 10 cm), the torque due to this weight is 10g * 35 cm. The meter stick's own weight acts at its center of gravity, which for a uniform stick is at the midway point, 50 cm from the ends. The stick balances at 45 cm, meaning the stick's center of gravity is 5 cm away from the pivot (50 cm - 45 cm).
The balance of torques is expressed as:
10g * 35 cm = m * 5 cm
Dividing both sides by 5 cm we get:
70g = m
So, the mass of the meter stick is 70g, which is not one of the options provided. This indicates there might have been a typo in the question or the provided options. Therefore, based on the principle of torque balance, none of the options (a) 10g, (b) 20g, (c) 30g, (d) 40g is correct.