a uniform meter rule has masses 20g and 10g hang at 10cm and 30cm mark respectively, what the value of mass required to balance the meter rule horizontally
To balance the meter rule horizontally, we need to ensure that the torque (moment) on one side of the meter rule is equal to the torque on the other side. The torque is calculated by multiplying the mass by the distance from the fulcrum.
Let's assume that the fulcrum is at the 0cm mark. The 20g mass is 10cm away from the fulcrum, and the 10g mass is 30cm away from the fulcrum. The torque caused by the 20g mass is therefore 20g * 10cm = 200 g·cm (gram-centimeters), while the torque caused by the 10g mass is 10g * 30cm = 300 g·cm.
In order to balance the meter rule horizontally, we need a mass on the other side that creates a torque of -300 g·cm (the negative sign indicates that it should be in the opposite direction). This is achieved by placing a mass that is equal to 300g and placing it at a distance x from the fulcrum, where the negative sign for x indicates the opposite direction.
Hence, the value of the mass required to balance the meter rule horizontally is 300g, which is equal to the sum of the torques created by the 20g and 10g masses.