Question

image shows a triple - beam balance (ohaus, capacity 610 g) with a yellow cylindrical object on the pan. there are number markings (0 - 9) above the balance, and the beams with sliding masses are visible.

Explanation:

Step1: Identify the balance type

This is a triple - beam balance. The middle beam (usually the main beam) has a rider, and the other beams (front and back) have standard masses. For a triple - beam balance, the mass of the object is the sum of the masses indicated by the riders on each beam.

Looking at the diagram, assume the front beam (the one with the 0 - 9 scale at the top) has a rider, and the other beams (the ones with the 0 - 100, 0 - 500 etc. scales) have their own contributions. Wait, actually, in a typical triple - beam balance, there are three beams: one with 0 - 10 g (usually the front, with a rider that moves in 0.1 g increments), one with 0 - 100 g (middle, rider in 10 g increments), and one with 0 - 500 g (back, rider in 100 g increments). But from the given diagram, the top scale is 0 - 9, and the lower scales seem to be for the other beams. Wait, maybe this is a simplified diagram. Let's assume that the main beam (the one with the rider) and the other beams. Wait, the yellow object is on the pan. Let's look at the rider positions.

Wait, maybe the scale at the top (0 - 9) is the front beam (0 - 10 g, with the rider at 2.4? Wait, no, the arrow is at 2.4? Wait, the top numbers are 0,1,2,3,4,5,6,7,8,9. The rider on the front beam (the one with the 0 - 9 top scale) is at 2.4? Wait, maybe the middle beam (the one with the 0 - 100, 0 - 200 etc. scale) has a rider at 200? No, maybe I misinterpret. Wait, actually, in a triple - beam balance, the mass is calculated as:

Mass = mass from the back beam + mass from the middle beam + mass from the front beam.

Looking at the diagram, the back beam (the one with the largest scale) - maybe it's at 0? The middle beam (the one with the 0 - 500? No, the lower scale has 0,100,200,300,400. Wait, the rider on the middle beam (the one with the 0 - 400 scale) is at 200? And the front beam (the one with the 0 - 9 top scale) has a rider at 2.4? Wait, no, the top scale is 0 - 9, and the rider is at 2.4? Wait, maybe the front beam is 0 - 10 g (with 0.1 g increments), middle beam 0 - 100 g (10 g increments), back beam 0 - 500 g (100 g increments). But in the diagram, the middle beam (the one with the 0 - 400 scale) - maybe it's a different balance. Wait, the balance is labeled "Triple - beam Balance, Capacity 610 g". So total capacity is 610 g, which is 500 (back) + 100 (middle) + 10 (front) = 610 g. So back beam: 0 - 500 g (100 g increments), middle beam: 0 - 100 g (10 g increments), front beam: 0 - 10 g (0.1 g increments).

Looking at the diagram, the back beam rider: maybe at 0 (since no visible rider there). Middle beam rider: looking at the middle scale (the one with 0,100,200,300,400), the rider is at 200? Wait, no, the middle beam (100 g scale) would have 0 - 100 g in 10 g increments. Wait, maybe the diagram is simplified. Let's look at the top scale (0 - 9) - that's the front beam (0 - 10 g, 0.1 g per division). The rider on the front beam is at 2.4? And the middle beam (the one with the 0 - 200? Wait, the lower scale has 0,100,200,300,400. Wait, maybe the middle beam is 0 - 500 g? No, capacity is 610, so 500 + 100 + 10 = 610. So back beam: 0 - 500 (100 g steps), middle: 0 - 100 (10 g steps), front: 0 - 10 (0.1 g steps).

Wait, in the diagram, the rider on the middle beam (the one with the 0 - 400 scale) - maybe it's at 200? No, that would be 200 g, but middle beam should be 0 - 100. Wait, maybe the diagram is using a different scale. Alternatively, maybe the balance is a two - beam balance? No, it's labeled triple - beam.

Wait, maybe the top scale (0 - 9) is the front beam, and the rider is at 2.4 g, and the other beam (the one with the 0 - 200 scale) has a rider at 200 g? But that would be 200 + 2.4 = 202.4 g. Wait, but the capacity is 610 g, so maybe the back beam is at 0, middle at 200, front at 2.4?

Wait, perhaps the correct way is: the mass of the object is equal to the sum of the masses indicated by each rider. Let's assume that the front beam (the one with the 0 - 9 top scale) has a rider at 2.4 g (since the arrow is at 2.4), and the middle beam (the one with the 0 - 200? No, the lower scale has 0,100,200,300,400) - maybe the middle beam is 0 - 500 g, but that doesn't fit. Wait, maybe the diagram is a simplified triple - beam balance where the back beam is at 0, the middle beam is at 200 g, and the front beam is at 2.4 g. So total mass = 200 + 2.4 = 202.4 g.

Step2: Calculate the mass

Assuming the middle beam (the one with the 0 - 400 scale) has a rider at 200 g (since the rider is at the 200 mark) and the front beam (the one with the 0 - 9 top scale) has a rider at 2.4 g (the arrow is between 2 and 3, closer to 2.4). So the total mass \( m=200 + 2.4=202.4\space g \)

Answer:

The mass of the yellow object is \(\boldsymbol{202.4\space g}\) (assuming the middle beam rider is at 200 g and front beam rider at 2.4 g)