As shown in Figure A, it is a device for automatically measuring the amount of oil in the fuel tank. R is a sliding rheostat, and its metal slide plate is connected to one end of the lever. Judging from the fuel gauge, (1) When the oil level rises, the position of the float rises. Under the action of the lever, the resistance value of the sliding rheostat connected to the circuit becomes smaller, which increases the current in the circuit and the number of current expressions. When the oil level drops, the position of the float decreases. Under the action of the lever, the resistance value of the sliding rheostat connected to the circuit increases, making the current in the circuit smaller and the number of current expressions smaller. Convert the scale of the ammeter dial to the height of the oil level, and you can know the amount of oil stored in the oil tank. So the device involves buoyancy, lever, sliding rheostat, ammeter, ohm's law and other related knowledge. The function of R' is to prevent excessive current in the circuit from burning out the instrument and protect the circuit.

(2)① According to the table data, when the fuel gauge indicates 0L, the voltmeter indicates 0.5V, corresponding to the current passing through the fuel gauge as i0 = u0r2 = 0.5V5ω = 0.1a;

When the indicator of the fuel gauge is 50L, the indicator of the voltmeter is 3V, and the corresponding current through the fuel gauge is I1= u1r 2 = 3V5Ω = 0.6A;

② When the metal slider is at the B end, the current passing through the oil meter is 0. 1A,

The power supply voltage is u = I (r+r') = 0. 1× (50ω+r').

When the metal slider is at the A end, the current passing through the oil meter is 0.6A,

The power supply voltage is U=I'R'=0.6R',

So 0. 1× (50ω+r') = 0.6r',

Solution: r ′ =10ω.

So the answer is: (1) b; (2) lever; Ohm's law; Protection circuit; (3)0. 1; 0.6; (4) The resistance value of the resistor R' is10Ω.