BMAT SECTION 2 PHYSICS EXAM QUESTIONS AND ANSWERS

Electrostatics is used when paint spraying the panels of a car. The tiny paint droplets are positively charged as they leave the spray gun. The panels of the car are earthed. Which of the following statements about electrostatic paint spraying is/are correct? 1) Droplets of paint are attracted to one another. 2) Droplets of paint are attracted to the panels. 3) Droplets of paint can reach parts of the panels that are not directly in line with the spray gun. 4) The panels of the car body gradually become positively charged. All the droplets have the same (positive) charge so they repel one another. This prevents them clumping together and leads to a smooth coat of paint on the car, so statement 1) is incorrect. As the droplets approach the car, they induce a negative charge on the surface and this attracts them to it, so statement 2) is correct. Droplets that pass behind the car body can be attracted back to it by induction, so statement 3) is correct. The car is earthed so, as positive charge is added by the droplets, electrons flow from earth to neutralise them and the car body remains neutral. Statement 4) is incorrect. A charge of 6000 μμC passes through a component in 50 minutes. What is the average current in the component? CURRENT = CHARGE/ TIME = 6000 *10^-6 / (50*60) = 6000 *10^-6/ 3000 = 2*10^-6 A A voltmeter has very .................. resistance, otherwise it would tend to '............ ............' the component across which it was connected (because there would be a significant amount of current in the voltmeter instead of in the component). An ammeter has very .................... resistance, otherwise it would tend to ............... the .................. that it was being used to measure. high short circuit low reduce current There is a current of 60 mA in a resistor when it is connected to a 12 V battery. What is the resistance of the resistor? RESISTANCE = VOLATGE/CURRENT = 12 /60*10^-3 = 1 /5*10^-3 = 0.2 *10^ +3 = 2*10 ^ 2 Ω = 200 Ω A 72 MΩ resistor is connected across a 1.8 kV power supply. What is the current in the resistor? CURRENT = VOLTAGE/ REISTSTANCE =1.8*10^3 / 72 *10^6 = 18/ 72*10^4 =1/4*10^4 =0.25*10^-4 =2.5*10^-5 A OR = 25 μA In a closed circuit a thermistor is connected to a voltmeter in parallel and an ammeter in series, with a fixed resistor also connected in series. What the reading of these two meters when the temperature of the thermistor decreases? As temperature increases the resistance decreases. On the ammeter: the overall resistance of the circuit increases therefore the current (and the reading on the ammeter)decreases On the Voltmeter: The reduced current in the circuit results in there being a smaller voltage across the fixed resistor (from V=I×RV=I×R, where RR is constant). This causes the voltage across the thermistor to increase, because the total voltage supplied by the battery has not changed, so the thermistor now has a larger 'share' of the battery's voltage. Note that this means that the effect of the increased resistance of the thermistor is greater than the effect of the decreased current in it.