Sunday, June 7, 2015

Physics 9702 Doubts | Help Page 166

  • Physics 9702 Doubts | Help Page 166


Question 830: [Current of Electricity > Resistance]
Copper wire is stretched so that its diameter is reduced from 1.0 mm to a uniform 0.5 mm.
Resistance of the unstretched copper wire is 0.2 Ω.
What will be the resistance of the stretched wire?
A 0.4 Ω                       B 0.8 Ω                       C 1.6 Ω                       D 3.2 Ω

Reference: Past Exam Paper – November 2012 Paper 13 Q33



Solution 830:
Answer: D.
Resistance R of a wire = ρL / A

Cross-sectional area A = πd2 / 4
The resistance R is inversely proportional to the area, and thus is also inversely proportional to the diameter squared.

When the wire is stretched, the diameter is reduced by half. This causes the area to be reduced by 4. So, the resistance R is increased by a factor of 4 due to the cross-sectional area A.

But, the volume of the wire is always constant.
Volume V = AL          where L is the length of the wire
The stretched wire has ¼ the area of the unstretched wire. For its volume to remain constant, its length must have been increased by a factor of 4.
Volume V = (A/4) (4L)

So, the length of the stretched wire is 4L.

Resistance R of unstretched wire = ρL / A = 0.2 Ω
Resistance R of unstretched wire = ρ(4L) / (A/4) = 16 (ρL / A) = 16 (0.2) = 3.2 Ω

In such a question, sketching a diagram would be useful.










Question 831: [Work, Energy and Power > Efficiency]
A crane is being used to lift containers off a ship. One container has mass of 14 000 kg and is being lifted vertically with a speed of 3.2 m s–1.
Electric motor being used to supply the power to lift the container is using a current of 240 A at a potential difference of 2200 V.
What is the efficiency of the system?
A 8.1 %                       B 8.5 %                       C 48 %                        D 83 %

Reference: Past Exam Paper – November 2012 Paper 12 Q21



Solution 831:
Answer: D.
Power P = VI
(Input) Power from electric motor = VI = 2200 × 240 = 5.28×105 W

Power = Energy / time = Force × distance / time = Force × speed
Force = Weight (= mg)
(Output) Power from crane = (14000 × 9.81) × 3.2 = 4.39488×105 W

Efficiency = Output power / Input power = (4.39488×105) / (5.28×105) = 0.83 = 83%








Question 832: [Dynamics + Energy]
Two trolleys are placed together on horizontal runway with a compressed spring between them.

When they are released, the 2 kg trolley moves to the left at 2 m s–1.
How much energy was stored in the spring?
A 4 J                            B 6 J                            C 8 J                            D 12 J

Reference: Past Exam Paper – November 2009 Paper 11 Q15 & Paper 12 Q14



Solution 832:
Answer: D.
From the conservation of momentum,
Sum of momentum before collision = Sum of momentum after collision

Before collision, the sum of momentum is zero since the trolleys are stationary.

After collision,
2kg (2ms-1) + 1kg (v) = 0
Velocity v = -4ms-1 = 4ms-1 (considering magnitude only).

As the trolleys move, they have kinetic energy. This kinetic energy was initially stored in the spring.

Kinetic energy = ½ mv2
Energy stored = Sum of kinetic energies of trolleys = ½ (2)(2)2 + ½ (1)(4)2 = 12J










Question 833: [Work, Energy and Power]
An escalator is 60 m long and lifts passengers through vertical height of 30 m, as shown.

To drive the escalator against forces of friction when there are no passengers requires a power of 2.0 kW.
The escalator is used by passengers of average mass 60 kg and the power to overcome friction remains constant.
How much power is required to drive the escalator when it is carrying 20 passengers and is travelling at 0.75 m s–1?
A 4.4 kW                    B 6.4 kW                     C 8.8 kW                     D 10.8 kW

Reference: Past Exam Paper – June 2014 Paper 12 Q16



Solution 833:
Answer: B.
EITHER
Total weight of passengers = W = 20(mg) = 20 × 60 × 9.81 N

Each passenger are moved a distance of 60m as the escalator goes up. The weight of the passengers acts vertically downwards. There will be a component of weight acting down the slope of the escalator.
Component of weight along slope of the escalator = W sinθ
where θ is the angle the escalator makes with the horizontal

sinθ = O / H = 30 / 60 = 0.5

Work needs to be done against forces. Additionally, work should be done against the component of weight acting down the slope.

Power = Energy / time
Power = 2000 + (W.D/t) = 2000+ Fv
Power = 2000 + (20×60×9.81×0.5)(0.75) = 6414.5W = 6.4kW


OR
The question can also be worked out by considering the gain in potential energy of the 20 passengers.
Total weight of passengers = W = 20(mg) = (20 × 60 × 9.81) N

Gain in potential energy = mgh
Gain in potential energy per second = 20(mgv)

The gain in potential energy per second depends on vertical distance moved per second. 
sinθ = O / H = 30 / 60 = 0.5
where θ is the angle the escalator makes with the horizontal

Vertical component of velocity, v = 0.75 × sinθ = 0.75 × 0.5 = 0.375 ms-1
Gain in potential energy per second = (20 × 60 × 9.81) × 0.375 = 4414.5W = 4.4 kW

Total power required = Power to move against friction + Gain in potential energy per second

Total power required = 2.0 + 4.4 = 6.4 kW


Here, 2 common mistakes are to use 60 m rather than 30 m vertical distance and forgetting to add on the 2.0 kW.











Question 834: [Measurement > Graphs]
Diagram shows a calibration curve for a thermistor, drawn with an unusual scale on the vertical axis.

What is the thermistor resistance corresponding to a temperature of 40 °C?
A 130 Ω                      B 150 Ω                      C 400 Ω                      D 940 Ω

Reference: Past Exam Paper – June 2013 Paper 12 Q5



Solution 834:
Answer: C.
We need to divide the range of values between the successive values displayed (on the y-axis) by number of lines in the range, including those corresponding to the 2 values.

Consider the mark ‘100’ and ‘1000’. Including the lines at 100 and 1000, we have 10 lines. Between these 10 lines, there are 9 intervals – one between each 2 lines.
9 intervals represent a range of 1000 – 100 = 900
1 interval correspond to 900 / 9 = 100 Ω

So, the horizontal temperature axis (the bottom-most horizontal line) corresponds to 100 Ω. The line just above it corresponds to 200 Ω. The next line is 300 Ω. Then, 400, 500, 600, ….

At 40°, the curve is at the 4th line from the bottom.
EITHER This corresponds to 400 Ω. 
OR There are 3 intervals from that line to the bottom-most horizontal line.
So, reading = 100 + 3(100) = 400 Ω





4 comments:

  1. physics paper may june 2013 paper 12 question 6

    ReplyDelete
    Replies
    1. Go to
      http://physics-ref.blogspot.com/2014/07/9702-june-2013-paper-12-worked.html

      Delete
  2. For Q830, why cant I directly use this equation 0.2(22/7)(1/4)= R(22/7)(0.25/4) R=0.8 ? Thank you.

    ReplyDelete
    Replies
    1. Could you explain the formula. How you reached there?

      Also, the correct solution is 3.2, not 0.8.

      Delete

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