Thursday, November 5, 2015

Physics 9702 Doubts | Help Page 224

  • Physics 9702 Doubts | Help Page 224



Question 1064: [Work, Energy and Power]
A small electric motor is mounted on a bench, as shown. The motor is connected to a 6.0 V supply and the current in the motor is 0.50 A. The motor is 50% efficient.

What is the time taken to lift a mass of 200 g up through a height of 90 cm?
A 0.59 s                       B 0.85 s                       C 1.2 s                         D 2.7 s

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q15



Solution 1064:
Answer: C.
Input (electrical) Power = VI = 6.0 × 0.50 = 3.0 W

Efficiency = Output power / Input power = 50%
Output power = 0.5 × 3.0 = 1.5 W

Power = Energy / time
Gravitational potential energy required to lift mass = mgh
Gravitational potential energy required to lift mass = 0.200 × 9.81 × 0.9 = 1.7658 J
{taking g as 10ms-2 would give the exact answer}

Time = 1.7658 / 1.5 = 1.1772 = 1.2 s












Question 1065: [Waves]
A sound wave moves with a speed of 320 m s–1 through air. The variation with time of the displacement of an air particle due to this wave is shown in the graph.

Which statement about the sound wave is correct?
A The frequency of the wave is 500 Hz.
B The graph shows that sound is a transverse wave.
C The intensity of the wave will be doubled if its amplitude is increased to 0.4 mm.
D The wavelength of the sound wave is 1.28 m.

Reference: Past Exam Paper – June 2015 Paper 11 Q24



Solution 1065:
Answer: D.
The graph shown is a displacement-time graph of sound (which is a  longitudinal wave). From this graph, we can obtain the period T, but not (directly) the wavelength.
2T = 8 ms
Period T = 4 ms

Frequency = 1 / T = 1 / 4×10-3 = 250 Hz

Speed v = fλ
Wavelength λ = 320 / 250 = 1.28 m

The intensity of a wave depends on the square of the amplitude. Doubling the amplitude causes the intensity to increase by a factor of 4.











Question 1066: [Waves > Diffraction + EM spectrum]
Electromagnetic waves from an unknown source in space were found to be significantly diffracted when passing through gaps of the order of 10–5 m.
Which type of wave are they most likely to be?
A radio waves
B microwaves
C infra-red waves
D ultraviolet waves

Reference: Past Exam Paper – June 2010 Paper 11 Q22 & Paper 12 Q24 & Paper 13 Q23



Solution 1066:
Answer: C.
For significant diffraction, the wavelength of the EM wave should be of the same order as the gap (which is of order of 10-5 m).

Wavelength of infra-red: 700 nm (frequency 430 THz) to 1 mm (300 GHz)










Question 1067: [Current of Electricity]
In the circuit shown, all the resistors are identical.

The reading on voltmeter V1 is 8.0 V and the reading on voltmeter V2 is 1.0 V.
What are the readings on the other voltmeters?

reading on voltmeter V3 / V                reading on voltmeter V4 / V
A                                 1.5                                           1.0
B                                 3.0                                           2.0
C                                 4.5                                           3.0
D                                 6.0                                           4.0

Reference: Past Exam Paper – June 2014 Paper 12 Q36



Solution 1067:
Answer: B.
Let the resistance of each resistor be R.

V2 = 1.0 V. This is the p.d. across a resistor R. The p.d. across the resistor just left of V2 is also 1.0V since the resistance is the same and they are in the same branch (the same current flows).

Total resistance in the lower branch of the parallel combination = 1.0 + 1.0 = 2.0 V

The total p.d. across any branch of a parallel combination is the same (from Kirchhoff’s law). So, V4 = 2.0 V [B is correct]

The complete circuit may be described as a series combination of a resistor R (vertical on the left), another resistor R (with p.d. = V3) and the parallel combination.
The p.d. across the vertical resistor is the same as V3 since they have the same resistance.

V1 = total e.m.f. in circuit = 8V
The sum of p.d. in any loop is equal to the e.m.f. in the circuit.
V3 + V3 + 2.0 = 8.0
2V3 = 6.0
V3 = 3.0 V










Question 1068: [Measurements > Uncertainty]
A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm. This measurement is made using a digital caliper with a zero error of (−0.20 ± 0.02) mm.
What is the percentage uncertainty in the calculated thickness of a single sheet?
A 1.0%                        B 2.0%                        C 4.0%                        D 6.7%

Reference: Past Exam Paper – June 2015 Paper 12 Q6



Solution 1068:
Answer: C.
This challenging question involves both random and systematic error.

The zero error (−0.20) can be removed but its uncertainty (± 0.02) must be added to the measurement uncertainty.
True value of total thickness = 0.80 + 0.20 = 1.0 mm  
Total uncertainty in measurement = ± (0.02 + 0.02) = ± 0.04

So the four sheets have a true thickness of (1.00 ± 0.04) mm.
A single sheet would have a thickness of (0.25 ± 0.01) mm.

Percentage error = (0.01 / 0.25) × 100% = 4%



6 comments:

  1. Assalamualaikum,
    in solution 1067 , how can we know that the separate resistor has the same p.d as V3

    ReplyDelete
    Replies
    1. Wslm. The resistors have the same resistance R, and since they are in series, the same current I flows through them.

      From Ohm's law, p.d V = IR. so V is also the same as I and R are the same.

      Delete
  2. Can you please post solution for June 2010 p11 Q33
    Thank you!

    ReplyDelete
    Replies
    1. See solution 387 at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-71.html

      Delete
  3. These solutions have helped me a lot. Thank youu
    Just wanted to know if you have posted all of May/june and Oct/nov 15 solutions?

    ReplyDelete
    Replies
    1. No problem.

      May be I have already solved some of them but they are scattered.

      Delete

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