• 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%