# 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 = fÎ»

Wavelength Î» = 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 V

_{1}is 8.0 V and the reading on voltmeter V_{2}is 1.0 V.
What are the readings on the other voltmeters?

reading on voltmeter V

_{3}/ V reading on voltmeter V_{4}/ 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.

V

_{2}= 1.0 V. This is the p.d. across a resistor R. The p.d. across the resistor just left of V_{2}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, V

_{4}= 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. = V

_{3}) and the parallel combination.
The p.d. across the vertical
resistor is the same as V

_{3}since they have the same resistance.
V

_{1}= 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.

V

_{3}+ V_{3}+ 2.0 = 8.0
2V

_{3}= 6.0
V

_{3}= 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%

Assalamualaikum,

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

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

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

Can you please post solution for June 2010 p11 Q33

ReplyDeleteThank you!

See solution 387 at

Deletehttp://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-71.html

These solutions have helped me a lot. Thank youu

ReplyDeleteJust wanted to know if you have posted all of May/june and Oct/nov 15 solutions?

No problem.

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